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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% The Project Gutenberg EBook of Spherical Trigonometry, by I. Todhunter  %
+%                                                                         %
+% This eBook is for the use of anyone anywhere in the United States and   %
+% most other parts of the world at no cost and with almost no restrictions%
+% whatsoever.  You may copy it, give it away or re-use it under the terms %
+% of the Project Gutenberg License included with this eBook or online at  %
+% www.gutenberg.org.  If you are not located in the United States, you'll %
+% have to check the laws of the country where you are located before using%
+% this ebook.                                                             %
+%                                                                         %
+%                                                                         %
+%                                                                         %
+% Title: Spherical Trigonometry                                           %
+%        For the Use of Colleges and Schools                              %
+%                                                                         %
+% Author: I. Todhunter                                                    %
+%                                                                         %
+% Release Date: August 19, 2020 [EBook #19770]                            %
+%                                                                         %
+% Language: English                                                       %
+%                                                                         %
+% Character set encoding: ISO-8859-1                                      %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK SPHERICAL TRIGONOMETRY ***    %
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
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+%%                                                                    %%
+%% The Project Gutenberg eBook of Spherical Trigonometry, by          %%
+%% I. Todhunter                                                       %%
+%%                                                                    %%
+%% This eBook is for the use of anyone anywhere at no cost and with   %%
+%% almost no restrictions whatsoever. You may copy it, give it away   %%
+%% or re-use it under the terms of the Project Gutenberg License      %%
+%% included with this eBook or online at www.gutenberg.net            %%
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+%% Longtables (Art. 86, PDF page 58/59: aligned                       %%
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+%% PDF Pages: 182                                                     %%
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+%%                                                                    %%
+%% Compile history:                                                   %%
+%%                                                                    %%
+%% August 13, 2020: (Andrew D. Hwang)                                 %%
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+%% 16th September 2006: LW compiled with pdflatex (tetex under        %%
+%% MacOSX)                                                            %%
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+%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{document}
+
+\thispagestyle{empty}
+\pagenumbering{gobble}
+\small
+\begin{verbatim}
+
+The Project Gutenberg EBook of Spherical Trigonometry, by I. Todhunter
+
+This eBook is for the use of anyone anywhere in the United States and
+most other parts of the world at no cost and with almost no restrictions
+whatsoever.  You may copy it, give it away or re-use it under the terms
+of the Project Gutenberg License included with this eBook or online at
+www.gutenberg.org.  If you are not located in the United States, you'll
+have to check the laws of the country where you are located before using
+this ebook.
+
+
+
+Title: Spherical Trigonometry
+       For the Use of Colleges and Schools
+
+Author: I. Todhunter
+
+Release Date: August 19, 2020 [EBook #19770]
+
+Language: English
+
+Character set encoding: ISO-8859-1
+
+*** START OF THIS PROJECT GUTENBERG EBOOK SPHERICAL TRIGONOMETRY ***
+Credit: K.F. Greiner, Berj Zamanian, Joshua Hutchinson andthe Online
+Distributed Proofreading Team at http://www.pgdp.net(This file was
+produced from images generously made availableby Cornell University
+Digital Collections
+
+\end{verbatim}
+\normalsize
+\newpage
+
+%-----File: 001.png------------------------------------------------
+%-----File: 002.png------------------------------------------------
+%-----File: 003.png------------------------------------------------
+
+\begin{frontmatter}
+\begin{titlepage}
+\begin{center}
+\vspace*{\stretch{1}}
+\Huge \textbf{SPHERICAL
+TRIGONOMETRY.}
+\vspace*{\stretch{2}}
+\end{center}
+\end{titlepage}
+%-----File: 004.png------------------------------------------------
+%-----File: 005.png------------------------------------------------
+\begin{titlepage}
+\begin{center}
+\large
+\bigskip
+{\huge SPHERICAL TRIGONOMETRY}
+\bigskip
+\bigskip
+
+{\LARGE $\mathfrak{For\ the\ Use\ of\ Colleges\ and\ Schools.}$}
+\bigskip
+\bigskip
+
+WITH NUMEROUS EXAMPLES.
+\bigskip
+\bigskip
+\vfill
+
+{\normalsize BY}\\
+\bigskip
+
+{\LARGE I.~TODHUNTER, M.A., F.R.S.,}\\
+\bigskip
+
+{\normalsize HONORARY FELLOW OF ST JOHN'S COLLEGE,\\
+CAMBRIDGE.}
+\bigskip
+\bigskip
+
+\vfill
+\textit{FIFTH EDITION.}
+\bigskip
+\bigskip
+
+\vfill
+$\mathfrak{London:}$
+
+MACMILLAN AND CO.
+
+1886\\
+
+[\textit{All Rights reserved.}]
+\end{center}
+\end{titlepage}
+%-----File: 006.png------------------------------------------------
+
+\begin{center}
+\vspace*{\stretch{1}}
+\large
+$\mathfrak{Cambridge}$:
+
+\normalsize
+PRINTED BY C.~J.\ CLAY, M.A.\ AND SON,
+
+AT THE UNIVERSITY PRESS.
+\vspace*{\stretch{2}}
+\end{center}
+
+%-----File: 007.png------------------------------------------------
+
+\chapter*{PREFACE}
+
+The present work is constructed on the same plan as my
+treatise on Plane Trigonometry, to which it is intended as a
+sequel; it contains all the propositions usually included under
+the head of Spherical Trigonometry, together with a large
+collection of examples for exercise. In the course of the work
+reference is made to preceding writers from whom assistance
+has been obtained; besides these writers I have consulted the
+treatises on Trigonometry by Lardner, Lefebure de Fourcy,
+and Snowball, and the treatise on Geometry published in the
+Library of Useful Knowledge. The examples have been
+chiefly selected from the University and College Examination
+Papers.
+
+In the account of Napier's Rules of Circular Parts an
+explanation has been given of a method of proof devised by
+Napier, which seems to have been overlooked by most modern
+writers on the subject. I have had the advantage of access to
+an unprinted Memoir on this point by the late R.~L. Ellis of
+Trinity College; Mr Ellis had in fact rediscovered for himself
+Napier's own method. For the use of this Memoir and for
+some valuable references on the subject I am indebted to the
+Dean of Ely.
+
+Considerable labour has been bestowed on the text in
+order to render it comprehensive and accurate, and the examples
+have all been carefully verified; and thus I venture
+to hope that the work will be found useful by Students and
+Teachers.
+
+\begin{flushright}
+{\large I. TODHUNTER. \mbox\qquad}
+\end{flushright}
+
+\begin{small}
+\textsc{St John's College,}\\
+\mbox{\hspace{5em}}\textit{August} 15, 1859.
+\end{small}
+\newpage
+%-----File: 008.png------------------------------------------------
+
+In the third edition I have made some additions which I
+hope will be found valuable. I have considerably enlarged
+the discussion on the connexion of Formul\ae\ in Plane and
+Spherical Trigonometry; so as to include an account of the
+properties in Spherical Trigonometry which are analogous to
+those of the Nine Points Circle in Plane Geometry. The
+mode of investigation is more elementary than those hitherto
+employed; and perhaps some of the results are new. The
+fourteenth Chapter is almost entirely original, and may deserve
+attention from the nature of the propositions themselves
+and of the demonstrations which are given.
+\bigskip
+
+\begin{small}
+\textsc{Cambridge},\\
+\mbox{\hspace{5em}}\textit{July}, 1871.
+\end{small}
+%-----File: 009.png------------------------------------------------
+
+
+%-----File: 010.png------------------------------------------------
+%-----File: 011.png------------------------------------------------
+
+\tableofcontents
+\end{frontmatter}
+\begin{mainmatter}
+\headsep =.5 in
+\begin{spacing}{1.3}
+
+\chapter[Great and Small Circles.]{GREAT AND SMALL CIRCLES.}
+
+\paragraph{1.} \textsc{A sphere} is a solid bounded by a surface every point of
+which is equally distant from a fixed point which is called the
+\textit{centre} of the sphere. The straight line which joins any point of
+the surface with the centre is called a \textit{radius}. A straight line
+drawn through the centre and terminated both ways by the surface
+is called a \textit{diameter}.
+
+\paragraph{2.} \textit{The section of the surface of a sphere made by any plane
+is a circle.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5cm]{images/011fc}
+\end{figure}
+
+Let $AB$ be the section of the surface of a sphere made by any
+plane, $O$ the centre of the sphere. Draw $OC$ perpendicular to the
+plane; take any point $D$ in the section and join $OD$, $CD$. Since
+$OC$ is perpendicular to the plane, the angle $OCD$ is a right angle;
+therefore $CD=\surd(OD^2-OC^2)$. Now $O$ and $C$ are fixed points, so
+that $OC$ is constant; and $OD$ is constant, being the radius of the
+%-----File: 012.png------------------------------------------------
+sphere; hence $CD$ is constant. Thus all points in the plane section
+are equally distant from the fixed point $C$; therefore the section
+is a circle of which $C$ is the centre.
+
+\paragraph{3.} The section of the surface of a sphere by a plane is called
+a \textit{great circle} if the plane passes through the centre of the sphere,
+and a \textit{small circle} if the plane does not pass through the centre of
+the sphere. Thus the radius of a great circle is equal to the
+radius of the sphere.
+
+\paragraph{4.} Through the centre of a sphere and any two points on the
+surface a plane can be drawn; and only one plane can be drawn,
+except when the two points are the extremities of a diameter of
+the sphere, and then an infinite number of such planes can be
+drawn. Hence only one great circle can be drawn through two
+given points on the surface of a sphere, except when the points are
+the extremities of a diameter of the sphere. When only one great
+circle can be drawn through two given points, the great circle is
+unequally divided at the two points; we shall for brevity speak of
+the shorter of the two arcs as \textit{the} arc of a great circle joining the
+two points.
+
+\paragraph{5.} The \textit{axis} of any circle of a sphere is that diameter of the
+sphere which is perpendicular to the plane of the circle; the extremities
+of the axis are called the \textit{poles} of the circle. The poles
+of a great circle are equally distant from the plane of the circle.
+The poles of a small circle are not equally distant from the plane
+of the circle; they may be called respectively the \textit{nearer} and \textit{further}
+pole; sometimes the nearer pole is for brevity called \textit{the} pole.
+
+\paragraph{6.} \textit{A pole of a circle is equally distant from every point of the
+circumference of the circle.}
+
+Let $O$ be the centre of the sphere, $AB$ any circle of the sphere,
+$C$ the centre of the circle, $P$ and $P'$ the poles of the circle. Take
+any point $D$ in the circumference of the circle; join $CD$, $OD$, $PD$.
+Then $PD=\surd(PC^2+CD^2)$; and $PC$ and $CD$ are constant, therefore
+$PD$ is constant. Suppose a great circle to pass through the points
+$P$ and $D$; then the chord $PD$ is constant, and therefore the arc of
+%-----File: 013.png------------------------------------------------
+a great circle intercepted between $P$ and $D$ is constant for all
+positions of $D$ on the circle $AB$.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5cm]{images/013f1c}
+\end{figure}
+
+Thus the distance of a pole of a circle from every point of the
+circumference of the circle is constant, whether that distance be
+measured by the straight line joining the points, or by the arc of
+a great circle intercepted between the points.
+
+\paragraph{7.} \textit{The arc of a great circle which is drawn from a pole of a
+great circle to any point in its circumference is a quadrant.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5cm]{images/013f2c}
+\end{figure}
+
+Let $P$ be a pole of the great circle $ABC$; then the arc $PA$ is a
+quadrant.
+
+For let $O$ be the centre of the sphere, and draw $PO$. Then
+$PO$ is at right angles to the plane $ABC$, because $P$ is the pole of
+$ABC$, therefore $POA$ is a right angle, and the arc $PA$ is a quadrant.
+%-----File: 014.png------------------------------------------------
+
+\paragraph{8.} \textit{The angle subtended at the centre of a sphere by the arc of
+a great circle which joins the poles of two great circles is equal to the
+inclination of the planes of the great circles.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5cm]{images/014fc}
+\end{figure}
+
+Let $O$ be the centre of the sphere, $CD$, $CE$ the great circles intersecting
+at $C$, $A$ and $B$ the poles of $CD$ and $CE$ respectively.
+
+Draw a great circle through $A$ and $B$, meeting $CD$ and $CE$ at
+$M$ and $N$ respectively. Then $AO$ is perpendicular to $OC$, which is
+a straight line in the plane $OCD$; and $BO$ is perpendicular to $OC$,
+which is a straight line in the plane $OCE$; therefore $OC$ is perpendicular
+to the plane $AOB$ (Euclid, \textsc{xi}.~4); and therefore $OC$ is
+perpendicular to the straight lines $OM$ and $ON$, which are in the
+plane $AOB$. Hence $MON$ is the angle of inclination of the planes
+$OCD$ and $OCE$. And the angle
+\[
+AOB = AOM - BOM = BON - BOM = MON.
+\]
+
+\paragraph{9.} By the angle between two great circles is meant \textit{the angle
+of inclination of the planes of the circles}. Thus, in the figure of
+the preceding Article, the angle between the great circles $CD$ and
+$CE$ is the angle $MON$.
+
+In the figure to Art.~6, since $PO$ is perpendicular to the plane
+$ACB$, every plane which contains $PO$ is at right angles to the
+plane $ACB$. Hence the angle between the plane of any circle
+and the plane of a great circle which passes through its poles is
+a right angle.
+%-----File: 015.png------------------------------------------------
+
+\paragraph{10.} \textit{Two great circles bisect each other.}
+
+For since the plane of each great circle passes through the
+centre of the sphere, the line of intersection of these planes is a
+diameter of the sphere, and therefore also a diameter of each great
+circle; therefore the great circles are bisected at the points where
+they meet.
+
+\paragraph{11.} \textit{If the arcs of great circles joining a point $\mathrm{P}$ on the surface
+of a sphere with two other points $\mathrm{A}$ and $\mathrm{C}$ on the surface of the
+sphere, which are not at opposite extremities of a diameter, be each of
+them equal to a quadrant, $\mathrm{P}$ is a pole of the great circle through
+$\mathrm{A}$ and $\mathrm{C}$.} (See the figure of Art.~7.)
+
+For suppose $PA$ and $PC$ to be quadrants, and $O$ the centre of
+the sphere; then since $PA$ and $PC$ are quadrants, the angles $POC$
+and $POA$ are right angles. Hence $PO$ is at right angles to the
+plane $AOC$, and $P$ is a pole of the great circle $AC$.
+
+\paragraph{12.} Great circles which pass through the poles of a great
+circle are called \textit{secondaries} to that circle. Thus, in the figure of
+Art.~8 the point $C$ is a pole of $ABMN$, and therefore $CM$ and $CN$
+are parts of secondaries to $ABMN$. And the angle between $CM$
+and $CN$ is measured by $MN$; that is, \textit{the angle between any two
+great circles is measured by the arc they intercept on the great circle
+to which they are secondaries.}
+
+\paragraph{13.} \textit{If from a point on the surface of a sphere there can be
+drawn two arcs of great circles, not parts of the same great circle,
+the planes of which are at right angles to the plane of a given circle,
+that point is a pole of the given circle.}
+
+For, since the planes of these arcs are at right angles to the
+plane of the given circle, the line in which they intersect is perpendicular
+to the plane of the given circle, and is therefore the
+axis of the given circle; hence the point from which the arcs are
+drawn is a pole of the circle.
+%-----File: 016.png------------------------------------------------
+
+\paragraph{14.} \textit{To compare the arc of a small circle subtending any angle
+at the centre of the circle with the arc of a great circle subtending
+the same angle at its centre.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/016fc}
+\end{figure}
+
+Let $ab$ be the arc of a small circle, $C$ the centre of the circle,
+$P$ the pole of the circle, $O$ the centre of the sphere. Through $P$
+draw the great circles $PaA$ and $PbB$, meeting the great circle
+of which $P$ is a pole, at $A$ and $B$ respectively; draw $Ca$, $Cb$, $OA$,
+$OB$. Then $Ca$, $Cb$, $OA$, $OB$ are all perpendicular to $OP$, because
+the planes $aCb$ and $AOB$ are perpendicular to $OP$; therefore $Ca$
+is parallel to $OA$, and $Cb$ is parallel to $OB$. Therefore the angle
+$aCb=\text{the angle }AOB$ (Euclid, \textsc{xi}.\ 10). Hence,
+\[
+  \frac{\operatorname{arc} ab} {\operatorname{radius} Ca}=\frac{\operatorname{arc} AB} {\operatorname{radius} OA},
+  \text{ (\textit{Plane Trigonometry}, Art.\ 18)};
+\]
+therefore,\hfill
+$\displaystyle \frac{\operatorname{arc} ab}{\operatorname{arc} AB}=\frac{Ca}{OA}
+=\frac{Ca}{Oa}=\sin POa$.\hfill\phantom{\text{therefore,}}
+
+\chapter[Spherical Triangles.]{SPHERICAL TRIANGLES.}
+
+\paragraph{15.} Spherical Trigonometry investigates the relations which
+subsist between the angles of the plane faces which form a solid
+angle and the angles at which the plane faces are inclined to each
+other.
+%-----File: 017.png------------------------------------------------
+
+\paragraph{16.} Suppose that the angular point of a solid angle is made
+the centre of a sphere; then the planes which form the solid angle
+will cut the sphere in arcs of great circles. Thus a figure will be
+formed on the surface of the sphere which is called a \textit{spherical
+triangle} if it is bounded by \textit{three} arcs of great circles; this will be
+the case when the solid angle is formed by the meeting of \textit{three}
+plane angles. If the solid angle be formed by the meeting of
+\textit{more than three} plane angles, the corresponding figure on the
+surface of the sphere is bounded by more than three arcs of great
+circles, and is called a \textit{spherical polygon}.
+
+\paragraph{17.} The three arcs of great circles which form a spherical
+triangle are called the \textit{sides} of the spherical triangle; the angles
+formed by the arcs at the points where they meet are called the
+\textit{angles} of the spherical triangle. (See Art.~9.)
+
+\paragraph{18.} Thus, let $O$ be the centre of a sphere, and suppose a solid
+angle formed at $O$ by the meeting of three plane angles. Let
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/017fc}
+\end{figure}
+$AB$, $BC$, $CA$ be the arcs of great circles in which the planes cut
+the sphere; then $ABC$ is a spherical triangle, and the arcs $AB$,
+$BC$, $CA$ are its sides. Suppose $Ab$ the tangent at $A$ to the arc
+$AB$, and $Ac$ the tangent at $A$ to the arc $AC$, the tangents being
+drawn from $A$ \textit{towards} $B$ and $C$ respectively; then the angle $bAc$
+is one of the angles of the spherical triangle. Similarly angles
+formed in like manner at $B$ and $C$ are the other angles of the
+spherical triangle.
+%-----File: 018.png------------------------------------------------
+
+\paragraph{19.} The principal part of a treatise on Spherical Trigonometry
+consists of theorems relating to spherical triangles; it is therefore
+necessary to obtain an accurate conception of a spherical triangle
+and its parts.
+
+It will be seen that what are called \textit{sides} of a spherical
+triangle are really \textit{arcs} of great circles, and these arcs are proportional
+to the three plane angles which form the solid angle
+corresponding to the spherical triangle. Thus, in the figure of
+the preceding Article, the arc $AB$ forms one side of the spherical
+triangle $ABC$, and the plane angle $AOB$ is measured by the fraction
+$\dfrac{\operatorname{arc} AB} {\operatorname{radius} OA}$; and thus the arc $AB$ is proportional to the angle
+$AOB$ so long as we keep to the same sphere.
+
+The \textit{angles} of a spherical triangle are the inclinations of the
+plane faces which form the solid angle; for since $Ab$ and $Ac$ are
+both perpendicular to $OA$, the angle $bAc$ is the angle of inclination
+of the planes $OAB$ and $OAC$.
+
+\paragraph{20.} The letters $A$, $B$, $C$ are generally used to denote the
+\textit{angles} of a spherical triangle, and the letters $a$, $b$, $c$ are used to
+denote the \textit{sides}. As in the case of plane triangles, $A$, $B$, and $C$
+may be used to denote the numerical values of the angles expressed
+in \textit{terms of any unit}, provided we understand distinctly what the
+unit is. Thus, if the angle $C$ be a right angle, we may say that
+$C = 90^\circ$, or that $C = \dfrac{\pi}{2}$, according as we adopt for the unit a degree
+or the angle subtended at the centre by an arc equal to the
+radius. So also, as the sides of a spherical triangle are proportional
+to the angles subtended at the centre of the sphere, we
+may use $a$, $b$, $c$ to denote the numerical values of those angles
+in terms of any unit. We shall usually suppose both the angles
+and sides of a spherical triangle expressed in \textit{circular measure}.
+(\textit{Plane Trigonometry}, Art.\ 20.)
+%-----File: 019.png------------------------------------------------
+
+\paragraph{21.} In future, unless the contrary be distinctly stated, any
+arc drawn on the surface of a sphere will be supposed to be an arc
+of a \textit{great} circle.
+
+\paragraph{22.} In spherical triangles each side is restricted to be less
+than a semicircle; this is of course a \textit{convention}, and it is adopted
+because it is found convenient.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/019fc}
+\end{figure}
+
+Thus, in the figure, the arc $ADEB$ is greater than a semicircumference,
+and we might, if we pleased, consider $ADEB$, $AC$,
+and $BC$ as forming a triangle, having its angular points at $A$, $B$,
+and $C$. But we agree to exclude such triangles from our consideration;
+and the triangle having its angular points at $A$, $B$,
+and $C$, will be understood to be that formed by $AFB$, $BC$, and $CA$.
+
+\paragraph{23.} From the restriction of the preceding Article it will
+follow that \textit{any angle of a spherical triangle is less than two right
+angles}.
+
+For suppose a triangle formed by $BC$, $CA$, and $BEDA$, having
+the angle $BCA$ greater than two right angles. Then suppose $D$
+to denote the point at which the arc $BC$, if produced, will meet
+$AE$; then $BED$ is a semicircle by Art.\ 10, and therefore $BEA$
+is greater than a semicircle; thus the proposed triangle is not one
+of those which we consider.
+%-----File: 020.png------------------------------------------------
+
+\chapter[Spherical Geometry.]{SPHERICAL GEOMETRY.}
+
+\paragraph{24.} The relations between the sides and angles of a Spherical
+Triangle, which are investigated in treatises on Spherical Trigonometry,
+are chiefly such as involve the \textit{Trigonometrical Functions}
+of the sides and angles. Before proceeding to these, however, we
+shall collect, under the head of Spherical Geometry, some theorems
+which involve the sides and angles \textit{themselves}, and not their trigonometrical
+ratios.
+
+\paragraph{25.} \textit{Polar triangle}. Let $ABC$ be any spherical triangle, and
+let the points $A'$, $B'$, $C'$ be those poles of the arcs $BC$, $CA$, $AB$
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/020fc}
+\end{figure}
+respectively which lie on the same sides of them as the opposite
+angles $A$, $B$, $C$; then the triangle $A'B'C'$ is said to be the \textit{polar
+triangle} of the triangle $ABC$.
+
+Since there are two poles for each side of a spherical triangle,
+\textit{eight} triangles can be formed having for their angular points poles
+of the sides of the given triangle; but there is only one triangle in
+which these poles $A'$, $B'$, $C'$ lie towards the same parts with the
+corresponding angles $A$, $B$, $C$; and this is the triangle which is
+known under the name of the \textit{polar triangle}.
+
+The triangle $ABC$ is called the \textit{primitive} triangle with respect
+to the triangle $A'B'C'$.
+%-----File: 021.png------------------------------------------------
+
+\paragraph{26.} \textit{If one triangle be the polar triangle of another, the latter
+will be the polar triangle of the former.}
+
+Let $ABC$ be any triangle, $A'B'C'$ the polar triangle: then $ABC$
+will be the polar triangle of $A'B'C'$.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/021fc}
+\end{figure}
+
+For since $B'$ is a pole of $AC$, the arc $AB'$ is a quadrant, and
+since $C'$ is a pole of $BA$, the arc $AC'$ is a quadrant (Art.\ 7); therefore
+$A$ is a pole of $B'C'$ (Art.\ 11). Also $A$ and $A'$ are on the same
+side of $B'C'$; for $A$ and $A'$ are by hypothesis on the same side of
+$BC$, therefore $A'A$ is less than a quadrant; and since $A$ is a pole
+of $B'C'$, and $AA'$ is less than a quadrant, $A$ and $A'$ are on the
+same side of $B'C'$.
+
+Similarly it may be shewn that $B$ is a pole of $C'A'$, and that $B$
+and $B'$ are on the same side of $C'A'$; also that $C$ is a pole of $A'B'$,
+and that $C$ and $C'$ are on the same side of $A'B'$. Thus $ABC$ is the
+polar triangle of $A'B'C'$.
+
+\paragraph{27.} \textit{The sides and angles of the polar triangle are respectively
+the supplements of the angles and sides of the primitive triangle.}
+
+For let the arc $B'C'$, produced if necessary, meet the arcs $AB$,
+$AC$, produced if necessary, at the points $D$ and $E$ respectively;
+then since $A$ is a pole of $B'C'$, the spherical angle $A$ is measured by
+the arc $DE$ (Art.\ 12). But $B'E$ and $C'D$ are each quadrants;
+therefore $DE$ and $B'C'$ are together equal to a semicircle; that is,
+the angle subtended by $B'C'$ at the centre of the sphere is the
+%-----File: 022.png------------------------------------------------
+supplement of the angle $A$. This we may express for shortness
+thus; $B'C'$ is the supplement of $A$. Similarly it may be shewn
+that $C'A'$ is the supplement of $B$, and $A'B'$ the supplement of $C$.
+
+And since $ABC$ is the polar triangle of $A'B'C'$, it follows that
+$BC$, $CA$, $AB$ are respectively the supplements of $A'$, $B'$, $C'$; that
+is, $A'$, $B'$, $C'$ are respectively the supplements of $BC$, $CA$, $AB$.
+
+From these properties a primitive triangle and its polar triangle
+are sometimes called \textit{supplemental triangles}.
+
+Thus, if $A$, $B$, $C$, $a$, $b$, $c$ denote respectively the angles and
+the sides of a spherical triangle, all expressed in circular measure,
+and $A'$, $B'$, $C'$, $a'$, $b'$, $c'$ those of the polar triangle, we have
+\begin{align*}
+A' &= \pi - a,& B' &= \pi - b,& C' &= \pi - c,\\
+a' &= \pi - A,& b' &= \pi - B,& c' &= \pi - C.
+\end{align*}
+
+\paragraph{28.} The preceding result is of great importance; for if any
+general theorem be demonstrated with respect to the sides and the
+angles of any spherical triangle it holds of course for the polar
+triangle also. \textit{Thus any such theorem will remain true when the
+angles are changed into the supplements of the corresponding sides
+and the sides into the supplements of the corresponding angles.} We
+shall see several examples of this principle in the next Chapter.
+
+\paragraph{29.} \textit{Any two sides of a spherical triangle are together greater
+than the third side.} (See the figure of Art.\ 18.)
+
+For any two of the three plane angles which form the solid
+angle at $O$ are together greater than the third (Euclid, \textsc{xi}.\ 20).
+Therefore any two of the arcs $AB$, $BC$, $CA$, are together greater
+than the third.
+
+From this proposition it is obvious that any side of a spherical
+triangle is greater than the difference of the other two.
+
+\paragraph{30.} \textit{The sum of the three sides of a spherical triangle is less than
+the circumference of a great circle.} (See the figure of Art.\ 18.)
+%-----File: 023.png------------------------------------------------
+
+For the sum of the three plane angles which form the solid
+angle at $O$ is less than four right angles (Euclid, \textsc{xi.}~21); therefore
+\[
+\dfrac{AB}{OA} + \dfrac{BC}{OA} + \dfrac{CA}{OA} \text{ is less than } 2\pi,
+\]
+therefore,\hfill
+$\displaystyle AB+BC+CD \text{ is less than } 2\pi\times OA;$\hfill\phantom{\text{therefore}}\\
+that is, the sum of the arcs is less than the circumference of a
+great circle.
+
+\paragraph{31.} The propositions contained in the preceding two Articles
+may be extended. Thus, if there be any polygon which has each
+of its angles less than two right angles, \textit{any one side is less than the
+sum of all the others.} This may be proved by repeated use of
+Art.~29. Suppose, for example, that the figure has four sides, and
+let the angular points be denoted by $A$, $B$, $C$, $D$. Then
+\[
+AD+BC\text{ is greater than }AC;
+\]
+therefore,\hfill
+$\displaystyle AB+BC+CD \text{ is greater than } AC+CD,$\hfill\phantom{\text{therefore}}\\
+and \textit{{\`a} fortiori} greater than $AD$.
+
+Again, if there be any polygon which has each of its angles
+less than two right angles, \textit{the sum of its sides will be less than the
+circumference of a great circle}. This follows from Euclid, \textsc{xi.}~21,
+in the manner shewn in Art.~30.
+
+\paragraph{32.} \textit{The three angles of a spherical triangle are together greater
+than two right angles and less than six right angles.}
+
+Let $A$, $B$, $C$ be the \textit{angles} of a spherical triangle; let $a'$, $b'$, $c'$
+be the \textit{sides} of the polar triangle. Then by Art.~30,
+\[
+a'+b'+c'\text{ is less than }2\pi,
+\]
+that is,\hfill
+$\displaystyle \pi-A+\pi-B+\pi-C\text{ is less than }2\pi;$\hfill{\phantom{that is}}\\
+therefore,\hfill
+$\displaystyle A+B+C\text{ is greater than }\pi$.\hfill\phantom{\text{therefore}}\\
+
+And since each of the angles $A$, $B$, $C$ is less than $\pi$, the sum
+$A+B+C$ is less than $3\pi$.
+%-----File: 024.png------------------------------------------------
+
+\paragraph{33.} \textit{The angles at the base of an isosceles spherical triangle are
+equal.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/024fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle having $AC=BC$; let $O$ be the
+centre of the sphere. Draw tangents at the points $A$ and $B$ to the
+arcs $AC$ and $BC$ respectively; these will meet $OC$ produced at the
+same point $S$, and $AS$ will be equal to $BS$.
+
+Draw tangents $AT$, $BT$ at the points $A$, $B$ to the arc $AB$; then
+$AT=TB$; join $TS$. In the two triangles $SAT$, $SBT$ the sides
+$SA$, $AT$, $TS$ are equal to $SB$, $BT$, $TS$ respectively; therefore the
+angle $SAT$ is equal to the angle $SBT$; and these are the angles at
+the base of the spherical triangle.
+
+The figure supposes $AC$ and $BC$ to be less than quadrants; if
+they are greater than quadrants the tangents to $AC$ and $BC$ will
+meet on $CO$ produced through $O$ instead of through $C$, and the
+demonstration may be completed as before. If $AC$ and $BC$ are
+quadrants, the angles at the base are right angles by Arts.~11
+and 9.
+
+\paragraph{34.} \textit{If two angles of a spherical triangle are equal, the opposite
+sides are equal.}
+
+Since the primitive triangle has two equal angles, the polar
+triangle has two equal sides; therefore in the polar triangle the
+angles opposite the equal sides are equal by Art.~33. Hence in
+the primitive triangle the sides opposite the equal angles are
+equal.
+%-----File: 025.png------------------------------------------------
+
+\paragraph{35.} \textit{If one angle of a spherical triangle be greater than another,
+the side opposite the greater angle is greater than the side
+opposite the less angle.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/025fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle, and let the angle $ABC$ be
+greater than the angle $BAC$: then the side $AC$ will be greater
+than the side $BC$. At $B$ make the angle $ABD$ equal to the angle
+$BAD$; then $BD$ is equal to $AD$ (Art.\ 34), and $BD+DC$ is greater
+than $BC$ (Art.\ 29); therefore $AD+DC$ is greater than $BC$; that
+is, $AC$ is greater than $BC$.
+
+\paragraph{36.} \textit{If one side of a spherical triangle be greater than another,
+the angle opposite the greater side is greater than the angle opposite
+the less side.}
+
+This follows from the preceding Article by means of the polar
+triangle.
+
+Or thus; suppose the side $AC$ greater than the side $BC$, then
+the angle $ABC$ will be greater than the angle $BAC$. For the
+angle $ABC$ cannot be less than the angle $BAC$ by Art.\ 35, and
+the angle $ABC$ cannot be equal to the angle $BAC$ by Art.\ 34;
+therefore the angle $ABC$ must be greater than the angle $BAC$.
+
+This Chapter might be extended; but it is unnecessary to do
+so because the Trigonometrical formul\ae\ of the next Chapter supply
+an easy method of investigating the theorems of Spherical
+Geometry. See Arts.\ 56, 57, and 58.
+%-----File: 026.png------------------------------------------------
+
+\chapter{Relations between the Trigonometrical
+Functions of the Sides and the Angles
+of a Spherical Triangle.}
+\chaptermark{RELATIONS BETWEEN THE FUNCTIONS.}
+
+\paragraph{37.} \textit{To express the cosine of an angle of a triangle in terms of
+sines and cosines of the sides.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/026fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle, $O$ the centre of the sphere.
+Let the tangent at $A$ to the arc $AC$ meet $OC$ produced at $E$, and
+let the tangent at $A$ to the arc $AB$ meet $OB$ produced at $D$; join
+$ED$. Thus the angle $EAD$ is the angle $A$ of the spherical triangle,
+and the angle $EOD$ measures the side $a$.
+
+From the triangles $ADE$ and $ODE$ we have
+\begin{align*}
+DE^2 &= AD^2 + AE^2 - 2AD \centerdot AE \cos A,\\
+DE^2 &= OD^2 + OE^2 - 2OD \centerdot OE \cos a;
+\end{align*}
+also the angles $OAD$ and $OAE$ are right angles, so that
+$OD^2 = OA^2 + AD^2$ and $OE^2 = OA^2 + AE^2$. Hence by subtraction
+we have
+\begin{flalign*}
+\multispan6{\hfil$0 = 2OA^2 + 2AD \centerdot AE \cos A - 2OD \centerdot OE \cos a$;\hfil}\\[1ex]
+&\text{therefore}& \cos a
+&= \dfrac{OA}{OE} \centerdot \dfrac{OA}{OD}
+ + \dfrac{AE}{OE} \centerdot \dfrac{AD}{OD} \cos A;
+&\phantom{\text{therefore}}&\\[1ex]
+%
+&\text{that is}& \cos a &= \cos b \cos c + \sin b \sin c \cos A.\\[1ex]
+%
+&\text{\indent Therefore}& \multispan{2}{\hfil$\cos A = \dfrac{\cos a - \cos b \cos c}{\sin b \sin c}$.\hfil}
+\end{flalign*}
+%-----File: 027.png------------------------------------------------
+
+\paragraph{38.} We have supposed, in the construction of the preceding
+Article, that the sides which contain the angle $A$ are less than
+quadrants, for we have assumed that the tangents at $A$ meet $OB$
+and $OC$ respectively produced. We must now shew that the
+formul\ae\ obtained is true when these sides are not less than quadrants.
+This we shall do by special examination of the cases in
+which one side or each side is greater than a quadrant or equal to
+a quadrant.
+
+(1) Suppose only one of the sides which contain the angle $A$
+to be greater than a quadrant, for example, $AB$. Produce $BA$
+and $BC$ to meet at $B'$; and put $AB' = c'$, $CB' = a'$.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/027f1c}
+\end{figure}
+
+Then we have from the triangle $AB'C$, by what has been
+already proved,
+\[
+  \cos a' = \cos b \cos c' + \sin b \sin c' \cos B'AC;
+\]
+but $a' = \pi - a$, $c' = \pi - c$, $B'AC = \pi - A$; thus
+\[
+  \cos a = \cos b \cos c + \sin b \sin c \cos A.
+\]
+
+(2) Suppose both the sides which contain the angle $A$ to be
+greater than quadrants. Produce $AB$ and $AC$ to meet at $A'$; put
+$A'B = c', A'C = b'$; then from the triangle $A'BC$, as before,
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/027f2c}
+\end{figure}
+\[
+  \cos a = \cos b' \cos c' + \sin b' \sin c' \cos A';
+\]
+but $b' = \pi - b$, $c' = \pi - c$, $A' = A$; thus
+\[
+  \cos a = \cos b \cos c + \sin b \sin c \cos A.
+\]
+%-----File: 028.png------------------------------------------------
+
+(3) Suppose that one of the sides which contain the angle $A$
+is a quadrant, for example, $AB$; on $AC$, produced if necessary,
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/028fc}
+\end{figure}
+take $AD$ equal to a quadrant and draw $BD$. If $BD$ is a quadrant
+$B$ is a pole of $AC$ (Art.\ 11); in this case $a = \dfrac{\pi}{2}$ and $A = \dfrac{\pi}{2}$ as well
+as $c = \dfrac{\pi}{2}$. Thus the formula to be verified reduces to the identity
+$0 = 0$. If $BD$ be not a quadrant, the triangle $BDC$ gives
+\[
+\cos a = \cos CD\cos BD + \sin CD\sin BD \cos CDB ,
+\]
+and \hfill$\ \cos CDB = 0,\ \
+  \cos CD = \cos \left(\dfrac{\pi}{2} - b\right) = \sin b ,\ \
+ \cos BD = \cos A $;\hfill\phantom{and}\\[1ex]
+thus \hfill$ \cos a = \sin b \cos A ; $\hfill \phantom{thus}\\[2ex]
+and this is what the formula in Art.\ 37 becomes when $c = \dfrac{\pi}{2}$.
+
+(4) Suppose that both the sides which contain the angle $A$
+are quadrants. The formula then becomes $\cos a = \cos A $; and this
+is obviously true, for $A$ is now the pole of $BC$, and thus $A = a$.
+
+Thus the formula in Art.\ 37 is proved to be universally true.
+\paragraph{39.} The formula in Art.\ 37 may be applied to express the
+cosine of any angle of a triangle in terms of sines and cosines of
+the sides; thus we have the three formul\ae,
+\begin{align*}
+\cos a &= \cos b \cos c + \sin b \sin c \cos A,\\
+\cos b &= \cos c \cos a + \sin c \sin a \cos B,\\
+\cos c &= \cos a \cos b + \sin a \sin b \cos C.
+\end{align*}
+%-----File: 029.png------------------------------------------------
+These may be considered as the fundamental equations of Spherical
+Trigonometry; we shall proceed to deduce various formul\ae\
+from them.
+
+\paragraph{40.} \textit{To express the sine of an angle of a spherical triangle in
+terms of trigonometrical functions of the sides.}
+\begin{flalign*}
+&\text{\indent We have}& \cos A&=\dfrac{\cos a-\cos b\cos c}{\sin b\sin c};\\[1.5ex]
+%
+&\text{therefore}& \sin^2 A&=1-\left(\dfrac{\cos a-\cos b\cos c}{\sin b\sin c}\right)^2\\[1.5ex]
+%
+&&&=\dfrac{(1-\cos^2 b)(1-\cos^2 c)-(\cos a-\cos b\cos c)^2}{\sin^2 b\sin^2 c}\\[1.5ex]
+%
+&&&=\dfrac{1-\cos^2 a-\cos^2 b-\cos^2 c+2\cos a\cos b\cos c}{\sin^2 b \sin^2 c};&\phantom{therefore}\\[1.5ex]
+%
+&\text{therefore} &\sin A&=\dfrac{\surd(1-\cos^2 a-\cos^2 b-\cos^2 c+2\cos a\cos b\cos c)}{\sin b \sin c}.
+\end{flalign*}
+The radical on the right-hand side must be taken with the positive
+sign, because $\sin b$, $\sin c$, and $\sin A$ are all positive.
+
+\paragraph{41.} From the value of $\sin A$ in the preceding Article it follows
+that
+\begin{gather*}
+\dfrac{\sin A}{\sin a}=\dfrac{\sin B}{\sin b}=\dfrac{\sin C}{\sin c},\\
+%
+\intertext{for each of these is equal to the same expression, namely,}
+%
+\dfrac{\surd(1-\cos^2 a-\cos^2 b-\cos^2 c+2\cos a\cos b\cos c)}{\sin a\sin b\sin c}.
+\end{gather*}
+Thus \textit{the sines of the angles of a spherical triangle are proportional
+to the sines of the opposite sides}. We will give an independent
+proof of this proposition in the following Article.
+
+\paragraph{42.} \textit{The sines of the angles of a spherical triangle are proportional
+to the sines of the opposite sides.}
+%-----File: 030.png------------------------------------------------
+
+Let $ABC$ be a spherical triangle, $O$ the centre of the sphere.
+Take any point $P$ in $OA$, draw $PD$ perpendicular to the plane
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/030fc}
+\end{figure}
+$BOC$, and from $D$ draw $DE$, $DF$ perpendicular to $OB$, $OC$ respectively;
+join $PE$, $PF$, $OD$.
+
+Since $PD$ is perpendicular to the plane $BOC$, it makes right
+angles with every straight line meeting it in that plane; hence
+\[
+PE^2 = PD^2 + DE^2 = PO^2-OD^2 + DE^2 = PO^2-OE^2;
+\]
+thus $PEO$ is a right angle. Therefore $PE = OP\sin POE= OP \sin c$;
+and $PD = PE \sin PED = PE \sin B = OP \sin c \sin B$.
+
+Similarly, $PD = OP \sin b \sin C$; therefore
+\begin{flalign*}
+&&OP \sin c \sin B &= OP \sin b \sin C;\\
+&\text{therefore} &\dfrac{\sin B}{\sin C}&=\dfrac{\sin b}{\sin c}.&\phantom{\text{therefore}}&
+\end{flalign*}
+
+The figure supposes $b$, $c$, $B$, and $C$ each less than a right angle;
+it will be found on examination that the proof will hold when the
+figure is modified to meet any case which can occur. If, for
+instance, $B$ alone is greater than a right angle, the point $D$ will
+fall beyond $OB$ instead of between $OB$ and $OC$; then $PED$ will
+be the \textit{supplement} of $B$, and thus $\sin PED$ is still equal to $\sin B$.
+
+\paragraph{43.} \textit{To shew that} $\cot a \sin b = \cot A \sin C + \cos b \cos C$.
+\begin{flalign*}
+&\text{\indent We have} &\cos a &= \cos b \cos c + \sin b \sin c \cos A,\\
+&&\cos c &= \cos a \cos b + \sin a \sin b \cos C,\\
+&&\sin c &= \sin a\,\dfrac{\sin C}{\sin A}.&\phantom{We have}
+\end{flalign*}
+%-----File: 031.png------------------------------------------------
+
+Substitute the values of $\cos c$ and $\sin c$ in the first equation;
+thus
+\[
+\cos a = (\cos a \cos b + \sin a \sin b \cos C) \cos b +
+\dfrac{\sin a \sin b \cos A \sin C}{\sin A};
+\]
+by transposition
+\[
+\cos a \sin^2 b = \sin a \sin b \cos b \cos C + \sin a \sin b \cot A \sin C;
+\]
+divide by $\sin a \sin b$; thus
+\[
+\cot a \sin b = \cos b \cos C + \cot A \sin C.
+\]
+
+\paragraph{44.} By interchanging the letters five other formul\ae\ may be
+obtained like that in the preceding Article; the whole six formul\ae\
+will be as follows:
+\begin{align*}
+&\cot a \sin b = \cot A \sin C + \cos b \cos C, \\
+&\cot b \sin a = \cot B \sin C + \cos a \cos C, \\
+&\cot b \sin c = \cot B \sin A + \cos c \cos A, \\
+&\cot c \sin b = \cot C \sin A + \cos b \cos A, \\
+&\cot c \sin a = \cot C \sin B + \cos a \cos B, \\
+&\cot a \sin c = \cot A \sin B + \cos c \cos B.
+\end{align*}
+
+\paragraph{45.} \textit{To express the sine, cosine, and tangent, of half an angle
+of a triangle as functions of the sides.}
+
+We have, by Art.~37, $\cos A = \dfrac{\cos a - \cos b \cos c}{\sin b \sin c}$;\\[1.5ex]
+therefore\hfill
+$\displaystyle 1 - \cos A = 1 - \dfrac{\cos a - \cos b \cos c}{\sin b \sin c} =
+\dfrac{\cos (b - c) - \cos a}{\sin b \sin c};$\hfill\phantom{\text{therefore}}\\[2ex]
+therefore\hfill
+$\displaystyle \sin^2 \dfrac{A}{2} = \dfrac{\sin \tfrac{1}{2}(a+b-c)\sin\tfrac{1}{2}(a-b+c)}{\sin b \sin c}$.\hfill\phantom{\text{therefore}}\\
+
+Let $2s = a + b + c$, so that $s$ is half the sum of the sides of
+the triangle; then
+\[
+a + b - c = 2s - 2c = 2(s - c),\quad
+a - b + c = 2s - 2b = 2(s - b);
+\]
+\begin{flalign*}
+&\text{thus,} &
+\sin^2 \dfrac{A}{2} &= \dfrac{\sin(s - b)\sin(s - c)}{\sin b \sin c}, &\phantom{\text{thus,}}\\
+%-----File: 032.png------------------------------------------------
+&\text{and}&
+\sin\dfrac{A}{2}&=\Surd{\left\{
+  \dfrac{\sin(s-b)\sin(s-c)}{\sin b \sin c}
+  \right\}}
+  &\phantom{\text{Also,}}\\[1ex]
+&\text{Also,}&
+1+\cos A &= 1+\dfrac{\cos a-\cos b\cos c}{\sin b \sin c}=
+  \dfrac{\cos a-\cos(b+c)}{\sin b \sin c};
+\end{flalign*}
+therefore
+\begin{flalign*}
+&&
+\cos^2\dfrac{A}{2} &=
+  \dfrac{\sin\tfrac{1}{2}(a+b+c)\sin\tfrac{1}{2}(b+c-a)}{\sin b\sin c} =
+  \dfrac{\sin s \sin(s-a)}{\sin b \sin c},&\phantom{\text{and}}\\[1.5ex]
+&\text{and}&
+\cos\dfrac{A}{2}&=\Surd{\left\{
+  \dfrac{\sin s \sin(s-a)}{\sin b \sin c}\right\}}
+\end{flalign*}
+
+From the expressions for $\sin\dfrac{A}{2}$ and $\cos\dfrac{A}{2}$ we deduce
+\[
+\tan\dfrac{A}{2}=\Surd{\left\{
+  \dfrac{\sin(s-b)\sin(s-c)}{\sin s \sin(s-a)} \right\}}.
+\]
+
+The positive sign must be given to the radicals which occur in
+this Article, because $\dfrac{A}{2}$ is less than a right angle, and therefore its
+sine, cosine, and tangent are all positive.
+
+\paragraph{46.} Since $\sin A = 2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}$, we obtain
+\[
+\sin A = \dfrac{2}{\sin b \sin c}\{\sin s \sin(s-a)\sin(s-b)\sin(s-c)\}^{\tfrac{1}{2}}.
+\]
+
+It may be shewn that the expression for $\sin A$ in Art.\ 40
+agrees with the present expression by putting the numerator of
+that expression in factors, as in \textit{Plane Trigonometry}, Art.\ 115.
+We shall find it convenient to use a symbol for the radical in the
+value of $\sin A$; we shall denote it by $n$, so that
+\begin{flalign*}
+&&n^2&=\sin s \sin(s-a)\sin(s-b)\sin(s-c),&\phantom{\text{and}}\\
+&\text{and} &4n^2&=1-\cos^2 a-\cos^2 b - \cos^2 c + 2\cos a\cos b\cos c.
+\end{flalign*}
+%-----File: 033.png------------------------------------------------
+
+\paragraph{47.} \textit{To express the cosine of a side of a triangle in terms of
+sines and cosines of the angles.}
+
+In the formula of Art.\ 37 we may, by Art.\ 28, change the
+sides into the supplements of the corresponding angles and the
+angle into the supplement of the corresponding side; thus
+\begin{flalign*}
+&&\multispan{2}{\hfil\clap{$\cos (\pi-A) = \cos (\pi-B) \cos (\pi-C)+\sin (\pi-B)\sin(\pi-C) \cos(\pi-a)$,}\hfil}\\
+&\text{that is,}&\cos A &=-\cos B \cos C + \sin B \sin C \cos a.\\
+&\text{Similarly}&\cos B &=-\cos C \cos A + \sin C \sin A \cos b,&\phantom{Similarly}&\\
+&\text{and}&\cos C &=-\cos A \cos B + \sin A \sin B \cos c.
+\end{flalign*}
+
+\paragraph{48.} The formul\ae\ in Art.\ 44 will of course remain true when
+the angles and sides are changed into the supplements of the corresponding
+sides and angles respectively; it will be found, however,
+that no \textit{new} formul\ae\ are thus obtained, but only the \textit{same}
+formul\ae\ over again. This consideration will furnish some assistance
+in retaining those formul\ae\ accurately in the memory.
+
+\paragraph{49.} \textit{To express the sine, cosine, and tangent, of half a side of a
+triangle as functions of the angles.}
+
+We have, by Art.\ 47, $\cos a=\dfrac{\cos A + \cos B \cos C}{\sin B\sin C}$;
+\begin{flalign*}
+&\text{therefore}&\\
+&&\multispan{2}{\hfil\clap{$\displaystyle1-\cos a= 1-\dfrac{\cos A + \cos B\cos C}{\sin B\sin C}= -\dfrac{\cos A+\cos(B+C)}{\sin B\sin C}$;}\hfil}\\[2ex]
+&\text{therefore}&\sin^{2}\dfrac{a}{2}&= -\frac{\cos\tfrac{1}{2}(A+B+C)\cos\tfrac{1}{2}(B+C-A)}{\sin B\sin C}.&\phantom{therefore}&\\[2ex]
+&\rlap{\indent Let $2S=A+B+C$; then $B+C-A=2(S-A)$, therefore}&\vphantom{\dfrac a2}\\
+&&\multispan{2}{\hfil$\displaystyle\sin^{2}\frac{a}{2}=-\frac{\cos S\cos(S-A)}{\sin B\sin C}$,\hfil}\\[2ex]
+&\text{and}&\multispan{2}{\hfil$\displaystyle\sin\frac{a}{2}=\Surd{\left\{-\dfrac{\cos S\cos(S-A)}{\sin B\sin C}\right\}}$.\hfil}
+\end{flalign*}
+%-----File: 034.png------------------------------------------------
+Also\ \hfil$1+\cos a=1+\dfrac{\cos A + \cos B \cos C}{\sin B\sin C}=\dfrac{\cos A +\cos (B-C)}{\sin B\sin C}$;\hfil\break\\
+therefore
+\[
+\cos^{2}\dfrac{a}{2}=
+\dfrac{\cos\tfrac{1}{2}(A-B+C)\cos\tfrac{1}{2}(A+B-C)}{\sin B\sin C}=
+\dfrac{\cos(S-B)\cos(S-C)}{\sin B\sin C},
+\]
+\begin{flalign*}
+&\text{and}&\cos\dfrac{a}{2}&=\Surd{\left\{\dfrac{\cos(S-B)\cos(S-C)}{\sin B\sin C}\right\}}.\\[2ex]
+&\text{Hence }&\tan\dfrac{a}{2}&=\Surd{\left\{-\dfrac{\cos S\cos(S-A)}{\cos(S-B)\cos(S-C)}\right\}}.&\phantom{Hence}&
+\end{flalign*}
+
+The positive sign must be given to the radicals which occur in
+this Article, because $\dfrac{a}{2}$ is less than a right angle.
+
+\paragraph{50.} The expressions in the preceding Article may also be
+obtained immediately from those given in Art.\ 45 by means of
+Art.\ 28.
+
+It may be remarked that the values of $\sin\dfrac{a}{2}$, $\cos\dfrac{a}{2}$, and $\tan\dfrac{a}{2}$
+are \textit{real}. For $S$ is greater than one right angle and less than three
+right angles by Art.\ 32; therefore $\cos S$ is \textit{negative}. And in the
+polar triangle any side is less than the sum of the other two; thus
+$\pi-A$ is less than $\pi-B + \pi-C$; therefore $B + C - A$ is less than
+$\pi$; therefore $S - A$ is less than $\dfrac{\pi}{2}$, and $B + C - A$ is algebraically
+greater than $-\pi$, so that $S-A$ is algebraically greater than $-\dfrac{\pi}{2}$;
+therefore $\cos (S-A)$ is \textit{positive}. Similarly also $\cos (S-B)$ and
+$\cos (S-C)$ are positive. Hence the values of $\sin\dfrac{a}{2}$, $\cos\dfrac{a}{2}$, and $\tan\dfrac{a}{2}$
+are real.
+
+\paragraph{51.} Since $\sin a= 2 \sin\dfrac{a}{2}\cos\dfrac{a}{2}$, we obtain
+\[
+\sin a=\dfrac{2}{\sin B\sin C}\left\{-\cos S\cos (S-A)\cos(S-B)\cos (S-C)\right\}^{\tfrac{1}{2}}.
+\]
+
+We shall use $N$ for $\left\{-\cos S\cos (S-A)\cos(S-B)\cos (S-C)\right\}^{\tfrac{1}{2}}$.
+%-----File: 035.png------------------------------------------------
+
+\paragraph{52.} \textit{To demonstrate Napier's Analogies}.
+\begin{flalign*}
+&\rlap{\indent We have}&\multispan{2}{\hfil$\displaystyle\dfrac{\sin A}{\sin a} = \dfrac{\sin B}{\sin b} = m$ suppose;\hfil}\\
+&\rlap{then, by a theorem of Algebra,}\\
+&&m &= \dfrac{\sin A + \sin B}{\sin a + \sin b},\tag{1}\\[1ex]
+&\rlap{and also}&m &= \dfrac{\sin A - \sin B}{\sin a - \sin b}.\tag{2}\\
+&\text{Now}&\cos A + \cos B \cos C &= \sin B \sin C \cos a = m \sin C \sin b \cos a,&\phantom{Now}\\
+&\text{and}&\cos B + \cos A \cos C &= \sin A \sin C \cos b = m \sin C \sin a \cos b,
+\end{flalign*}
+therefore, by addition,
+\[
+(\cos A + \cos B)(1 + \cos C) = m\sin C \sin (a + b);\tag{3}
+\]
+therefore by (1) we have
+\begin{flalign*}
+&&\dfrac{\sin A + \sin B}{\cos A + \cos B} &= \dfrac{\sin a + \sin b}{\sin(a + b)}\dfrac{1 + \cos C}{\sin C},\\[1ex]
+&\text{that is,}&\tan\tfrac{1}{2}(A + B) &= \frac{\cos\tfrac{1}{2}(a - b)}{\cos\tfrac{1}{2}(a + b)}\cot\frac{C}{2}.\tag{4}&\phantom{that is,}&
+\end{flalign*}
+
+Similarly from (3) and (2) we have
+\begin{flalign*}
+&&\dfrac{\sin A - \sin B}
+     {\cos A + \cos B} &=
+\dfrac{\sin a - \sin b}
+     {\sin(a + b)}\,
+\dfrac{1 + \cos C}
+     {\sin C},\\[1ex]
+&\text{that is,}&\tan\frac{1}{2}(A - B) &=
+\frac{\sin\tfrac{1}{2}(a - b)}
+     {\sin\tfrac{1}{2}(a + b)}
+      \cot\frac{C}{2}.\tag{5}&\phantom{that is,}
+\end{flalign*}
+
+By writing $\pi-A$ for $a$, and so on in (4) and (5) we obtain
+\begin{align*}
+\tan\tfrac{1}{2}(a + b) &=
+\frac{\cos\tfrac{1}{2}(A - B)}
+     {\cos\tfrac{1}{2}(A + B)}
+      \tan\frac{c}{2},			\tag{6}\\[1ex]
+\tan\tfrac{1}{2}(a - b) &=
+\frac{\sin\tfrac{1}{2}(A - B)}
+     {\sin\tfrac{1}{2}(A + B)}
+      \tan\frac{c}{2}.			\tag{7}
+\end{align*}
+
+The formul\ae\ (4), (5), (6), (7) may be put in the form of proportions
+or analogies, and are called from their discoverer \textit{Napier's
+%-----File: 036.png------------------------------------------------
+Analogies:} the last two may be demonstrated without recurring
+to the polar triangle by starting with the formul\ae\ in Art.~39.
+
+\paragraph{53.} In equation (4) of the preceding Article, $\cos\tfrac{1}{2}(a - b)$ and
+$\cot\dfrac{C}{2}$ are necessarily positive quantities; hence the equation
+shews that $\tan\tfrac{1}{2}(A + B)$ and $\cos\tfrac{1}{2}(a + b)$ are of the same sign;
+thus $\tfrac{1}{2}(A+B)$ and $\tfrac{1}{2}(a + b)$ are either both less than a right angle or both greater than a right angle. This is expressed by saying
+that $\tfrac{1}{2}(A+B)$ and $\tfrac{1}{2}(a + b)$ \textit{are of the same affection}.
+
+\paragraph{54.} \textit{To demonstrate Delambre's Analogies.}
+
+We have $\cos c = \cos a \cos b + \sin a \sin b \cos C$; therefore
+\begin{align*}
+1 + \cos c &= 1 + \cos a \cos b +
+                  \sin a \sin b (\cos^2\tfrac{1}{2}C -
+                                 \sin^2\tfrac{1}{2}C)\\
+%
+           &= \{1 + \cos (a - b)\} \cos^2\tfrac{1}{2}C +
+              \{1 + \cos (a + b)\} \sin^2\tfrac{1}{2}C;
+\end{align*}
+therefore $\displaystyle%
+    \cos^2\tfrac{1}{2}c =
+    \cos^2\tfrac{1}{2}(a - b)
+    \cos^2\tfrac{1}{2}C +
+    \cos^2\tfrac{1}{2}(a + b)
+    \sin^2\tfrac{1}{2}C$.
+\\
+Similarly, $\displaystyle%
+    \sin^2\tfrac{1}{2}c =
+    \sin^2\tfrac{1}{2}(a - b)
+    \cos^2\tfrac{1}{2}C +
+    \sin^2\tfrac{1}{2}(a + b)
+    \sin^2\tfrac{1}{2}C$.
+
+Now add unity to the square of each member of Napier's first
+two analogies; hence by the formul\ae\ just proved
+\begin{align*}
+    \sec^{2}\tfrac{1}{2}(A + B) &=
+        \frac{\cos^{2}\tfrac{1}{2}c}
+        {\cos^{2}\tfrac{1}{2}(a + b)
+        \sin^{2}\tfrac{1}{2}C},\\[1ex]
+%
+    \sec^{2}\tfrac{1}{2}(A - B) &=
+        \frac{\sin^{2}\tfrac{1}{2}c}
+        {\sin^{2}\tfrac{1}{2}(a + b)
+        \sin^{2}\tfrac{1}{2}C}.
+\end{align*}
+
+Extract the square roots; thus, since $\frac{1}{2}(A + B)$ and $\frac{1}{2}(a + b)$
+are of the same affection, we obtain
+\begin{align*}
+    \cos\tfrac{1}{2}(A + B)
+    \cos\tfrac{1}{2}c &=
+    \cos\tfrac{1}{2}(a + b)
+    \sin\tfrac{1}{2}C, \tag{1}\\[2ex]
+%
+    \cos\tfrac{1}{2}(A - B)
+    \sin\tfrac{1}{2}c &=
+    \sin\tfrac{1}{2}(a + b)
+    \sin\tfrac{1}{2}C. \tag{2}
+\end{align*}
+
+Multiply the first two of Napier's analogies respectively by
+these results; thus
+\begin{align*}
+    \sin\tfrac{1}{2}(A + B)
+    \cos\tfrac{1}{2}c &=
+    \cos\tfrac{1}{2}(a - b)
+    \cos\tfrac{1}{2}C, \tag{3}\\[2ex]
+%
+    \sin\tfrac{1}{2}(A - B)
+    \sin\tfrac{1}{2}c &=
+    \sin\tfrac{1}{2}(a - b)
+    \cos\tfrac{1}{2}C. \tag{4}
+\end{align*}
+%-----File: 037.png------------------------------------------------
+
+The last four formul\ae\ are commonly, but improperly, called
+\textit{Gauss's Theorems}; they were first given by Delambre in the
+\textit{Connaissance des Tems} for 1809, page~445. See the \textit{Philosophical
+Magazine} for February, 1873.
+
+\paragraph{55.} The properties of supplemental triangles were proved
+geometrically in Art.~27, and by means of these properties the
+formul\ae\ in Art.~47 were obtained; but these formul\ae\ may be
+deduced analytically from those in Art.~39, and thus the whole
+subject may be made to depend on the formul\ae\ of Art.~39.
+
+For from Art.~39 we obtain expressions for $\cos A$, $\cos B$, $\cos C$;
+and from these we find
+\begin{multline*}
+\cos A + \cos B \cos C \\
+= \dfrac{(\cos a - \cos b \cos c) \sin^2 a + (\cos b - \cos a \cos c)(\cos c - \cos a \cos b)}
+{\sin^2 a \sin b \sin c}.
+\end{multline*}
+In the numerator of this fraction write $1 - \cos^2 a$ for $\sin^2 a$; thus
+the numerator will be found to reduce to
+\[
+\cos a (1 - \cos^2 a - \cos^2 b - \cos^2 c + 2\cos a \cos b \cos c),
+\]
+and this is equal to $\cos a \sin B \sin C \sin^2 a \sin b \sin c$, (Art.~41);\\
+therefore \hfill $\cos A + \cos B \cos C - \cos a \sin B \sin C$.
+\hfill\phantom{therefore}\\
+Similarly the other two corresponding formul\ae\ may be proved.
+
+Thus the formul\ae\ in Art.~47 are established; and therefore,
+without assuming the existence and properties of the Polar Triangle,
+we deduce the following theorem: \textit{If the sides and angles
+of a spherical triangle be changed respectively into the supplements
+of the corresponding angles and sides, the fundamental formul\ae\ of
+Art.~39 hold good, and therefore also all results deducible from them.}
+
+\paragraph{56.} The formul\ae\ in the present Chapter may be applied to
+establish analytically various propositions respecting spherical triangles
+which either have been proved geometrically in the preceding
+Chapter, or may be so proved. Thus, for example, the
+second of Napier's analogies is
+\[
+\tan \tfrac{1}{2}(A-B) =
+\dfrac{\sin\tfrac{1}{2}(a-b)}{\sin\tfrac{1}{2}(a+b)}
+\cot\dfrac{C}{2};
+\]
+%-----File: 038.png------------------------------------------------
+this shews that $\tfrac{1}{2}(A - B)$ is positive, negative, or zero, according
+as $\dfrac{1}{2}(a - b)$ is positive, negative, or zero; thus we obtain all the
+results included in Arts.~33\ldots36.
+
+\paragraph{57.} \textit{If two triangles have two sides of the one equal to two
+sides of the other, each to each, and likewise the included angles
+equal, then their other angles will be equal, each to each, and likewise
+their bases will be equal.}
+
+We may shew that the bases are equal by applying the first
+formula in Art.~39 to each triangle, supposing $b$, $c$, and $A$ the
+same in the two triangles; then the remaining two formul\ae\ of
+Art.~39 will shew that $B$ and $C$ are the same in the two triangles.
+
+It should be observed that the two triangles in this case are
+\textit{not} necessarily such that one may be made to \textit{coincide with the
+other by superposition}. The sides of one may be equal to those of
+the other, each to each, but in a reverse order, as in the following
+figures.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/038fc}
+\end{figure}
+
+Two triangles which are equal in this manner are said to be
+\textit{symmetrically} equal; when they are equal so as to admit of superposition
+they are said to be \textit{absolutely} equal.
+
+\paragraph{58.} \textit{If two spherical triangles have two sides of the one equal to
+two sides of the other, each to each, but the angle which is contained
+by the two sides of the one greater than the angle which is contained
+by the two sides which are equal to them of the other, the base of that
+%-----File: 039.png------------------------------------------------
+which has the greater angle will be greater than the base of the
+other; and conversely}.
+
+Let $b$ and $c$ denote the sides which are equal in the two triangles;
+let $a$ be the base and $A$ the opposite angle of one triangle,
+and $a'$ and $A'$ similar quantities for the other. Then
+\begin{align*}
+  \cos a  &= \cos b \cos c + \sin b \sin c \cos A,\\
+  \cos a' &= \cos b \cos c + \sin b \sin c \cos A';
+\end{align*}
+therefore \hfill
+$\cos a - \cos a' + \sin b \sin c (\cos A - \cos A');$
+\hfill\phantom{therefore}\\
+that is,
+\[
+  \sin\tfrac{1}{2} (a + a')
+  \sin\tfrac{1}{2} (a - a') =
+  \sin b \sin c \sin\tfrac{1}{2} (A + A')
+                \sin\tfrac{1}{2} (A - A');
+\]
+this shews that $\tfrac{1}{2} (a - a')$ and $\tfrac{1}{2} (A - A')$ are of the same sign.
+
+\paragraph{59.} \textit{If on a sphere any point be taken within a circle which
+is not its pole, of all the arcs which can be drawn from that point
+to the circumference, the greatest is that in which the pole is, and the
+other part of that produced is the least; and of any others, that which
+is nearer to the greatest is always greater than one more remote; and
+from the same point to the circumference there can be drawn only
+two arcs which are equal to each other, and these make equal angles
+with the shortest arc on opposite sides of it.}
+
+This follows readily from the preceding three Articles.
+
+\paragraph{60.} We will give another proof of the fundamental formul\ae\
+in Art.~39, which is very simple, requiring only a knowledge of
+the elements of Co-ordinate Geometry.
+
+Suppose $ABC$ any spherical triangle, $O$ the centre of the
+sphere, take $O$ as the origin of co-ordinates, and let the axis of $z$
+pass through $C$. Let $x_1$, $y_1$, $z_1$ be the co-ordinates of $A$, and $x_2$,
+$y_2$, $z_2$ those of $B$; let $r$ be the radius of the sphere. Then the
+square on the straight line $AB$ is equal to
+\[
+  (x_1 - x_2)^2 +
+  (y_1 - y_2)^2 +
+  (z_1 - z_2)^2,
+\]
+and also to \hfill $r^2 + r^2 - 2r^2 \cos AOB;$
+\hfill\phantom{and also to}\\[1ex]
+%-----File: 040.png------------------------------------------------
+and \ $x_1^2 + y_1^2 + z_1^2 = r^2$, $x_2^2 + y_2^2 + z_2^2 = r^2$, thus
+\[
+x_1x_2 + y_1y_2 + z_1z_2 = r^2 \cos AOB.
+\]
+
+Now make the usual substitutions in passing from rectangular
+to polar co-ordinates, namely,
+\begin{align*}
+z_1 &= r \cos \theta_1, &x_1 &= r \sin \theta_1 \cos \phi_1, &y_1 &= r \sin \theta_1 \sin \phi_1, \\
+z_2 &= r \cos \theta_2, &x_2 &= r \sin \theta_2 \cos \phi_2, &y_2 &= r \sin \theta_2 \sin \phi_2;
+\end{align*}
+thus we obtain
+\[
+\cos \theta_2 \cos \theta_1 + \sin \theta_2 \sin \theta_1 \cos (\phi_1 - \phi_2) = \cos AOB,
+\]
+that is, in the ordinary notation of Spherical Trigonometry,
+\[
+\cos a \cos b + \sin a \sin b \cos C = \cos c.
+\]
+
+This method has the advantage of giving a \textit{perfectly general
+proof}, as all the equations used are universally true.
+
+\section*{\centering\normalsize EXAMPLES.}
+
+\indent 1. If $A = a$, shew that $B$ and $b$ are equal or supplemental, as
+also $C$ and $c$.
+\medskip
+
+2. If one angle of a triangle be equal to the sum of the other
+two, the greatest side is double of the distance of its middle point
+from the opposite angle.
+\medskip
+
+3. When does the polar triangle coincide with the primitive
+triangle?
+\medskip
+
+4. If $D$ be the middle point of $AB$, shew that
+\[
+\cos AC + \cos BC = 2 \cos \tfrac{1}{2} AB \cos CD.
+\]
+
+5. If two angles of a spherical triangle be respectively equal
+to the sides opposite to them, shew that the remaining side is the
+supplement of the remaining angle; or else that the triangle has
+two quadrants and two right angles, and then the remaining side
+is equal to the remaining angle.
+%-----File: 041.png------------------------------------------------
+\medskip
+
+6. In an equilateral triangle, shew that $2 \cos \dfrac a2 \sin \dfrac A2 = 1$.
+\medskip
+
+7. In an equilateral triangle, shew that $\tan^2 \dfrac a2 = 1 - 2 \cos A$;\\
+hence deduce the limits between which the sides and the angles of
+an equilateral triangle are restricted.
+\medskip
+
+8. In an equilateral triangle, shew that $\sec A = 1 + \sec a$.
+\medskip
+
+9. If the three sides of a spherical triangle be halved and
+a new triangle formed, the angle $\theta$ between the new sides $\dfrac b2$ and $\dfrac c2$
+is given by $\cos \theta = \cos A + \tfrac 12 \tan \dfrac b2 \tan \dfrac c2 \sin^2 \theta$.
+\medskip
+
+10. $AB$, $CD$ are quadrants on the surface of a sphere intersecting
+at $E$, the extremities being joined by great circles: shew
+that
+\[
+\cos AEC = \cos AC \cos BD - \cos BC \cos AD.
+\]
+
+11. If $b + c = \pi$, shew that $\sin 2B + \sin 2C = 0$.
+\medskip
+
+12. If $DE$ be an arc of a great circle bisecting the sides $AB$,
+$AC$ of a spherical triangle at $D$ and $E$, $P$ a pole of $DE$, and $PB$,
+$PD$, $PE$, $PC$ be joined by arcs of great circles, shew that the angle
+$BPC =$ twice the angle $DPE$.
+\medskip
+
+13. In a spherical triangle shew that
+\[
+\sin b \sin c + \cos b \cos c \cos A =
+\sin B \sin C - \cos B \cos C \cos a.
+\]
+
+14. If $D$ be any point in the side $BC$ of a triangle, shew that
+\[
+\cos AD \sin BC = \cos AB \sin DC + \cos AC \sin BD.
+\]
+
+15. In a spherical triangle shew that $\theta$, $\phi$, $\psi$ be the lengths
+of arcs of great circles drawn from $A$, $B$, $C$ perpendicular to the
+opposite sides,
+\begin{gather*}
+  \sin a \sin \theta =
+  \sin b \sin \phi =
+  \sin c \sin \psi \\
+= \surd(1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c).
+\end{gather*}
+%-----File: 042.png------------------------------------------------
+
+16. In a spherical triangle, if $\theta$, $\phi$, $\psi$ be the arcs bisecting the
+angles $A$, $B$, $C$ respectively and terminated by the opposite sides,
+shew that
+\[
+  \cot\theta\cos\dfrac A2 +
+  \cot\phi  \cos\dfrac B2 +
+  \cot\psi  \cos\dfrac C2 =
+  \cot a + \cot b + \cot c.
+\]
+
+17. Two ports are in the same parallel of latitude, their common
+latitude being $l$ and their difference of longitude $2\lambda$: shew
+that the saving of distance in sailing from one to the other on the
+great circle, instead of sailing due East or West, is
+\[
+  2r\{\lambda \cos l - \sin^{-1}(\sin \lambda \cos l)\},
+\]
+$\lambda$ being expressed in circular measure, and $r$ being the radius of
+the Earth.
+\medskip
+
+18. If a ship be proceeding uniformly along a great circle and
+the observed latitudes be $l_1$, $l_2$, $l_3$, at equal intervals of time, in
+each of which the distance traversed is $s$, shew that
+\[
+  s = r\cos^{-1}
+      \frac{\sin\tfrac 12(l_1 + l_3)\cos\tfrac 12(l_1 - l_3)}
+           {\sin l_2},
+\]
+$r$ denoting the Earth's radius: and shew that the change of longitude
+may also be found in terms of the three latitudes.
+
+\chapter[Solution of Right-angled Triangles.]{SOLUTION OF RIGHT-ANGLED TRIANGLES.}
+
+\paragraph{61.} In every spherical triangle there are six elements, namely,
+the three sides and the three angles, besides the radius of the
+sphere, which is supposed constant. The solution of spherical triangles
+is the process by which, when the values of a sufficient
+number of the six elements are given, we calculate the values of
+the remaining elements. It will appear, as we proceed, that when
+the values of three of the elements are given, those of the remaining
+three can generally be found. We begin with the right-angled
+triangle: here two elements, in addition to the right angle, will be
+supposed known.
+%-----File: 043.png------------------------------------------------
+
+\paragraph{62.} The formul\ae\ requisite for the solution of right-angled
+triangles may be obtained from the preceding Chapter by supposing
+one of the angles a right angle, as $C$ for example. They
+may also be obtained very easily in an independent manner, as
+we will now shew.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/043fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle having a right angle at $C$;
+let $O$ be the centre of the sphere. From any point $P$ in $OA$ draw
+$PM$ perpendicular to $OC$, and from $M$ draw $MN$ perpendicular to
+$OB$, and join $PN$. Then $PM$ is perpendicular to $MN$, because the
+plane $AOC$ is perpendicular to the plane $BOC$; hence
+\[
+PN^2=PM^2+MN^2=OP^2-OM^2+OM^2-ON^2=OP^2-ON^2;
+\]
+therefore $PNO$ is a right angle. And
+\begin{flalign*}
+&&&\left.\begin{aligned}
+&\rule{2em}{0pt}
+\llap{$ \dfrac{ON}{OP}$}
+\rlap{ $= \dfrac{ON}{OM} \centerdot\dfrac{OM}{OP}$, }
+\rule[1em]{7em}{0pt}
+\rule{6em}{0pt}
+\llap{ that is, $\cos c$}
+\rlap{ $=\cos a\cos b$,}
+\rule[1em]{6em}{0pt}
+\end{aligned}
+\right.
+&(1)&
+\\[1ex]
+&&&\left.\begin{aligned}
+&\rule{2em}{0pt}
+\llap{$ \dfrac{PM}{OP}$}
+\rlap{ $= \dfrac{PM}{PN} \centerdot\dfrac{PN}{OP}$, }
+\rule[1em]{7em}{0pt}
+\rule{6em}{0pt}
+\llap{ that is, $\sin b$}
+\rlap{ $=\sin B\sin c$}
+\rule[1em]{6em}{0pt}
+\\
+&\rule{15em}{0pt}
+\llap{ Similarly,\qquad $\sin a$}
+\rlap{ $=\sin A \sin c$}
+\rule[1em]{5em}{0pt}
+\end{aligned}
+\right\},
+&(2)&
+\\[1ex]
+&&&\left.\begin{aligned}
+&\rule{2em}{0pt}
+\llap{$ \dfrac{MN}{ON}$}
+\rlap{ $= \dfrac{MN}{PN} \centerdot\dfrac{PN}{ON}$, }
+\rule[1em]{7em}{0pt}
+\rule{6em}{0pt}
+\llap{ that is, $\tan a$}
+\rlap{ $=\cos B\tan c$}
+\rule[1em]{6em}{0pt}
+\\
+&\rule{15em}{0pt}
+\llap{ Similarly,\qquad $\tan b$}
+\rlap{ $= \cos A \tan c$}
+\rule[1em]{5em}{0pt}
+\end{aligned}
+\right\},
+&(3)&
+\\[1ex]
+&&&\left.\begin{aligned}
+&\rule{2em}{0pt}
+\llap{$ \dfrac{PM}{OM}$}
+\rlap{ $= \dfrac{PM}{MN} \centerdot\dfrac{MN}{OM}$, }
+\rule[1em]{7em}{0pt}
+\rule{6em}{0pt}
+\llap{ that is, $\tan b$}
+\rlap{ $=\tan B\sin a$}
+\rule[1em]{6em}{0pt}
+\\
+&\rule{15em}{0pt}
+\llap{ Similarly,\qquad $\tan a$}
+\rlap{ $= \tan A\sin b$}
+\rule[1em]{5em}{0pt}
+\end{aligned}
+\right\}.
+&(4)&
+\end{flalign*}
+%-----File: 044.png------------------------------------------------
+
+Multiply together the two formul\ae\ (4); thus,
+\[
+  \tan A \tan B =
+\dfrac{\tan a \tan b}{\sin a \sin b} =
+\frac{1}{\cos a \cos b} =
+\frac{1}{\cos c}\text{ by (1);}
+\]
+therefore \hfill$ \cos c = \cot A \cot B$.\hfill(5)
+\\
+Multiply crosswise the second formula in (2) and the first
+in (3); thus \linebreak[4]$\sin a \cos B \tan c = \tan a \sin A \sin c$;
+\begin{flalign*}
+&\phantom{therefore}&&&\phantom{therefore}&\\[-\baselineskip]
+&\rlap{therefore}&\multispan{2}{\hfil$\displaystyle
+\cos B = \dfrac{\sin A \cos c}{\cos a} =
+\sin A \cos b$ by (1).\hfil}\\[1ex]
+&\begin{aligned}
+&\text{Thus}\\
+&\text{Similarly}
+\end{aligned}&\multispan2{\hfil$\displaystyle
+\left.\begin{aligned}
+\cos B &= \sin A \cos b \\
+\cos A &= \sin B \cos a
+\end{aligned}\right\}$.\hfil}&\text{(6)}
+\end{flalign*}
+
+These six formul\ae\ comprise ten equations; and thus we can
+solve every case of right-angled triangles. For every one of these
+ten equations is a distinct combination involving three out of the
+five quantities $a$, $b$, $c$, $A$, $B$; and out of five quantities only ten
+combinations of three can be formed. Thus any two of the five
+quantities being given and a third required, some one of the preceding
+ten equations will serve to determine that third quantity.
+
+\paragraph{63.} As we have stated, the above six formul\ae\ may be obtained
+from those given in the preceding Chapter by supposing $C$ a
+right angle. Thus (1) follows from Art.~39, (2) from Art.~41,
+(3) from the fourth and fifth equations of Art.~44, (4) from the
+first and second equations of Art.~44, (5) from the third equation
+of Art.~47, (6) from the first and second equations of Art.~47.
+
+Since the six formul\ae\ may be obtained from those given in
+the preceding Chapter which have been proved to be universally
+true, we do not stop to shew that the demonstration of Art.~62
+may be applied to every case which can occur; the student may
+for exercise investigate the modifications which will be necessary
+when we suppose one or more of the quantities $a$, $b$, $c$, $A$, $B$ equal
+to a right angle or greater than a right angle.
+%-----File: 045.png------------------------------------------------
+
+\paragraph{64.} Certain properties of right-angled triangles are deducible
+from the formul\ae\ of Art.~62.
+
+From (1) it follows that $\cos c$ has the same sign as the product
+$\cos a\cos b$; hence either all the cosines are positive, or else only
+one is positive. Therefore \textit{in a right-angled triangle either all the
+three sides are less than quadrants, or else one side is less than a
+quadrant and the other two sides are greater than quadrants.}
+
+From (4) it follows that $\tan a$ has the same sign as $\tan A$.
+Therefore $A$ and $a$ are either both greater than $\dfrac{\pi}{2}$, or both less
+than $\dfrac{\pi}{2}$; this is expressed by saying that $A$ and $a$ are of the \textit{same
+affection}. Similarly $B$ and $b$ are of the same affection.
+
+\paragraph{65.} The formul\ae\ of Art.~62 are comprised in the following
+enunciations, which the student will find it useful to remember;
+the results are distinguished by the same numbers as have been
+already applied to them in Art.~62; the side opposite the right
+angle is called the \textit{hypotenuse:}
+\begin{center}
+\begin{tabular}{r@{\ }l@{}l}
+  Cos hyp &= product of cosines of sides \dotfill&(1),\\[1ex]
+  Cos hyp &= product of cotangents of angles \dotfill&(5),\\[1ex]
+  Sine side &= sine of opposite angle $\times$ sine hyp \dotfill&(2),\\[1ex]
+  Tan side &= tan hyp $\times$ cos included angle \dotfill&(3),\\[1ex]
+  Tan side &= tan opposite angle $\times$ sine of other side \ldots\dotfill&(4),\\[1ex]
+  Cos angle &= cos opposite side $\times$ sine of other angle \dotfill&(6).
+\end{tabular}
+\end{center}
+
+\paragraph{66.} \textit{Napier's Rules}. The formul\ae\ of Art.~62 are comprised
+in two rules, which are called, from their inventor, \textit{Napier's Rules
+of Circular Parts}. Napier was also the inventor of Logarithms,
+and the Rules of Circular Parts were first published by him in a
+work entitled \textit{Mirifici Logarithmorum Canonis Descriptio}\dots\dots
+Edinburgh, 1614. These rules we will now explain.
+%-----File: 046.png------------------------------------------------
+
+The right angle is left out of consideration; the two sides
+which include the right angle, the complement of the hypotenuse,
+and the complements of the other angles are called the \textit{circular
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/046fc}
+\end{figure}
+parts} of the triangle. Thus there are \textit{five} circular parts, namely,
+$a$, $b$, $\dfrac{\pi}{2}-A$, $\dfrac{\pi}{2}-c$, $\dfrac{\pi}{2}-B$; and these are supposed to be ranged
+round a circle in the order in which they naturally occur with
+respect to the triangle.
+
+Any one of the five parts may be selected and called the
+\textit{middle part}, then the two parts next to it are called \textit{adjacent
+parts}, and the remaining two parts are called \textit{opposite parts}. For
+example, if $\dfrac{\pi}{2}-B$ is selected as the middle part, then the adjacent
+parts are $a$ and $\dfrac{\pi}{2}-c$, and the opposite parts are $b$ and $\dfrac{\pi}{2}-A$.
+
+Then Napier's Rules are the following:\\
+sine of the middle part $=$ product of tangents of adjacent parts, \\
+sine of the middle part $=$ product of cosines of opposite parts.
+
+\paragraph{67.} Napier's Rules may be demonstrated by shewing that
+they agree with the results already established. The following
+table shews the required agreement: in the first column are given
+the \textit{middle parts}, in the second column the results of Napier's
+Rules, and in the third column the same results expressed as in
+Art.~62, with the number for reference used in that Article.
+%-----File: 047.png------------------------------------------------
+
+\[
+\begin{array}{@{}r@{\ } r@{\;}l@{ } r@{\;}l@{.}r}
+   \dfrac{\pi}{2}-c
+&  \sin\left(\dfrac{\pi}{2}-c\right)
+&= \tan\left(\dfrac{\pi}{2}-A\right)
+   \tan\left(\dfrac{\pi}{2}-B\right)
+& \cos c &= \cot A \cot B \dotfill & (5),
+\\[3ex]
+&  \sin\left(\dfrac{\pi}{2}-c\right)
+&= \cos a \cos b
+&  \cos c &= \cos a \cos b \dotfill & (1),
+\\[3ex]
+   \dfrac{\pi}{2}-B
+&         \sin\left(\dfrac{\pi}{2}-B\right)
+&= \tan a \tan\left(\dfrac{\pi}{2}-c\right)
+&  \cos B &= \tan a \cot c \dotfill & (3),
+\\[3ex]
+&         \sin\left(\dfrac{\pi}{2}-B\right)
+&= \cos b \cos\left(\dfrac{\pi}{2}-A\right)
+& \cos B &= \cos b \sin A \dotfill & (6),
+\\[3ex]
+a& \sin a
+&= \tan b \tan \left(\dfrac{\pi}{2}-B\right)
+&  \sin a &= \tan b \cot B \ldots\dotfill & (4),
+\\[3ex]
+&  \sin a
+&= \cos\left(\dfrac{\pi}{2}-A\right)
+   \cos\left(\dfrac{\pi}{2}-c\right)
+&  \sin a &= \sin A \sin c \dotfill & (2),
+\\[3ex]
+b& \sin b
+&= \tan\left(\dfrac{\pi}{2}-A\right) \tan a
+&  \sin b &= \cot A\tan a \dotfill & (4),
+\\[3ex]
+&  \sin b
+&= \cos\left(\dfrac{\pi}{2}-B\right)
+   \cos\left(\dfrac{\pi}{2}-c\right)
+&  \sin b &=\sin B \sin c \dotfill & (2),
+\\[3ex]
+   \dfrac{\pi}{2}-A
+&        \sin\left(\dfrac{\pi}{2}-A\right)
+&= \tan b \tan\left(\dfrac{\pi}{2}-c\right)
+& \cos A &= \tan b\cot c \dotfill & (3),
+\\[3ex]
+&        \sin\left(\dfrac{\pi}{2}-A\right)
+&= \cos a \cos\left(\dfrac{\pi}{2}-B\right)
+& \cos A &= \cos a \sin B \dotfill & (6).
+\end{array}
+\]
+
+The last four cases need not have been given, since it is obvious
+that they are only repetitions of what had previously been given;
+the seventh and eighth are repetitions of the fifth and sixth, and
+the ninth and tenth are repetitions of the third and fourth.
+
+\paragraph{68.} It has been sometimes stated that the method of the
+preceding Article is the only one by which Napier's Rules can be
+demonstrated; this statement, however, is inaccurate, since besides
+this method Napier himself indicated another method of proof in
+his \textit{Mirifici Logarithmorum Canonis Descriptio}, pp.\ 32, 35. This
+we will now briefly explain.
+%-----File: 048.png------------------------------------------------
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/048fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle right-angled at $C$; with $B$
+as pole describe a great circle $DEFG$, and with $A$ as pole describe
+a great circle $HFKL$, and produce the sides of the original triangle
+$ABC$ to meet these great circles. Then since $B$ is a pole of $DEFG$
+the angles at $D$ and $G$ are right angles, and since $A$ is a pole of
+$HFKL$ the angles at $H$ and $L$ are right angles. Hence the five
+triangles $BAC$, $AED$, $EFH$, $FKG$, $KBL$ are all \textit{right-angled}; and
+moreover it will be found on examination that, although the elements
+of these triangles are different, yet \textit{their circular parts are
+the same}. We will consider, for example, the triangle $AED$; the
+angle $EAD$ is equal to the angle $BAC$, the side $AD$ is the complement
+of $AB$; as the angles at $C$ and $G$ are right angles $E$ is a
+pole of $GC$ (Art.~13), therefore $EA$ is the complement of $AC$; as
+$B$ is a pole of $DE$ the angle $BED$ is a right angle, therefore the
+angle $AED$ is the complement of the angle $BEC$, that is, the
+angle $AED$ is the complement of the side $BC$ (Art.~12); and similarly
+the side $DE$ is equal to the angle $DBE$, and is therefore the
+complement of the angle $ABC$. Hence, if we denote the elements
+of the triangle $ABC$ as usual by $a$, $b$, $c$, $A$, $B$, we have in the
+triangle $AED$ the hypotenuse equal to $\dfrac{\pi}{2} - b$, the angles equal to
+$A$ and $\dfrac{\pi}{2} - a$, and the sides respectively opposite these angles equal
+to $\dfrac{\pi}{2} - B$ and $\dfrac{\pi}{2} - c$. The \textit{circular parts} of $AED$ are therefore the
+%-----File: 049.png------------------------------------------------
+same as those of $ABC$. Similarly the remaining three of the five
+right-angled triangles may be shewn to have the same circular
+parts as the triangle $ABC$ has.
+
+Now take \textit{two} of the theorems in Art.~65, for example (1) and
+(3); then the truth of the \textit{ten} cases comprised in Napier's Rules
+will be found to follow from applying the two theorems in succession
+to the five triangles formed in the preceding figure. Thus
+this method of considering Napier's Rules regards each Rule, not
+as the statement of dissimilar properties of one triangle, but as the
+statement of similar properties of five allied triangles.
+
+\paragraph{69.} In Napier's work a figure is given of which that in the
+preceding Article is a copy, except that different letters are used;
+Napier briefly intimates that the truth of the Rules can be easily
+seen by means of this figure, as well as by the method of induction
+from consideration of all the cases which can occur. The late
+T.~S. Davies, in his edition of Dr Hutton's \textit{Course of Mathematics},
+drew attention to Napier's own views and expanded the demonstration
+by a systematic examination of the figure of the preceding
+Article.
+
+It is however easy to evade the necessity of examining the
+whole figure; all that is wanted is to observe the connexion
+between the triangle $AED$ and the triangle $BAC$. For let $a_1$, $a_2$,
+$a_3$, $a_4$, $a_5$ represent the elements of the triangle $BAC$ taken in
+order, beginning with the hypotenuse and omitting the right
+angle; then the elements of the triangle $AED$ taken in order,
+beginning with the hypotenuse and omitting the right angle, are
+$\dfrac{\pi}{2} - a_3$, $\dfrac{\pi}{2} - a_4$, $\dfrac{\pi}{2} - a_5$,
+$\dfrac{\pi}{2} - a_1$, and $a_2$. If, therefore, to characterise
+the former we introduce a new set of quantities $p_1$, $p_2$, $p_3$, $p_4$, $p_5$,
+such that $a_1 + p_1 = a_2 + p_2 = a_5 + p_5 = \dfrac{\pi}{2}$, and that $p_3 = a_3$ and $p_4 = a_4$,
+then the original triangle being characterised by $p_1$, $p_2$, $p_3$, $p_4$, $p_5$,
+the second triangle will be similarly characterised by $p_3$, $p_4$, $p_5$,
+$p_1$, $p_2$. As the second triangle can give rise to a third in like
+manner, and so on, we see that every right-angled triangle is one
+%-----File: 050.png------------------------------------------------
+of a system of five such triangles which are all characterised by
+the quantities $p_1$, $p_2$, $p_3$, $p_4$, $p_5$, always taken in order, each
+quantity in its turn standing first.
+
+The late R.~L.\ Ellis pointed out this connexion between the
+five triangles, and thus gave the true significance of Napier's
+Rules. The memoir containing Mr Ellis's investigations, which
+was unpublished when the first edition of the present work appeared,
+will be found in pages 328\ldots335 of \textit{The Mathematical and
+other writings of Robert Leslie Ellis}\dots Cambridge, 1863.
+
+Napier's own method of considering his Rules was neglected
+by writers on the subject until the late T.~S.\ Davies drew attention
+to it. Hence, as we have already remarked in Art.\ 68, an
+erroneous statement was made respecting the Rules. For instance,
+Woodhouse says, in his \textit{Trigonometry}: ``There is no separate
+and independent proof of these rules;\ldots.'' Airy says, in the
+treatise on Trigonometry in the \textit{Encyclop\ae dia Metropolitana}:
+``These rules are proved to be true only by showing that they comprehend
+all the equations which we have just found.''
+
+\paragraph{70.} Opinions have differed with respect to the \textit{utility} of
+Napier's Rules in practice. Thus Woodhouse says, ``In the whole
+compass of mathematical science there cannot be found, perhaps,
+rules which more completely attain that which is the proper
+object of rules, namely, facility and brevity of computation.''
+(\textit{Trigonometry}, chap.\ \textsc{x}.) On the other hand may be set the following
+sentence from Airy's Trigonometry (\textit{Encyclop\ae dia Metropolitana}):
+``In the opinion of Delambre (and no one was better
+qualified by experience to give an opinion) these theorems are best
+recollected by the practical calculator in their unconnected form.''
+See Delambre's \textit{Astronomie}, vol.~\textsc{i}.\ p.~205. Professor De Morgan
+strongly objects to Napier's Rules, and says (\textit{Spherical Trigonometry},
+Art.~17): ``There are certain mnemonical formul\ae\ called
+\textit{Napier's Rules of Circular Parts}, which are generally explained.
+We do not give them, because we are convinced that they only
+create confusion instead of assisting the memory.''
+%-----File: 051.png------------------------------------------------
+
+\paragraph{71.} We shall now proceed to apply the formul\ae\ of Art.\ 62
+to the solution of right-angled triangles. We shall assume that
+the given quantities are subject to the limitations which are stated
+in Arts.\ 22 and 23, that is, a given side must be less than the
+semicircumference of a great circle, and a given angle less than
+two right angles. There will be six cases to consider.
+
+\paragraph{72.} \textit{Having given the hypotenuse $\mathrm{c}$ and an angle $\mathrm{A}$.}
+
+Here we have from (3), (5) and (2) of Art.\ 62,
+\[
+\tan b = \tan c \cos A,\quad \cot B = \cos c \tan A,\quad \sin a = \sin c \sin A.
+\]
+
+Thus $b$ and $B$ are determined immediately without ambiguity;
+and as $a$ must be of the same affection as $A$ (Art.\ 64), $a$ also is
+determined without ambiguity.
+
+It is obvious from the formul\ae\ of solution, that in this case
+the triangle is always possible.
+
+If $c$ and $A$ are both right angles, $a$ is a right angle, and $b$ and
+$B$ are indeterminate.
+
+\paragraph{73.} \textit{Having given a side $\mathrm{b}$ and the adjacent angle $\mathrm{A}$.}
+
+Here we have from (3), (4) and (6) of Art.\ 62,
+\[
+\tan c=\dfrac{\tan b}{\cos A},\quad \tan a = \tan A \sin b,\quad \cos B = \cos b \sin A.
+\]
+
+Here $c$, $a$, $B$ are determined without ambiguity, and the triangle
+is always possible.
+
+\paragraph{74.} \textit{Having given the two sides $\mathrm{a}$ and $\mathrm{b}$.}
+
+Here we have from (1) and (4) of Art.\ 62,
+\[
+\cos c = \cos a \cos b,\quad \cot A = \cot a \sin b,\quad \cot B = \cot b \sin a.
+\]
+
+Here $c$, $A$, $B$ are determined without ambiguity, and the triangle
+is always possible.
+
+\paragraph{75.} \textit{Having given the hypotenuse $\mathrm{c}$ and a side $\mathrm{a}$.}
+
+Here we have from (1), (3) and (2) of Art.\ 62,
+\[
+\cos b=\dfrac{\cos c}{\cos a},\quad \cos B=\dfrac{\tan a}{\tan c},\quad \sin A=\dfrac{\sin a}{\sin c}.
+\]
+%-----File: 052.png------------------------------------------------
+
+Here $b$, $B$, $A$ are determined without ambiguity, since $A$ must
+be of the same affection as $a$. It will be seen from these formul\ae\
+that there are limitations of the data in order to insure a possible
+triangle; in fact, $c$ must lie between $a$ and $\pi - a$ in order that the
+values found for $\cos b$, $\cos B$, and $\sin A$ may be numerically not
+greater than unity.
+
+If $c$ and $a$ are right angles, $A$ is a right angle, and $b$ and $B$ are
+indeterminate.
+
+\paragraph{76.} \textit{Having given the two angles $\mathrm{A}$ and $\mathrm{B}$.}
+
+Here we have from (5) and (6) of Art.\ 62,
+\[
+\cos c = \cot A \cot B, \qquad
+\cos a = \frac{\cos A}{\sin B}, \qquad
+\cos b = \frac{\cos B}{\sin A}.
+\]
+
+Here $c$, $a$, $b$ are determined without ambiguity. There are
+limitations of the data in order to insure a possible triangle. First
+suppose $A$ less than $\dfrac{\pi}{2}$, then $B$ must lie between $\dfrac{\pi}{2} - A$ and $\dfrac{\pi}{2} + A$;
+next suppose $A$ greater than $\dfrac{\pi}{2}$, then $B$ must lie between
+$\dfrac{\pi}{2} - (\pi - A)$ and
+$\dfrac{\pi}{2} + (\pi - A)$, that is, between
+$A - \dfrac{\pi}{2}$ and $\dfrac{3\pi}{2} - A$.
+
+\paragraph{77.} \textit{Having given a side $\mathrm a$ and the opposite angle $\mathrm A$.}
+
+Here we have from (2), (4) and (6) of Art.\ 62,
+\[
+\sin c = \dfrac{\sin a}{\sin A}, \qquad
+\sin b = \tan a\, \cot A, \qquad
+\sin B = \dfrac{\cos A}{\cos a}.
+\]
+
+Here there is an ambiguity, as the parts are determined from
+their sines. If $\sin a$ be less than $\sin A$, there are two values
+admissible for $c$; corresponding to each of these there will be
+\textit{in general} only one admissible value of $b$, since we must have
+$\cos c = \cos a\, \cos b$, and only one admissible value of $B$, since we
+must have $\cos c = \cot A \cot B$. Thus if one triangle exists with the
+given parts, there will be \textit{in general} two, and only two, triangles
+with the given parts. We say \textit{in general} in the preceding sentences,
+because if $a = A$ there will be only \textit{one} triangle, unless $a$
+%-----File: 053.png------------------------------------------------
+and $A$ are each right angles, and then $b$ and $B$ become indeterminate.
+
+It is easy to see from a figure that the ambiguity must occur
+in general.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/053fc}
+\end{figure}
+
+For, suppose $BAC$ to be a triangle which satisfies the given
+conditions; produce $AB$ and $AC$ to meet again at $A'$; then the
+triangle $A'BC$ also satisfies the given conditions, for it has a right
+angle at $C$, $BC$ the given side, and $A' = A$ the given angle.
+
+If $a = A$, then the formul\ae\ of solution shew that $c$, $b$, and $B$
+are right angles; in this case $A$ is the pole of $BC$, and the triangle
+$A'BC$ is symmetrically equal to the triangle $ABC$ (Art.\ 57).
+
+If $a$ and $A$ are both right angles, $B$ is the pole of $AC$; $B$ and $b$
+are then equal, but may have any value whatever.
+
+There are limitations of the data in order to insure a possible
+triangle. $A$ and $a$ must have the same affection by Art.\ 64; hence
+the formul\ae\ of solution shew that $a$ must be less than $A$ if both
+are acute, and greater than $A$ if both are obtuse.
+
+\section*{\centering\normalsize EXAMPLES.}
+
+If $ABC$ be a triangle in which the angle $C$ is a right angle,
+prove the following relations contained in Examples 1 to 5.
+\medskip
+
+1. $\operatorname{Sin}^2\dfrac{c}{2} =
+                      \sin^2 \dfrac{a}{2}\, \cos^2 \dfrac{b}{2} +
+                      \cos^2 \dfrac{a}{2}\, \sin^2 \dfrac{b}{2}$.
+\medskip
+
+2. $\operatorname{Tan}\tfrac{1}{2}(c + a)\,
+                      \tan \tfrac{1}{2}(c - a) =
+                      \tan^2 \dfrac{b}{2}$.
+\medskip
+
+3. $\operatorname{Sin}(c - b) = \tan^2 \dfrac{A}{2}\,\sin(c + b)$.
+%-----File: 054.png------------------------------------------------
+\medskip
+
+4. $\operatorname{Sin} a\, \tan \tfrac{1}{2}A
+                     - \sin b\, \tan \tfrac{1}{2}B = \sin (a - b)$.
+
+\begin{flalign*}
+\rlap{\indent 5.} && \operatorname{Sin} (c - a)
+&= \sin b\, \cos a\, \tan \tfrac{1}{2}B, &&\\
+&& \operatorname{Sin} (c - a)
+&= \tan b\, \cos c\, \tan \tfrac{1}{2}B. &&
+\end{flalign*}
+
+6. If $ABC$ be a spherical triangle, right-angled at $C$, and
+$\cos A = \cos^2 a$, shew that if $A$ be not a right angle $b + c = \tfrac{1}{2}\pi$ or
+$\dfrac{3}{2}\pi$, according as $b$ and $c$ are both less or both greater than $\dfrac{\pi}{2}$.
+\medskip
+
+7. If $\alpha$, $\beta$ be the arcs drawn from the right angle respectively
+perpendicular to and bisecting the hypotenuse $c$, shew that
+\[
+\sin^2 \dfrac{c}{2}\,(1 + \sin^2\alpha) = \sin^2\beta.
+\]
+
+8. In a triangle, if $C$ be a right angle and $D$ the middle point
+of $AB$, shew that
+\[
+4\cos^2\dfrac{c}{2}\, \sin^2 CD = \sin^2 a + \sin^2 b.
+\]
+
+9. In a right-angled triangle, if $\delta$ be the length of the arc
+drawn from $C$ perpendicular to the hypotenuse $AB$, shew that
+\[
+\cot\delta = \surd{(\cot^2a + \cot^2b)}.
+\]
+
+10. $OAA_1$ is a spherical triangle right-angled at $A_1$ and acute-angled
+at $A$; the arc $A_1A_2$ of a great circle is drawn perpendicular
+to $OA$, then $A_2A_3$ is drawn perpendicular to $OA_1$, and so on: shew
+that $A_nA_{n+1}$ vanishes when $n$ becomes infinite; and find the value
+of $\cos AA_1 \cos A_1A_2 \cos A_2A_3\ldots\ldots$ to infinity.
+\medskip
+
+11. $ABC$ is a right-angled spherical triangle, $A$ not being the
+right angle: shew that if $A = a$, then $c$ and $b$ are quadrants.
+\medskip
+
+12. If $\delta$ be the length of the arc drawn from $C$ perpendicular
+to $AB$ in \textit{any} triangle, shew that
+\[
+\cos\delta = \operatorname{cosec} c\, (\cos^2 a + \cos^2 b - 2 \cos a\, \cos b\, \cos c)^{\tfrac{1}{2}}.
+\]
+
+13. $ABC$ is a great circle of a sphere; $AA'$, $BB'$, $CC'$, are arcs
+of great circles drawn at right angles to $ABC$ and reckoned positive
+%-----File: 055.png------------------------------------------------
+when they lie on the same side of it: shew that the condition
+of $A'$, $B'$, $C'$ lying in a great circle is
+\[
+\tan AA'\, \sin BC + \tan BB'\, \sin CA + \tan CC'\, \sin AB = 0.
+\]
+
+14. Perpendiculars are drawn from the angles $A$, $B$, $C$ of any
+triangle meeting the opposite sides at $D$, $E$, $F$ respectively: shew
+that
+\[
+\tan BD\, \tan CE\, \tan AF = \tan DC\, \tan EA\, \tan FB.
+\]
+
+15. $Ox$, $Oy$ are two great circles of a sphere at right angles to
+each other, $P$ is any point in $AB$ another great circle. $OC = p$ is
+the arc perpendicular to $AB$ from $O$, making the angle $COx = a$
+with $Ox$. $PM$, $PN$ are arcs perpendicular to $Ox$, $Oy$ respectively:
+shew that if $OM = x$ and $ON = y$,
+\[
+\cos a\, \tan x + \sin a\, \tan y = \tan p.
+\]
+
+16. The position of a point on a sphere, with reference to two
+great circles at right angles to each other as axes, is determined
+by the portions $\theta$, $\phi$ of these circles cut off by great circles through
+the point, and through two points on the axes, each $\dfrac{\pi}{2}$ from their
+point of intersection: shew that if the three points ($\theta$, $\phi$), ($\theta'$, $\phi'$),
+($\theta''$, $\phi''$) lie on the same great circle
+\begin{gather*}
+  \tan \phi\, (\tan \theta' - \tan \theta'')
++ \tan \phi'\, (\tan \theta'' - \tan \theta)\\
++ \tan \phi''\, (\tan \theta - \tan \theta') = 0.
+\end{gather*}
+
+17. If a point on a sphere be referred to two great circles at
+right angles to each other as axes, by means of the portions of
+these axes cut off by great circles drawn through the point and
+two points on the axes each $90^\circ$ from their intersection, shew that
+the equation to a great circle is
+\[
+\tan \theta\, \cot \alpha + \tan \phi\, \cot \beta = 1.
+\]
+
+18. In a spherical triangle, if $A = \dfrac{\pi}{5}$, $B = \dfrac{\pi}{3}$, and, $C = \dfrac{\pi}{2}$, shew that\, $a + b + c = \dfrac{\pi}{2}$.
+%-----File: 056.png------------------------------------------------
+
+\chapter[Solution of Oblique-Angled Triangles.]{SOLUTION OF OBLIQUE-ANGLED TRIANGLES.}
+
+\paragraph{78.} The solution of oblique-angled triangles may be made in
+some cases to depend immediately on the solution of right-angled
+triangles; we will indicate these cases before considering the subject
+generally.
+
+(1) Suppose a triangle to have one of its given sides equal to
+a \textit{quadrant}. In this case the polar triangle has its corresponding
+angle a right angle; the polar triangle can therefore be solved by
+the rules of the preceding Chapter, and thus the elements of the
+primitive triangle become known.
+
+(2) Suppose among the given elements of a triangle there are
+two \textit{equal sides} or two \textit{equal angles}. By drawing an arc from the
+vertex to the middle point of the base, the triangle is divided into
+two equal \textit{right-angled} triangles; by the solution of one of these
+right-angled triangles the required elements can be found.
+
+(3) Suppose among the given elements of a triangle there
+are two sides, one of which is the supplement of the other, or two
+angles, one of which is the supplement of the other. Suppose, for
+example, that $b + c = \pi$, or else that $B + C = \pi$; produce $BA$ and
+$BC$ to meet at $B'$ (see the first figure to Art.\ 38); then the triangle
+$B'AC$ has two equal sides given, or else two equal angles given;
+and by the preceding case the solution of it can be made to depend
+on the solution of a right-angled triangle.
+
+\paragraph{79.} We now proceed to the solution of oblique-angled triangles
+in general. There will be six cases to consider.
+
+\paragraph{80.} \textit{Having given the three sides.}
+
+Here we have $\cos A = \dfrac{\cos a - \cos b \cos c}{\sin b \sin c}$, and similar formul\ae\
+for $\cos B$ and $\cos C$. Or if we wish to use formul\ae\ suited to logarithms,
+%-----File: 057.png------------------------------------------------
+we may take the formula for the sine, cosine, or tangent of
+half an angle given in Art.~45. In selecting a formula, attention
+should be paid to the remarks in \textit{Plane Trigonometry}, Chap.~XII.
+towards the end.
+
+\paragraph{81.} \textit{Having given the three angles.}
+
+Here we have $\cos a = \dfrac{\cos A + \cos B \cos C}{\sin B \sin C}$, and similar formul\ae\
+for $\cos b$ and $\cos c$. Or if we wish to use formul\ae\ suited to logarithms,
+we may take the formula for the sine, cosine, or tangent of
+half a side given in Art.~49.
+
+There is no ambiguity in the two preceding cases; the triangles
+however may be impossible with the given elements.
+
+\paragraph{82.} \textit{Having given two sides and the included angle} ($a$, $C$, $b$).
+
+By Napier's analogies
+\begin{align*}
+&\tan \tfrac{1}{2}(A + B) = \dfrac{\cos\tfrac{1}{2}(a - b)}{\cos\tfrac{1}{2}(a + b)}\cot\tfrac{1}{2}C,\\[1ex]
+&\tan \tfrac{1}{2}(A - B) = \dfrac{\sin\tfrac{1}{2}(a - b)}{\sin\tfrac{1}{2}(a + b)}\cot\tfrac{1}{2}C;
+\end{align*}
+these determine $\tfrac{1}{2}(A + B)$ and $\tfrac{1}{2}(A - B)$, and thence $A$ and $B$.
+
+Then $c$ may be found from the formula $\sin c = \dfrac{\sin a\, \sin C}{\sin A}$; in
+this case, since $c$ is found from its sine, it may be uncertain which
+of two values is to be given to it; the point may be sometimes
+settled by observing that the greater side of a triangle is opposite
+to the greater angle. Or we may determine $c$ from equation (1) of
+Art.~54, which is free from ambiguity.
+
+Or we may determine $c$, without previously determining $A$ and
+$B$, from the formula $\cos c = \cos a\, \cos b + \sin a\, \sin b\, \cos C$; this is
+free from ambiguity. This formula may be adapted to logarithms
+thus:
+\[
+\cos c = \cos b\, (\cos a + \sin a \tan b \cos C);
+\]
+%-----File: 058.png------------------------------------------------
+assume $\tan \theta = \tan b\, \cos C$; then
+\[
+\cos c = \cos b\, (\cos a + \sin a\, \tan \theta) = \dfrac{\cos b\, \cos (a - \theta)}{\cos \theta};
+\]
+this is adapted to logarithms.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/058fc}
+\end{figure}
+
+Or we may treat this case conveniently by resolving the triangle
+into the sum or difference of two right-angled triangles.
+From $A$ draw the arc $AD$ perpendicular to $CB$ or $CB$ produced;
+then, by Art.\ 62, $\tan CD = \tan b \cos C$, and this determines $CD$,
+and then $DB$ is known. Again, by Art.\ 62,
+\[
+\cos c = \cos AD \cos DB = \cos DB \dfrac{\cos b}{\cos CD};
+\]
+this finds $c$. It is obvious that $CD$ is what was denoted by $\theta$ in
+the former part of the Article.
+
+By Art.\ 62,
+\begin{gather*}
+  \tan AD = \tan C \sin CD,\text{ and }\tan AD = \tan ABD \sin DB; \\
+  \text{thus } \tan ABD \sin DB = \tan C \sin \theta,
+\end{gather*}
+where $DB = a - \theta$ or $\theta - a$, according as $D$ is on $CB$ or $CB$ produced,
+and $ABD$ is either $B$ or the supplement of $B$; this formula
+enables us to find $B$ independently of $A$.
+
+Thus, in the present case, there is no real ambiguity, and the
+triangle is always possible.
+%-----File: 059.png------------------------------------------------
+
+\paragraph{83.} \textit{Having given two angles and the included side $\mathrm{(A, c, B)}$.}
+
+By Napier's analogies,
+\begin{align*}
+   \tan \tfrac{1}{2} (a + b)
+&= \dfrac{\cos \tfrac{1}{2} (A - B)}{\cos \tfrac{1}{2} (A + B)} \tan \tfrac{1}{2} c, \\[1ex]
+   \tan \tfrac{1}{2} (a - b)
+&= \dfrac{\sin \tfrac{1}{2} (A - B)}{\sin \tfrac{1}{2} (A + B)} \tan \tfrac{1}{2} c;
+\end{align*}
+these determine $\tfrac{1}{2} (a + b)$ and $\tfrac{1}{2} (a - b)$, and thence $a$ and $b$.
+
+Then $C$ may be found from the formula $\sin C = \dfrac{\sin A \sin c}{\sin a}$; in
+this case, since $C$ is found from its sine, it may be uncertain which
+of two values is to be given to it; the point may be sometimes
+settled by observing that the greater angle of a triangle is opposite
+to the greater side. Or we may determine $C$ from equation (3) of
+Art.\ 54, which is free from ambiguity.
+
+Or we may determine $C$ without previously determining $a$ and
+$b$ from the formula $\cos C = -\cos A \cos B + \sin A \sin B \cos c$. This
+formula may be adapted to logarithms, thus:
+\[
+\cos C = \cos B (-\cos A + \sin A\, \tan B\, \cos c);
+\]
+assume $\cot \phi = \tan B\, \cos c$; then
+\[
+\cos C = \cos B (-\cos A + \cot \phi \sin A) = \dfrac{\cos B \sin (A-\phi)}{\sin \phi}\,;
+\]
+this is adapted to logarithms.
+
+Or we may treat this case conveniently by resolving the triangle
+into the sum or difference of two right-angled triangles.
+From $A$ draw the arc $AD$ perpendicular to $CB$ (see the right-hand
+figure of Art.\ 82); then, by Art.\ 62, $\cos c = \cot B \cot DAB$,
+and this determines $DAB$, and then $CAD$ is known. Again,
+by Art.\ 62,
+\[
+\cos AD \sin CAD = \cos C,\text{ and }\cos AD \sin BAD = \cos B;
+\]
+therefore $\dfrac{\cos C}{\sin CAD}
+= \dfrac{\cos B}{\sin BAD}$; this finds $C$.
+\\[1ex]
+It is obvious that $DAB$ is what was denoted by $\phi$ in the former
+part of the Article.
+%-----File: 060.png------------------------------------------------
+
+By Art.\ 62,
+\[
+  \tan AD = \tan AC \cos CAD,\text{ and }
+  \tan AD = \tan AB \cos BAD;
+\]
+thus \hfill$\tan b \cos CAD = \tan c \cos \phi$,\hfill\phantom{thus}\\[1ex]
+where $CAD = A - \phi$; this formula enables us to find $b$ independently
+of $a$.
+
+Similarly we may proceed when the perpendicular $AD$ falls on
+$CB$ \textit{produced}; (see the left-hand figure of Art.~82).
+
+Thus, in the present case, there is no real ambiguity; moreover
+the triangle is always possible.
+
+\paragraph{84.} \textit{Having given two sides and the angle opposite one of them
+$(\mathrm a$, $\mathrm b$, $\mathrm A)$.}
+
+The angle $B$ may be found from the formula
+\[
+  \sin B = \frac{\sin b}{\sin a} \sin A;
+\]
+and then $C$ and $c$ may be found from Napier's analogies,
+\begin{align*}
+  \tan \tfrac{1}{2} C
+&= \dfrac{\cos \tfrac{1}{2} (a - b)}
+        {\cos \tfrac{1}{2} (a + b)} \cot \tfrac{1}{2} (A + B),
+\\[1ex]
+  \tan \tfrac{1}{2} c
+&= \dfrac{\cos \tfrac{1}{2} (A + B)}
+        {\cos \tfrac{1}{2} (A - B)} \tan \tfrac{1}{2} (a + b).
+\end{align*}
+In this case, since $B$ is found from its sine, there will sometimes
+be two solutions; and sometimes there will be no solution at all,
+namely, when the value found for $\sin B$ is greater than unity. We
+will presently return to this point. (See Art.~86.)
+
+We may also determine $C$ and $c$ independently of $B$ by formul\ae\
+adapted to logarithms. For, by Art.~44,
+\[
+  \cot a\, \sin b
+= \cos b\, \cos C + \sin C\, \cot A
+= \cos b\, (\cos C + \dfrac{\cot A}{\cos b}\, \sin C);
+\]
+assume $\tan \phi = \dfrac{\cot A}{\cos b}$; thus
+\[
+  \cot a\, \sin b
+= \cos b (\cos C + \tan \phi\, \sin C)
+= \dfrac{\cos b\, \cos (C - \phi)}{\cos \phi};
+\]
+therefore \hfill $
+  \cos (C - \phi) = \cos \phi \cot a \tan b$; \hfill\phantom{therefore}\\
+%-----File: 061.png------------------------------------------------
+from this equation $C - \phi$ is to be found, and then $C$. The ambiguity
+still exists; for if the last equation leads to $C - \phi = \alpha$, it
+will be satisfied also by $\phi - C = \alpha$; so that we have two admissible
+values for $C$, if $\phi + \alpha$ is less than $\pi$, and $\phi - \alpha$ is positive.
+
+And
+\[
+\cos a = \cos b \cos c + \sin b \sin c \cos A = \cos b (\cos c + \sin c \tan b \cos A);
+\]
+assume $\tan\theta = \tan b \cos A$; thus
+\[
+\cos a = \cos b (\cos c + \sin c \tan \theta) = \dfrac{\cos b\cos(c -\theta)}{\cos\theta};
+\]
+therefore \hfill$
+\cos(c - \theta) = \dfrac{\cos a \cos\theta}{\cos b};
+$\hfill\phantom{therefore}\\[2ex]
+from this equation $c - \theta$ is to be found, and then $c$; and there may
+be an ambiguity as before.
+
+Or we may treat this case conveniently by resolving the triangle
+into the sum or difference of two right-angled triangles.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/061fc}
+\end{figure}
+
+Let $CA = b$, and let $CAE = \text{the given angle }A$; from $C$ draw
+$CD$ perpendicular to $AE$, and let $CB$ and $CB' = a$; thus the figure
+shews that there may be two triangles which have the given elements.
+Then, by Art.~62, $\cos b = \cot A \cot ACD$; this finds $ACD$.
+Again, by Art.~62,
+\begin{align*}
+&\tan CD = \tan AC \cos ACD, \\
+&\text{and } \tan CD = \tan CB \cos BCD, \text{ or }
+                       \tan CB' \cos B'CD,
+\end{align*}
+therefore $\tan AC \cos ACD = \tan CB \cos BCD$, or $\tan CB' \cos B'CD$;
+this finds $BCD$ or $B'CD$.
+
+It is obvious that $ACD$ is what was denoted by $\phi$ in the former
+part of the Article.
+%-----File: 062.png------------------------------------------------
+
+\begin{flalign*}
+\text{\indent Also, by Art.\ 62, }
+  \tan AD &= \tan AC \cos A; \text{ this finds $AD$. Then }&&\\
+  \cos AC &= \cos CD \cos AD, \\
+  \cos CB &= \cos CD \cos BD, \\
+  \text{ or }\cos CB' &= \cos CD \cos B'D;
+\end{flalign*}
+therefore \hfill$\displaystyle
+  \dfrac{\cos AC}{\cos AD}
+= \dfrac{\cos CB}{\cos BD} \text{ or }
+  \dfrac{\cos CB'}{\cos B'D}$; \hfill\phantom{therefore}\\[1ex]
+this finds $BD$ or $B'D$.
+
+It is obvious that $AD$ is what was denoted by $\theta$ in the former
+part of the Article.
+
+\paragraph{85.} \textit{Having given two angles and the side opposite one of them
+$(\mathrm A$, $\mathrm B$, $\mathrm a)$.}
+
+This case is analogous to that immediately preceding, and
+gives rise to the same ambiguities. The side $b$ may be found from
+the formula
+\[
+  \sin b=\dfrac{\sin B \sin a}{\sin A};
+\]
+and then $C$ and $c$ may be found from Napier's analogies,
+\begin{align*}
+  \tan \tfrac{1}{2}C
+&= \dfrac{\cos\tfrac{1}{2}(a-b)}
+        {\cos\tfrac{1}{2}(a+b)} \cot\tfrac{1}{2}(A+B),
+\\[1ex]
+  \tan \tfrac{1}{2}c
+&= \dfrac{\cos\tfrac{1}{2}(A+B)}
+        {\cos\tfrac{1}{2}(A-B)} \tan\tfrac{1}{2}(a+b),
+\end{align*}
+
+We may also determine $C$ and $c$ independently of $b$ by formul\ae\
+adapted to logarithms. For
+\begin{multline*}
+  \cos A= -\cos B\cos C + \sin B\sin C\cos a \\
+= \cos B (-\cos C + \tan B\sin C \cos a),
+\end{multline*}
+assume $\cot \phi=\tan B \cos a;$ thus
+\[
+  \cos A = \cos B(-\cos C + \sin C \cot \phi)
+= \dfrac{\cos B\sin(C - \phi)}{\sin \phi};
+\]
+$\rlap{therefore }\hfill\displaystyle
+  \sin(C - \phi)=\dfrac{\cos A \sin \phi}{\cos B}; \hfill$ \\[1ex]
+%-----File: 063.png------------------------------------------------
+from this equation $C-\phi$ is to be found and then $C$. Since $C-\phi$
+is found from its sine there may be an ambiguity. Again, by
+Art.~44,
+\[
+\cot A \sin B = \cot a \sin c - \cos c \cos B =
+\cos B \left(- \cos c + \dfrac{\cot a \sin c}{\cos B}\right),
+\]
+$\text{assume } \cot \theta = \dfrac{\cot a}{\cos B}; \text{ then}$
+\[
+\cot A \sin B = \cos B (- \cos c + \sin c \cot \theta) =
+\frac{\cos B \sin (c - \theta)}{\sin \theta};
+\]
+$\rlap{therefore }\hfill
+\sin (c - \theta) = \cot A \tan B \sin \theta;
+\hfill$\\[1ex]
+from this equation $c - \theta$ is to be found, and then $c$. Since $c - \theta$ is
+found from its sine there may be an ambiguity. As before, it may
+be shewn that these results agree with those obtained by resolving
+the triangle into two right-angled triangles; for if in the triangle
+$ACB'$ the arc $CD$ be drawn perpendicular to $AB'$, then $B'CD$
+will $=\phi$, and $B'D = \theta$.
+
+\paragraph{86.} We now return to the consideration of the ambiguity
+which may occur in the case of Art.\ 84, when two sides are given
+and the angle opposite one of them. The discussion is somewhat
+tedious from its length, but presents no difficulty.
+
+Before considering the problem generally, we will take the
+particular case in which $a = b$; then $A$ must $= B$. The first and
+third of Napier's analogies give
+\[
+\cot \tfrac{1}{2}C = \tan A \cos a,\qquad \tan \tfrac{1}{2}c = \tan a \cos A;
+\]
+now $\cot \tfrac{1}{2}C$ and $\tan \tfrac{1}{2}c$ must both be \textit{positive}, so that $A$ and $a$ must
+be of the same affection. Hence, when $a = b$, there will be no
+solution at all, unless $A$ and $a$ are of the same affection, and then
+there will be only one solution; except when $A$ and $a$ are both
+right angles, and then $\cot \tfrac{1}{2}C$ and $\tan\tfrac{1}{2}c$ are indeterminate, and
+there is an infinite number of solutions.
+
+We now proceed to the general discussion.
+
+If $\sin b \sin A$ be greater than $\sin a$, there is no triangle which
+satisfies the given conditions; if $\sin b \sin A$ is \textit{not} greater than
+%-----File: 064.png------------------------------------------------
+$\sin a$, the equation $\sin B = \dfrac{\sin b \sin A}{\sin a}$ furnishes two values of $B$,
+which we will denote by $\beta$ and $\beta'$, so that $\beta' = \pi - \beta$; we will suppose
+that $\beta$ is the one which is not greater than the other.
+
+Now, in order that these values of $B$ may be admissible, it is
+necessary and sufficient that the values of $\cot\frac{1}{2}C$ and of $\tan\frac{1}{2}c$
+should both be positive, that is, $A-B$ and $a-b$ must have the
+same sign by the second and fourth of Napier's analogies. We
+have therefore to compare the sign of $A-\beta$ and the sign of $A-\beta'$
+with that of $a-b$.
+
+We will suppose that $A$ is less than a right angle, and separate
+the corresponding discussion into three cases.\medskip
+
+I\@. Let $b$ be less than $\dfrac{\pi}{2}$.
+
+(1) Let $a$ be less than $b$; the formula $\sin B = \dfrac{\sin b}{\sin a}\sin A$ make
+$\beta$ greater than $A$, and \textit{\`a fortiori} $\beta'$ greater than $A$. Hence there
+are two solutions.
+
+(2) Let $a$ be equal to $b$; then there is one solution, as previously
+shewn.
+
+(3) Let $a$ be greater than $b$; we may have then $a + b$ less than
+$\pi$ or equal to $\pi$ or greater than $\pi$. If $a + b$ is less than $\pi$, then
+$\sin a$ is greater than $\sin b$; thus $\beta$ is less than $A$ and therefore
+admissible, and $\beta'$ is greater than $A$ and inadmissible. Hence there
+is one solution. If $a + b$ is equal to $\pi$, then $\beta$ is equal to $A$, and
+$\beta'$ greater than $A$, and both are inadmissible. Hence there is no
+solution. If $a+b$ is greater than $\pi$, then $\sin a$ is less than $\sin b$,
+and $\beta$ and $\beta'$ are both greater than $A$, and both inadmissible.
+Hence there is no solution.\medskip
+
+II\@. Let $b$ be equal to $\dfrac{\pi}{2}$.
+
+(1) Let $a$ be less than $b$; then $\beta$ and $\beta'$ are both greater than
+$A$, and both admissible. Hence there are two solutions.
+
+(2) Let $a$ be equal to $b$; then there is no solution, as previously
+shewn.
+%-----File: 065.png------------------------------------------------
+
+(3) Let $a$ be greater than $b$; then $\sin a$ is less than $\sin b$, and
+$\beta$ and $\beta'$ are both greater than $A$, and inadmissible. Hence there
+is no solution.\medskip
+
+III\@. Let $b$ be greater than $\dfrac{\pi}{2}$.
+
+(1) Let $a$ be less than $b$; we may have then $a + b$ less than
+$\pi$ or equal to $\pi$ or greater than $\pi$. If $a + b$ is less than $\pi$, then
+$\sin a$ is less than $\sin b$, and $\beta$ and $\beta'$ are both greater than $A$ and
+both admissible. Hence there are two solutions. If $a + b$ is equal
+to $\pi$, then $\beta$ is equal to $A$ and inadmissible, and $\beta'$ is greater
+than $A$ and admissible. Hence there is one solution. If $a + b$
+is greater than $\pi$, then $\sin a$ is greater than $\sin b$; $\beta$ is less
+than $A$ and admissible, and $\beta'$ is greater than $A$ and admissible.
+Hence there is one solution.
+
+(2) Let $a$ be equal to $b$; then there is no solution, as previously
+shewn.
+
+(3) Let $a$ be greater than $b$; then $\sin a$ is less than $\sin b$,
+and $\beta$ and $\beta'$ are both greater than $A$ and both inadmissible.
+Hence there is no solution.
+
+We have then the following results when $A$ \textit{is less than a
+right angle}.
+
+\begin{longtable}{@{}l@{}l@{}l}
+& \rule{.6\textwidth}{0pt} \\[-2ex]
+\multirow{4}{*}{$b < \dfrac{\pi}{2} \left\{\rule{0pt}{7.5ex}\right.$}
+& $a < b$ \dotfill& two solutions,\\*
+& $a = b$ \dotfill& one solution,\\*
+& $a > b$ and $a + b < \pi$ \dotfill& one solution,\\*
+& $a > b$ and $a + b = \pi$ or ${}>\pi$ \dotfill& no solution.
+\\[2ex]
+\multirow{2}{*}{$b = \dfrac{\pi}{2} \left\{\rule{0pt}{4ex}\right.$}
+& $a < b$            \dotfill& two solutions,\\*
+& $a = b$ or $a > b$ \dotfill& no solution.
+\\[2ex]
+\multirow{3}{*}{$b > \dfrac{\pi}{2} \left\{\rule{0pt}{5.5ex}\right.$}
+& $a < b$ and $a+b<\pi$ \dotfill& two solutions,\\*
+& $a < b$ and $a+b=\pi$ or ${}>\pi$ \dotfill& one solution,\\*
+& $a = b$ or ${} > b$ \dotfill& no solution.
+\end{longtable}
+%-----File: 066.png------------------------------------------------
+
+It must be remembered, however, that in the cases in which
+two solutions are indicated, there will be no solution at all if
+$\sin a$ be less than $\sin b\sin A$.
+
+In the same manner the cases in which $A$ is equal to a right
+angle or greater than a right angle may be discussed, and the
+following results obtained.
+\medskip
+
+\textit{When $\mathrm A$ is equal to a right angle,}
+
+\begin{longtable}{@{}l@{}l@{}l}
+& \rule{.6\textwidth}{0pt} \\[-1ex]
+\multirow{3}{*}{$b < \dfrac{\pi}{2} \left\{\rule{0pt}{5.5ex}\right.$}
+& $a < b$ or  $a = b$ \dotfill& no solution, \\*
+& $a > b$ and $a + b < \pi$ \dotfill& one solution, \\*
+& $a > b$ and $a + b = \pi$ or ${} >\pi$ \dotfill& no solution.
+\\[2ex]
+\multirow{2}{*}{$b = \dfrac{\pi}{2} \left\{\rule{0pt}{4ex}\right.$}
+& $a < b$ or $a > b$ \dotfill& no solution,\\*
+&\multicolumn{2}{@{}l}{ $a = b$ \dotfill infinite number of solutions.}
+\\[2ex]
+\multirow{3}{*}{$b > \dfrac{\pi}{2} \left\{\rule{0pt}{5.5ex}\right.$}
+& $a < b$ and $a + b > \pi$ \dotfill& one solution,\\*
+& $a < b$ and $a + b = \pi$ or ${}<\pi$ \dotfill& no solution,\\*
+& $a = b$ or $a > b$ \dotfill& no solution.
+\end{longtable}\medskip
+
+\textit{When $\mathrm A$ is greater than a right angle,}
+
+\begin{longtable}{@{}l@{}l@{}l}
+& \rule{.6\textwidth}{0pt} \\[-1ex]
+\multirow{3}{*}{$b < \dfrac{\pi}{2} \left\{\rule{0pt}{5.5ex}\right.$}
+& $a < b$ or $a = b$ \dotfill& no solution,\\*
+& $a > b$ and $a + b = \pi$ or ${}<\pi$ \dotfill& one solution,\\*
+& $a > b$ and $a + b > \pi$ \dotfill& two solutions.
+\\[2ex]
+\multirow{2}{*}{$b = \dfrac{\pi}{2} \left\{\rule{0pt}{4ex}\right.$}
+& $a < b$ or $a = b$ \dotfill& no solution, \\*
+& $a > b$ \dotfill& two solutions.
+\\[2ex]
+\multirow{4}{*}{$b > \dfrac{\pi}{2} \left\{\rule{0pt}{7.5ex}\right.$}
+& $a < b$ and $a + b > \pi$ \dotfill& one solution, \\*
+& $a < b$ and $a + b = \pi$ or ${}<\pi$ \dotfill& no solution, \\*
+& $a = b$ \dotfill& one solution, \\*
+& $a > b$ \dotfill& two solutions.
+\end{longtable}\medskip
+
+As before in the cases in which two solutions are indicated,
+there will be no solution at all if $\sin a$ be less than
+$\sin b \sin A$.
+
+It will be seen from the above investigations that if $a$ lies
+between $b$ and $\pi-b$, there will be one solution; if $a$ does not lie
+between $b$ and $\pi - b$ either there are two solutions or there is
+no solution; this enunciation is not meant to include the cases in
+which $a = b$ or $= \pi - b$.
+%-----File: 067.png------------------------------------------------
+
+\paragraph{87.} The results of the preceding Article may be illustrated by
+a figure.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/067fc}
+\end{figure}
+
+Let $ADA'E$ be a great circle; suppose $PA$ and $PA'$ the
+projections on the plane of this circle of arcs which are each
+equal to $b$ and inclined at an angle $A$ to $ADA'$; let $PD$ and
+$PE$ be the projections of the least and greatest distances of $P$
+from the great circle (see Art.\ 59). Thus the figure supposes
+$A$ and $b$ each less than $\dfrac{\pi}{2}$.
+
+If $a$ be less than the arc which is represented by $PD$ there is
+no triangle; if $a$ be between $PD$ and $PA$ in magnitude, there are
+two triangles, since $B$ will fall on $ADA'$, and we have two triangles
+$BPA$ and $BPA'$; if $a$ be between $PA$ and $PH$ there will be only
+one triangle, as $B$ will fall on $A'H$ or $AH'$, and the triangle will be
+either $APB$ with $B$ between $A'$ and $H$, or else $A'PB$ with $B$ between
+$A$ and $H'$; but these two triangles are symmetrically equal
+(Art.\ 57); if $a$ be greater than $PH$ there will be no triangle.
+The figure will easily serve for all the cases; thus if $A$ is greater
+than $\dfrac{\pi}{2}$, we can suppose $PAE$ and $PA'E$ to be equal to $A$; if
+$b$ is greater than $\dfrac{\pi}{2}$, we can take $PH$ and $PH'$ to represent $b$.
+%-----File: 068.png------------------------------------------------
+
+\paragraph{88.} The ambiguities which occur in the last case in the solution
+of oblique-angled triangles (Art.\ 85) may be discussed in the
+same manner as those in Art.\ 86; or, by means of the polar
+triangle, the last case may be deduced from that of Art.~86.
+
+\section*{\centering\normalsize EXAMPLES.}
+
+1. The sides of a triangle are $105^\circ$, $90^\circ$, and $75^\circ$ respectively:
+find the sines of all the angles.
+\medskip
+
+2. Shew that $\tan \tfrac{1}{2} A \tan \tfrac{1}{2} B= \dfrac{\sin(s-c)}{\sin s}$. Solve a triangle
+when a side, an adjacent angle, and the sum of the other two
+sides are given.
+\medskip
+
+3. Solve a triangle having given a side, an adjacent angle,
+and the sum of the other two angles.
+\medskip
+
+4. A triangle has the sum of two sides equal to a semicircumference:
+find the arc joining the vertex with the middle of
+the base.
+\medskip
+
+5. If $a$, $b$, $c$ are known, $c$ being a \textit{quadrant}, determine the
+angles: shew also that if $\delta$ be the perpendicular on $c$ from the
+opposite angle, $\cos^2 \delta = \cos^2 a + \cos^2 b$.
+\medskip
+
+6. If one side of a spherical triangle be divided into four
+equal parts, and $\theta_1$, $\theta_2$, $\theta_3$, $\theta_4$, be the angles subtended at the opposite
+angle by the parts taken in order, shew that
+\[
+\sin(\theta_1 + \theta_2) \sin \theta_2 \sin\theta_4 = \sin(\theta_3 + \theta_4) \sin\theta_1 \sin\theta_3.
+\]
+
+7. In a spherical triangle if $A = B = 2C$, shew that
+\[
+8 \sin\left(a + \dfrac{c}{2}\right) \sin^2 \dfrac{c}{2} \cos \dfrac{c}{2} = \sin^3 a.
+\]
+%-----File: 069.png------------------------------------------------
+
+8. In a spherical triangle if $A = B = 2C$, shew that
+\[
+8 \sin^2 \dfrac{C}{2} \left(\cos s + \sin \dfrac{C}{2}\right)
+\dfrac{\cos\dfrac{c}{2}}{\cos a} = 1.
+\]
+
+9. If the equal sides of an isosceles triangle $ABC$ be bisected
+by an arc $DE$, and $BC$ be the base, shew that
+\[
+\sin \dfrac{DE}{2} = \tfrac{1}{2} \sin \dfrac{BC}{2} \sec \dfrac{AC}{2}.
+\]
+
+10. If $c_1$, $c_2$ be the two values of the third side when $A$, $a$, $b$
+are given and the triangle is ambiguous, shew that
+\[
+  \tan \dfrac{c_1}{2} \tan \dfrac{c_2}{2}
+= \tan \tfrac{1}{2} (b - a) \tan \tfrac{1}{2} (b + a).
+\]
+
+\chapter[Circumscribed and Inscribed Circles.]{CIRCUMSCRIBED AND INSCRIBED CIRCLES.}
+
+\paragraph{89.} \textit{To find the angular radius of the small circle inscribed
+in a given triangle.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/069fc}
+\end{figure}
+
+Let $ABC$ be the triangle; bisect the angles $A$ and $B$ by arcs
+meeting at $P$; from $P$ draw $PD$, $PE$, $PF$ perpendicular to the
+sides. Then it may be shewn that $PD$, $PE$, $PF$ are all equal;
+also that $AE = AF$, $BF = BD$, $CD = CE$. Hence $BC + AF =$ half
+the sum of the sides $= s$; therefore $AF = s - a$. Let $PF = r$.
+\begin{flalign*}
+&\rlap{\indent Now }&
+  \tan PF &= \tan PAF \sin AF \text{ (Art.~62);} &&\\
+&\rlap{thus }&
+  \tan r  &= \tan\dfrac{A}{2} \sin (s-a). \tag{1}
+\end{flalign*}
+%-----File: 070.png------------------------------------------------
+
+The value of $\tan r$ may be expressed in various forms; thus
+from Art.~45, we obtain
+\[
+\tan \dfrac{A}{2} = \Surd {\frac{\sin (s - b) \sin (s - c)}{\sin s\, \sin (s - a)} } ;
+\]
+substitute this value in (1), thus
+\[
+\tan r = \Surd{\left\{
+  \dfrac{\sin (s - a) \sin (s - b) \sin (s - c)}{\sin s} \right\}}
+  = \dfrac{n}{\sin s} \text{ (Art.~46)}. \tag{2}
+\]
+
+Again
+\begin{align*}
+  \sin (s - a) &= \sin \{\tfrac{1}{2}(b + c) - \tfrac{1}{2}a\} \\[1ex]
+    &= \sin \tfrac{1}{2}(b + c) \cos \tfrac{1}{2}a - \cos \tfrac{1}{2}(b + c) \sin \tfrac{1}{2}a \\[1ex]
+    &= \dfrac{\sin \tfrac{1}{2}a \cos \tfrac{1}{2}a}{\sin \tfrac{1}{2}A}
+        \{\cos \tfrac{1}{2}(B - C) - \cos \tfrac{1}{2}(B + C)\},\text{ (Art.~54)} \\[1ex]
+    &= \dfrac{\sin a \sin \tfrac{1}{2}B \sin \tfrac{1}{2}C}{\sin \tfrac{1}{2}A};
+\end{align*}
+\begin{flalign*}
+\text{therefore from (1) }&
+\tan r = \dfrac{\sin\tfrac{1}{2}B \sin \tfrac{1}{2}C}{\cos \tfrac{1}{2}A} \sin a; \tag{3}&&
+\end{flalign*}
+hence, by Art.~51,
+\begin{multline*}
+\tan r =
+\dfrac{\surd\{-\cos S \cos (S - A) \cos (S - B) \cos (S - C)\}}{2 \cos \tfrac{1}{2}A \cos \tfrac{1}{2}B \cos \tfrac{1}{2}C} \\[1ex]
+=\frac{N}{2 \cos \tfrac{1}{2}A \cos \tfrac{1}{2}B \cos \tfrac{1}{2}C}. \tag{4}
+\end{multline*}
+
+It may be shewn by common trigonometrical formul\ae\ that
+\[
+4 \cos\tfrac{1}{2}A \cos\tfrac{1}{2}B \cos\tfrac{1}{2}C = \cos S + \cos (S - A) + \cos (S - B) + \cos (S - C);
+\]
+hence we have from (4)
+\[
+\cot r = \frac{1}{2N} \bigl\{\cos S + \cos (S - A) + \cos (S - B) + \cos (S - C)\bigr\}. \tag{5}
+\]
+%-----File: 071.png------------------------------------------------
+
+\paragraph{90.} \textit{To find the angular radius of the small circle described
+so as to touch one side of a given triangle, and the other sides
+produced.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/071fc}
+\end{figure}
+
+Let $ABC$ be the triangle; and suppose we require the radius
+of the small circle which touches $BC$, and $AB$ and $AC$ produced.
+Produce $AB$ and $AC$ to meet at $A'$; then we require the radius of
+the small \textit{circle inscribed in $\mathrm{A'BC}$}, and the sides of $A'BC$ are $a$,
+$\pi-b$, $\pi-c$ respectively. Hence if $r_1$ be the required radius, and
+$s$ denote as usual $\frac{1}{2} (a + b + c)$, we have from Art.~89,
+\[
+  \tan r_1 = \tan\dfrac{A}{2}\sin s. \tag{1}
+\]
+
+From this result we may derive other equivalent forms as in
+the preceding Article; or we may make use of those forms immediately,
+observing that the angles of the triangle $A'BC$ are $A$,
+$\pi-B$, $\pi-C$ respectively. Hence $s$ being $\frac{1}{2} (a + b + c)$ and $S$
+being $\frac{1}{2} (A + B + C)$ we shall obtain
+\begin{align*}
+\tan r_1 &= \Surd{\left\{\dfrac{\sin s \sin(s-b)\sin(s-c)}
+                         {\sin(s-a)}\right\}}
+          = \dfrac{n}{\sin(s-a)}, \tag{2}
+\\[1.5ex]
+\tan r_1 &= \dfrac{\cos\tfrac{1}{2}B \cos\tfrac{1}{2}C}
+                 {\cos\tfrac{1}{2}A} \sin a, \tag{3} \\[1ex]
+\tan r_1 &
+  \begin{gathered}[t]
+  = \dfrac{\surd{\{-\cos S \cos(S-A) \cos(S-B) \cos(S-C)\}}}
+         {2\cos\tfrac{1}{2}A \sin\tfrac{1}{2}B \sin\tfrac{1}{2}C} \notag
+\\[1.5ex]
+  = \dfrac{N}{2\cos\tfrac{1}{2}A \sin\tfrac{1}{2}B \sin\tfrac{1}{2}C},
+  \end{gathered} \tag{4}
+\\[1.5ex]
+\multispan{2}{$\cot r_1 = \dfrac{1}{2N} \{-c\cos S - \cos(S-A) + \cos(S-B) + \cos(S-C) \} \tag{5}$.}
+\end{align*}
+%-----File: 072.png------------------------------------------------
+
+These results may also be found independently by bisecting
+two of the angles of the triangle $A'BC,$ so as to determine the
+pole of the small circle, and proceeding as in Art.~89.
+
+\paragraph{91.} A circle which touches one side of a triangle and the
+other sides produced is called an \textit{escribed circle;} thus there are
+three escribed circles belonging to a given triangle. We may
+denote the radii of the escribed circles which touch $CA$ and $AB$
+respectively by $r_2$ and $r_3$, and values of $\tan r_2$ and $\tan r_3$ may
+be found from what has been already given with respect to
+$\tan r_1$ by appropriate changes in the letters which denote the
+sides and angles.
+
+In the preceding Article a triangle $A'BC$ was formed by producing
+$AB$ and $AC$ to meet again at $A'$; similarly another triangle
+may be formed by producing $BC$ and $BA$ to meet again, and
+another by producing $CA$ and $CB$ to meet again. The original
+triangle $ABC$ and the three formed from it have been called
+\textit{associated triangles}, $ABC$ being the fundamental triangle. Thus
+the inscribed and escribed circles of a given triangle are the same
+as the circles inscribed in the system of associated triangles of
+which the given triangle is the fundamental triangle.
+
+\paragraph{92.} \textit{To find the angular radius of the small circle described
+about a given triangle.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/072fc}
+\end{figure}
+
+Let $ABC$ be the given triangle; bisect the sides $CB$, $CA$ at
+$D$ and $E$ respectively, and draw from $D$ and $E$ arcs at right angles
+to $CB$ and $CA$ respectively, and let $P$ be the intersection of these
+%-----File: 073.png------------------------------------------------
+arcs. Then $P$ will be the pole of the small circle described about
+$ABC$. For draw $PA$, $PB$, $PC$; then from the right-angled
+triangles $PCD$ and $PBD$ it follows that $PB=PC$; and from
+the right-angled triangles $PCE$ and $PAE$ it follows that $PA = PC$;
+hence $PA = PB=PC$. Also the angle $PAB =$ the angle $PBA$,
+the angle $PBC =$ the angle $PCB$, and the angle $PCA =$ the angle
+$PAC$; therefore $PCB + A = \tfrac{1}{2} (A + B + C)$, and $PCB = S-A$.
+
+Let $PC = R$.
+\begin{flalign*}
+&\text{\indent Now }&
+  \tan CD &= \tan CP \cos PCD,\ \text{(Art.\ 62,)}
+  &\phantom{therefore }&\\[1ex]
+&\text{thus }&
+  \tan \tfrac{1}{2} a &= \tan R \cos (S-A),
+  &&\\
+&\text{therefore }&
+  \tan R &= \dfrac{\tan\tfrac{1}{2}a}{\cos(S-A)}.
+  \tag{1}
+\end{flalign*}
+
+The value of $\tan R$ may be expressed in various forms; thus
+if we substitute for $\tan\dfrac{a}{2}$ from Art.~49, we obtain
+\[
+\tag{2}
+\tan R
+= \Surd{\left\{\dfrac{-\cos S}{\cos(S-A)\cos(S-B)\cos(S-C)}\right\}}
+= \dfrac{\cos S}{N} \,.
+\]
+\begin{flalign*}
+&\text{Again }&
+  \cos(S-A) &= \cos\left\{\tfrac{1}{2}(B+C)-\tfrac{1}{2}A\right\}
+&\phantom{Again }&
+\\[1.5ex]
+&&&= \cos\tfrac{1}{2}(B+C) \cos\tfrac{1}{2}A
+   + \sin\tfrac{1}{2}(B+C) \sin\tfrac{1}{2}A
+&&
+\end{flalign*}
+\vspace{-3ex}
+\begin{align*}
+\mspace{50mu}
+&= \frac{\sin\tfrac{1}{2}A\cos\tfrac{1}{2}A}{\cos\tfrac{1}{2}a}
+   \left\{\cos\tfrac{1}{2}(b+c)
+         + \cos\tfrac{1}{2}(b-c) \right\},\ \text{(Art.\ 54,)}
+\\
+&= \dfrac{\sin A}{\cos\tfrac{1}{2}a} \cos\tfrac{1}{2}b\cos\tfrac{1}{2}c;
+\end{align*}
+therefore from (1)
+\[
+\tag{3}
+\tan R = \dfrac{\sin\tfrac{1}{2}a }
+              {\sin A\cos\tfrac{1}{2}b\cos\tfrac{1}{2}c } \,.
+\]
+
+Substitute in the last expression the value of $\sin A$ from
+Art.~46; thus
+\begin{align*}
+\tan R
+&= \dfrac{2\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin\tfrac{1}{2}c}
+       {\surd{\left\{\sin s\sin(s-a) \sin(s-b) \sin(s-c) \right\}}}
+\\[1.5ex]
+&= \dfrac{2\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin\tfrac{1}{2}c}{n}
+\,.
+\tag{4} \\[-3ex]
+\end{align*}
+%-----File: 074.png------------------------------------------------
+
+It may be shewn, by common trigonometrical formul\ae\, that
+\[
+  4\sin\tfrac{1}{2}a\sin\tfrac{1}{2}b\sin\tfrac{1}{2}c
+= \sin(s-a) + \sin(s-b) + \sin(s-c)-\sin s;
+\]
+hence we have from (4)
+\[
+\tag{5}
+\tan R=\dfrac{1}{2n}\{
+\sin(s-a)+\sin(s-b)+\sin(s-c)-\sin s
+\}.
+\]
+
+\paragraph{93.} \textit{To find the angular radii of the small circles described
+round the triangles associated with a given fundamental triangle.}
+
+Let $R_1$ denote the radius of the circle described round the
+triangle formed by producing $AB$ and $AC$ to meet again at $A'$;
+similarly let $R_2$ and $R_3$ denote the radii of the circles described
+round the other two triangles which are similarly formed. Then
+we may deduce expressions for $\tan R_1$, $\tan R_2$, and $\tan R_3$ from
+those found in Art.~92 for $\tan R$. The sides of the triangle $A'BC$
+are $a$, $\pi-b$, $\pi-c$, and its angles are $A$, $\pi-B$, $\pi-C$; hence if
+$s = \tfrac{1}{2} (a + b + c)$ and $S = \tfrac{1}{2} (A + B + C)$ we shall obtain from
+Art.~92
+\begin{align*}
+\tag{1}
+  \tan R_1 &= \frac{\tan\frac{1}{2}a}{-\cos S} \,,
+\\[1.5ex]
+\tag{2}
+  \tan R_1 &= \Surd{\left\{
+  \frac{\cos(S-A)}{-\cos S\cos(S-B)\cos(S-C)}\right\}}
+= \frac{\cos(S-A)}{N} \,,
+\\[1.5ex]
+\tag{3}
+  \tan R_1 &= \frac{\sin\frac{1}{2}a}
+                   {\sin A\sin\frac{1}{2}b\sin\frac{1}{2}c} \,,
+\\[1.5ex]
+\tag{4}
+  \tan R_1
+&= \frac{2\sin\frac{1}{2}a\cos\frac{1}{2}b\cos\frac{1}{2}c}
+        {\surd{\{\sin s\sin(s-a)\sin(s-b)\sin(s-c)\}}} \,,
+\\[1.5ex]
+\tag{5}
+  \tan R_1
+&=\frac{1}{2n}
+  \{\sin s - \sin(s-a) + \sin(s-b) + \sin(s-c) \}.
+\end{align*}
+
+Similarly we may find expressions for $\tan R_2$ and $\tan R_3$.
+
+\paragraph{94.} Many examples may be proposed involving properties of
+the circles inscribed in and described about the associated triangles.
+We will give one that will be of use hereafter.
+%-----File: 075.png------------------------------------------------
+
+To prove that
+\[(\cot r + \tan R)^2=\dfrac{1}{4n^2}(\sin a+\sin b+\sin c)^2 -1.\]
+
+We have
+\[
+  4n^2=1-\cos^2 a-\cos^2 b-\cos^2 c+2\cos a\cos b\cos c;
+\]
+therefore
+\[
+(\sin a + \sin b + \sin c)^2-4n^2
+\]
+$= 2\ ( 1+ \sin a \sin b + \sin b \sin c + \sin c \sin a - \cos a \cos b \cos c \ )$.\\
+Also $\cot r + \tan R
+= \dfrac{1}{2n}\Bigl\{\sin s + \sin(s-a)+\sin(s-b)+\sin (s-c)\Bigr\}$;
+and by squaring both members of this equation the required
+result will be obtained. For it may be shewn by reduction that
+\[
+\sin^2 s + \sin^2 (s-a) + \sin^2 (s-b) + \sin^2 (s-c) = 2-2 \cos a \cos b \cos c,
+\]
+and
+\begin{gather*}
+\sin s \sin (s-a) + \sin s \sin (s-b) + \sin s \sin (s-c)
+\\
+{}+ \sin (s-a) \sin (s-b) + \sin (s-b) \sin (s-c) + \sin (s-c) \sin (s-a)
+\\
+= \sin a \sin b + \sin b \sin c + \sin c \sin a.
+\end{gather*}
+
+Similarly we may prove that
+\[
+(\cot r_1-\tan R)^2=\dfrac{1}{4n^2}(\sin b+\sin c-\sin a)^2 -1.
+\]
+
+\paragraph{95.} In the figure to Art.\ 89, suppose $DP$ produced through
+$P$ to a point $A'$ such that $DA'$ is a quadrant, then $A'$ is a pole of
+$BC$, and $PA' = \dfrac{\pi}{2}-r$; similarly, suppose $EP$ produced through $P$
+to a point $B'$ such that $EB'$ is a quadrant, and $FP$ produced
+through $P$ to a point $C'$ such that $FC'$ is a quadrant. Then
+$A'B'C'$ is the polar triangle of $ABC$, and $PA' = PB' = PC' = \dfrac{\pi}{2}-r$.
+Thus $P$ is the pole of the small circle \textit{described round} the polar
+triangle, and the angular radius of the small circle described round
+the polar triangle is the complement of the angular radius of the
+%-----File: 076.png------------------------------------------------
+small circle inscribed in the primitive triangle. And in like
+manner the point which is the pole of the small circle inscribed in
+the polar triangle is also the pole of the small circle described
+round the primitive triangle, and the angular radii of the two
+circles are complementary.
+
+\section*{\centering\normalsize EXAMPLES.}
+
+In the following examples the notation of the Chapter is
+retained. Shew that in any triangle the following relations hold contained
+in Examples 1 to 7:
+\medskip
+
+1. $\operatorname{Tan} r_1 \tan r_2 \tan r_3 = \tan r \sin^2 s$.
+\medskip
+
+2. $\operatorname{Tan} R + \cot r = \tan R_1 + \cot r_1 = \tan R_2 + \cot r_2$\\
+\rightline{$= \tan R_3 + \cot r_3 = \tfrac{1}{2} (\cot r + \cot r_1 + \cot r_2 + \cot r_3)$.}
+\medskip
+
+3. $\operatorname{Tan}^2 R + \tan^2 R_1 + \tan^2 R_2 + \tan^2 R_3$\\
+\rightline{$ = \cot^2 r + \cot^2 r_1 + \cot^2 r_2 + \cot^2 r_3$.}
+\medskip
+
+4. $\dfrac{\operatorname{Tan} r_1 + \tan r_2 + \tan r_3 - \tan r}
+       {\cot r_1 + \cot r_2 + \cot r_3 - \cot r}
+= \tfrac{1}{2} (1 + \cos a + \cos b + \cos c)$.
+\medskip
+
+5. $\operatorname{Cosec}^2 r
+= \cot (s - a) \cot (s - b)
++ \cot (s - b) \cot (s - c)
++ \cot (s - c)      (s - a)$.
+\medskip
+
+6. $\operatorname{Cosec}^2 r_1
+= \cot (s - b) \cot (s - c)
+- \cot s \cot (s - b)
+- \cot s \cot (s - c)$.
+\medskip
+
+7. $\operatorname{Tan} R_1 \tan R_2 \tan R_3 = \tan R \sec^2 S$.
+\medskip
+
+8. Shew that in an equilateral triangle $\tan R = 2\tan r$.
+\medskip
+
+9. If $ABC$ be an equilateral spherical triangle, $P$ the pole of
+the circle circumscribing it, $Q$ any point on the sphere, shew that
+\[
+  \cos QA + \cos QB + \cos QC = 3\cos PA \cos PQ.
+\]
+
+10. If three small circles be inscribed in a spherical triangle
+having each of its angles $120^\circ$, so that each touches the other two
+as well as two sides of the triangle, shew that the radius of each
+of the small circles $= 30^\circ$, and that the centres of the three small
+circles coincide with the angular points of the polar triangle.
+%-----File: 077.png------------------------------------------------
+
+\chapter[Area of a Spherical Triangle. Spherical Excess.]{AREA OF A SPHERICAL TRIANGLE\@. SPHERICAL EXCESS.}
+
+\paragraph{96.} \textit{To find the area of a Lune.}
+
+A \textit{Lune} is that portion of the surface of a sphere which is
+comprised between two great semicircles.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/077fc}
+\end{figure}
+
+Let $ACBDA$, $ADBEA$ be two lunes having equal angles at $A$;
+then one of these lunes may be supposed placed on the other so as
+to coincide exactly with it; thus \textit{lunes having equal angles are
+equal.} Then by a process similar to that used in the first proposition
+of the Sixth Book of Euclid it may be shewn that \textit{lunes
+are proportional to their angles.} Hence since the whole surface of
+a sphere may be considered as a lune with an angle equal to four
+right angles, we have for a lune with an angle of which the
+circular measure is $A$,
+\[
+  \dfrac{\text{area of lune}}{\text{surface of sphere}}
+= \dfrac{A}{2\pi}\,.
+\]
+
+Suppose $r$ the radius of the sphere, then the surface is $4\pi r^2$
+(\textit{Integral Calculus}, Chap.~\textsc{vii.}); thus
+\[
+  \text{area of lune } = \dfrac{A}{2\pi} 4\pi r^2 = 2Ar^2.
+\]
+%-----File: 078.png------------------------------------------------
+
+\paragraph{97.} \textit{To find the area of a Spherical Triangle.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/078fc}
+\end{figure}
+
+Let $ABC$ be a spherical triangle; produce the arcs which form
+its sides until they meet again two and two, which will happen
+when each has become equal to the semicircumference. The
+triangle $ABC$ now forms a part of three lunes, namely, $ABDCA$,
+$BCEAB$, and $CAFBC$. Now the triangles $CDE$ and $AFB$ are
+subtended by vertically opposite solid angles at $O$, and \textit{we will
+assume} that their areas are equal; therefore the lune $CAFBC$ is
+equal to the sum of the two triangles $ABC$ and $CDE$. Hence if
+$A$, $B$, $C$ denote the circular measures of the angles of the triangle,
+we have
+\begin{align*}
+& \text{triangle } ABC+BGDC = \text{lune } ABDCA = 2Ar^2, \\
+& \text{triangle } ABC+AHEC = \text{lune } BCEAB = 2Br^2, \\
+& \text{triangle } ABC+\text{triangle } CDE
+= \text{lune } CAFBC = 2Cr^2;
+\end{align*}
+hence, by addition,
+\[
+  \text{twice triangle } ABC + \text{surface of hemisphere}
+= 2(A+B+C)r^2;
+\]
+$\rlap{therefore }\hfill
+  \text{triangle } ABC=(A+B+C-\pi)r^2.
+\hfill$
+
+The expression $A+B+C-\pi$ is called the \textit{spherical excess} of
+the triangle; and since
+\[
+  (A+B+C-\pi)r^2 = \dfrac{A+B+C-\pi}{2\pi}\, 2\pi r^2,
+\]
+%-----File: 079.png------------------------------------------------
+the result obtained may be thus enunciated: \textit{the area of a spherical
+triangle is the same fraction of half the surface of the sphere as the
+spherical excess is of four right angles.}
+
+\paragraph{98.} We have assumed, as is usually done, that the areas of
+the triangles $CDE$ and $AFB$ in the preceding Article are equal.
+The triangles are, however, not absolutely equal, but \textit{symmetrically}
+equal (Art.\ 57), so that one cannot be made to coincide
+with the other by superposition. It is, however, easy to decompose
+two such triangles into pieces which admit of superposition,
+and thus to prove that their areas are equal. For describe a
+small circle round each, then the angular radii of these circles
+will be equal by Art.~92. If the pole of the circumscribing circle
+falls inside each triangle, then each triangle is the sum of three
+isosceles triangles, and if the pole falls outside each triangle, then
+each triangle is the excess of two isosceles triangles over a third;
+and in each case the isosceles triangles of one set are respectively
+\textit{absolutely equal} to the corresponding isosceles triangles of the
+other set.
+
+\paragraph{99.} \textit{To find the area of a spherical polygon.}
+
+Let $n$ be the number of sides of the polygon, $\Sigma$ the sum of all
+its angles. Take any point within the polygon and join it with
+all the angular points; thus the figure is divided into $n$ triangles.
+Hence, by Art.~97,
+\[
+  \text{area of polygon
+= (sum of the angles of the triangles} - n\pi)r^2,
+\]
+and the sum of the angles of the triangles is equal to $\Sigma$ together
+with the four right angles which are formed round the common
+vertex; therefore
+\[
+  \text{area of polygon} = \Bigl\{\Sigma - (n-2)\pi \Bigr\} r^2.
+\]
+
+This expression is true even when the polygon has some of its
+angles greater than two right angles, provided it can be decomposed
+into triangles, of which each of the angles is less than two
+right angles.
+%-----File: 080.png------------------------------------------------
+
+\paragraph{100.} We shall now give some expressions for certain trigonometrical
+functions of the \textit{spherical excess} of a triangle. We denote
+the spherical excess by $E$, so that $E=A+B+C-\pi$.
+
+\paragraph{101.} \textit{Cagnoli's Theorem.} To shew that
+\[
+  \sin\tfrac{1}{2}E
+= \dfrac{\surd\{\sin s \sin(s-a) \sin(s-b) \sin(s-c) \} x}
+{2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }.
+\]
+\begin{align*}
+   \operatorname{Sin}\tfrac{1}{2}E
+&= \sin\tfrac{1}{2}(A+B+C-\pi)
+ = \sin\{\tfrac{1}{2}(A+B) - \tfrac{1}{2}(\pi-C) \}
+\\[1.5ex]
+&= \sin\tfrac{1}{2}(A+B) \sin\tfrac{1}{2}C
+ - \cos\tfrac{1}{2}(A+B) \cos\tfrac{1}{2}C
+\\[1.5ex]
+&= \dfrac{\sin\tfrac{1}{2}C \cos\tfrac{1}{2}C }{\cos\tfrac{1}{2}c }
+   \{\cos\tfrac{1}{2}(a-b) - \cos\tfrac{1}{2}(a-b) \},
+   \quad \text{(Art.~54),}
+\\[1.5ex]
+&= \dfrac{\sin C \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b }{\cos\tfrac{1}{2}c}
+\\[1.5ex]
+&= \frac{\sin\tfrac{1}{2}a\sin\tfrac{1}{2}b}{\cos\tfrac{1}{2}c}\centerdot
+   \frac{2}{\sin a\sin b} \centerdot
+   \surd\{\sin s\sin(s-a)\sin(s-b)\sin(s-c) \}\\
+&= \frac{\surd\{\sin s\sin(s-a)\sin(s-b)\sin(s-c) \} }
+        {2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }.
+\end{align*}
+
+\paragraph{102.} \textit{Lhuilier's Theorem.} To shew that
+\begin{flalign*}
+&&\tan\tfrac{1}{4}E
+&= \surd\{\tan\tfrac{1}{2}s \tan\tfrac{1}{2}(s-a)
+       \tan\tfrac{1}{2}(s-b) \tan\tfrac{1}{2}(s-c) \}.
+\\[3ex]
+&&\operatorname{Tan}\tfrac{1}{4}E
+&= \dfrac{\sin\tfrac{1}{2}(A+B+C-\pi) }
+        {\cos\tfrac{1}{4}(A+B+C-\pi) }
+\\[1.5ex]
+&&&= \dfrac{\sin\tfrac{1}{2}(A+B) - \sin\tfrac{1}{2}(\pi-C) }
+          {\cos\tfrac{1}{2}(A+B) + \cos\tfrac{1}{2}(\pi-C) }\,,
+&\llap{(\textit{Plane Trig.}~Art.~84),}&
+\\[1.5ex]
+&&&= \dfrac{\sin\tfrac{1}{2}(A+B) - \cos\tfrac{1}{2}C }
+          {\cos\tfrac{1}{2}(A+B) + \sin\tfrac{1}{2}C }
+\\[1.5ex]
+&&&= \dfrac{\cos\tfrac{1}{2}(a-b) - \cos\tfrac{1}{2}c }
+          {\cos\tfrac{1}{2}(a+b) + \cos\tfrac{1}{2}c } \centerdot
+     \dfrac{\cos\tfrac{1}{2}C }{\sin\tfrac{1}{2}C }\,,
+&\llap{(Art.~54)}.&
+\end{flalign*}
+
+Hence, by Art.~45, we obtain
+\begin{align*}
+\tan\tfrac{1}{4}E
+&= \dfrac{\sin\tfrac{1}{4}(c+a-b) \sin\tfrac{1}{4}(c+b-a) }
+        {\cos\tfrac{1}{4}(a+b+c) \cos\tfrac{1}{4}(a+b-c) }
+   \Surd{\left\{
+          \dfrac{\sin s\sin(s-c) }{\sin(s-a)\sin(s-b) } \right\} }
+\\[1.5ex]
+&= \surd\{\tan\tfrac{1}{2}s \tan\tfrac{1}{2}(s-a)
+           \tan\tfrac{1}{2}(s-b) \tan\tfrac{1}{2}(s-c) \}.
+\end{align*}
+%-----File: 081.png------------------------------------------------
+
+\paragraph{103.} We may obtain many other formul\ae\ involving trigonometrical
+functions of the spherical excess. Thus, for example,
+\begin{align*}
+\cos\tfrac{1}{2}E
+&= \cos\left\{\tfrac{1}{2}(A+B)
+ - \tfrac{1}{2}(\pi-C) \right\}
+\\[1.5ex]
+&= \cos\tfrac{1}{2}(A+B) \sin\tfrac{1}{2}C +
+   \sin\tfrac{1}{2}(A+B) \cos\tfrac{1}{2}C
+\\[1.5ex]
+&= \Bigl\{\cos\tfrac{1}{2}(a+b) \sin^2\tfrac{1}{2}C +
+           \cos\tfrac{1}{2}(a-b) \cos^2\tfrac{1}{2}C \Bigr\}
+   \sec\tfrac{1}{2}c,\ \text{(Art.\ 54),}
+\\[1.5ex]
+&= \Bigl\{\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b
+   (\cos^2C + \sin^2\tfrac{1}{2}C)
+\\
+&\mspace{100mu}
+{}+\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b
+   (\cos^2\tfrac{1}{2}C - \sin^2\tfrac{1}{2}C) \Bigr\}
+   \sec\tfrac{1}{2}c
+\\[1.5ex]
+&=\left\{\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b +
+          \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \cos C \right\}
+  \sec\tfrac{1}{2}c.
+\tag{1}
+\end{align*}
+
+Again, it was shewn in Art.\ 101, that
+\begin{flalign*}
+&&& \sin\tfrac{1}{2}E
+  = \sin C \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sec\tfrac{1}{2}c; &&
+\\[1.5ex]
+&\rlap{therefore}&
+\tag{2}
+& \tan\tfrac{1}{2}E =
+  \frac{\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin C }
+       {\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b
+       + \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \cos C } \,.
+\end{flalign*}
+
+Again, we have from above
+\[
+  \cos\tfrac{1}{2}E
+= \Bigl\{\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b
+        + \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \cos C \Bigr\}
+  \sec\tfrac{1}{2}c
+\]
+\begin{align*}
+&= \dfrac{(1+\cos a)(1+\cos b) + \sin a\sin b\cos C}
+        {4\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }
+\\[1.5ex]
+&= \dfrac{1+\cos a+\cos b\cos c }
+        {4\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }
+ = \frac{\cos^2\tfrac{1}{2}a + \cos^2\tfrac{1}{2}b
+        + \cos^2\tfrac{1}{2}c-1 }
+        {2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }
+\tag{3} \,.
+\end{align*}
+
+In (3) put $1-2\sin^2\frac{1}{4}E$ for $\cos \frac{1}{2}E$; thus
+\[
+\sin^2\tfrac{1}{4}E =
+  \dfrac{1 + 2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c
+       - \cos^2\tfrac{1}{2}a - \cos^2\tfrac{1}{2}b - \cos^2\tfrac{1}{2}c}
+       {4\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }\,.
+\]
+
+By ordinary development we can shew that the numerator of
+the above fraction is equal to
+\[
+  4\sin\tfrac{1}{2}s \sin\tfrac{1}{2}(s-a)
+   \sin\tfrac{1}{2}(s-b) \sin\tfrac{1}{2}(s-c);
+\]
+%-----File: 082.png------------------------------------------------
+therefore
+\begin{align*}
+\sin^2\tfrac{1}{4}E
+&= \frac{\sin\frac{1}{2}s \sin\frac{1}{2}(s-a)
+          \sin\frac{1}{2}(s-b) \sin\frac{1}{2}(s-c) }
+        {\cos\frac{1}{2}a \cos\frac{1}{2}b \cos\frac{1}{2}c }.\tag{4}\\
+\intertext{\indent Similarly}
+\cos^2\tfrac{1}{4}E
+&= \frac{\cos\frac{1}{2}s \cos\frac{1}{2}(s-a)
+          \cos\frac{1}{2}(s-b) \cos\frac{1}{2}(s-c) }
+        {\cos\frac{1}{2}a \cos\frac{1}{2}b \cos\frac{1}{2}c }.\tag{5}
+\end{align*}
+
+Hence by division we obtain Lhuilier's Theorem.
+
+Again,
+\begin{multline*}
+\frac{\sin(C-\tfrac12E)}{\sin\frac12E} = \sin C\cot\tfrac12E-\cos C\\
+\begin{aligned}
+&= \sin C \frac{\cos\frac12a \cos\frac12b
+               + \sin\frac12a \sin\frac12b \cos C}
+               {\sin\frac12a \sin\frac12b \sin C}
+ - \cos C, \text{ by (2),}
+\\
+&=\cos\tfrac12a \cot\tfrac12b;
+\end{aligned}
+\end{multline*}
+therefore, by Art.\ 101,
+\[
+  \sin(C-\tfrac12E)
+= \frac{\surd\{\sin s \sin(s-a) \sin(s-b) \sin(s-c) \} }
+       {2\sin\frac12a \sin\frac12b \cos\frac12c }.
+\]
+
+Again, $\cos(C-\frac12E)=\cos C\cos\frac12E+\sin C\sin\frac12E$
+\begin{align*}
+&= \frac{(1+\cos a)(1+\cos b)\cos C + \sin a\sin b\cos^2 C }
+        {4\cos\frac12a \cos\frac12b \cos\frac12c }
+ + \sin^2C \sin\tfrac12a \sin\tfrac12b \sec\tfrac12c
+\\[1.5ex]
+&= \frac{(1+\cos a)(1+\cos b)\cos C + \sin a\sin b }
+        {4\cos\frac12a \cos\frac12b \cos\frac12c }
+\\[1.5ex]
+&= \Bigl\{\cos\tfrac12a \cos\tfrac12b \cos C
+         + \sin\tfrac12a \sin\tfrac12b \Bigr\} \sec\tfrac12c
+\\[1.5ex]
+&= \frac{\sin a\sin b\cos C + 4\sin^2\frac12a \sin^2\frac12b }
+        {4\sin\frac12a \sin\frac12b \cos\frac12c }
+\\[1.5ex]
+&= \frac{\cos c - \cos a\cos b + (1-\cos a)(1-\cos b) }
+        {4\sin\frac12a \sin\frac12b \cos\frac12c }
+\\[1.5ex]
+&= \frac{1 + \cos c - \cos a - \cos b }
+        {4\sin\frac12a \sin\frac12b \cos\frac12c }
+ = \frac{\cos^2\frac12c - \cos^2\frac12a - \cos^2\frac12b + 1 }
+        {2\sin\frac12a \sin\frac12b \cos\frac12c }\,.\tag{6}
+\end{align*}
+%-----File: 083.png------------------------------------------------
+
+From this result we can deduce two other results, in the
+same manner as (4) and (5) were deduced from (3); or we may
+observe that the right-hand member of (6) can be obtained from
+the right-hand member of (3) by writing $\pi-a$ and $\pi-b$ for
+$a$ and $b$ respectively, and thus we may deduce the results more
+easily. We shall have then
+\begin{align*}
+\sin^2 (\tfrac{1}{2}C-\tfrac{1}{4}E) &=
+  \frac{\cos\frac{1}{2}s \sin\frac{1}{2}(s-a) \sin\frac{1}{2}(s-b) \cos\frac{1}{2}(s-c) }
+       {\sin\frac{1}{2}a \sin\frac{1}{2}b \cos\frac{1}{2}c },
+\\[1.5ex]
+\cos^2 (\tfrac{1}{2}C-\tfrac{1}{4}E) &=
+  \frac{\sin\frac{1}{2}s \cos\frac{1}{2}(s-a) \cos\frac{1}{2}(s-b) \sin\frac{1}{2}(s-c) }
+       {\sin\frac{1}{2}a \sin\frac{1}{2}b \cos\frac{1}{2}c }.
+\end{align*}
+
+\section*{\centering\normalsize EXAMPLES.}
+
+1. Find the angles and sides of an equilateral triangle whose
+area is one-fourth of that of the sphere on which it is described.
+\medskip
+
+2. Find the surface of an equilateral and equiangular spherical
+polygon of $n$ sides, and determine the value of each of the
+angles when the surface equals half the surface of the sphere.
+\medskip
+
+3. If $a=b=\dfrac{\pi}{3}$, and $c=\dfrac{\pi}{2}$, shew that $E=\cos^{-1}\dfrac{7}{9}$.
+\medskip
+
+4. If the angle $C$ of a spherical triangle be a right angle,
+shew that
+\[
+\sin \tfrac{1}{2} E= \sin \tfrac{1}{2} a \sin \tfrac{1}{2} b \sec \tfrac{1}{2} c, \quad
+\cos \tfrac{1}{2} E= \cos \tfrac{1}{2} a \cos \tfrac{1}{2} b \sec \tfrac{1}{2} c.
+\]
+
+5. If the angle $C$ be a right angle, shew that
+\[
+\frac{\sin^2 c}{\cos c}\cos E= \frac{\sin^2 a}{\cos a}+\frac{\sin^2 b}{\cos b}.
+\]
+
+6. If $a=b$ and $C=\dfrac{\pi}{2}$, shew that $\tan E=\dfrac{\sin^2 a}{2\cos a}$.
+\medskip
+
+7. The sum of the angles in a right-angled triangle is less
+than four right angles.
+\medskip
+
+8. Draw through a given point in the side of a spherical
+triangle an arc of a great circle cutting off a given part of the
+triangle.
+%-----File: 084.png------------------------------------------------
+\medskip
+
+9. In a spherical triangle if $\cos C=-\tan\dfrac{a}{2}\tan\dfrac{b}{2}$, then
+$C=A+B$.
+\medskip
+
+10. If the angles of a spherical triangle be together equal to
+four right angles
+\[
+  \cos^2\tfrac{1}{2}a + \cos^2\tfrac{1}{2}b + \cos^2\tfrac{1}{2}c = 1.
+\]
+
+11. If $r_1$, $r_2$, $r_3$ be the radii of three small circles of a
+sphere of radius $r$ which touch one another at $P$, $Q$, $R$, and
+$A$, $B$, $C$ be the angles of the spherical triangle formed by joining
+their centres,
+\[
+\text{area }PQR = (A\cos r_1 + B\cos r_2 + C\cos r_3 - \pi)r^2.
+\]
+
+12. Shew that
+\[
+\sin s
+= \frac{\Bigl\{\sin\frac{1}{2}E \sin(A-\frac{1}{2}E)
+         \sin(B-\frac{1}{2}E) \sin(C-\frac{1}{2}E)
+         \Bigr\}^{\frac{1}{2}} }
+       {2\sin\frac{1}{2}A \sin\frac{1}{2}B \sin\frac{1}{2}C } \,.
+\]
+
+13. Given two sides of a spherical triangle, determine when
+the area is a maximum.
+\medskip
+
+14. Find the area of a regular polygon of a given number of
+sides formed by arcs of great circles on the surface of a sphere;
+and hence deduce that, if $\alpha$ be the angular radius of a small
+circle, its area is to that of the whole surface of the sphere as
+$\operatorname{versin}\alpha$ is to 2.
+\medskip
+
+15. $A$, $B$, $C$ are the angular points of a spherical triangle;
+$A'$, $B'$, $C'$ are the middle points of the respectively opposite sides.
+If $E$ be the spherical excess of the triangle, shew that
+\[
+  \cos\tfrac{1}{2}E
+= \frac{\cos A'B'}{\cos \frac{1}{2}c}
+= \frac{\cos B'C'}{\cos \frac{1}{2}a}
+= \frac{\cos C'A'}{\cos \frac{1}{2}b}\,.
+\]
+
+16. If one of the arcs of great circles which join the middle
+points of the sides of a spherical triangle be a quadrant, shew that
+the other two are also quadrants.
+%-----File: 085.png------------------------------------------------
+
+\chapter[On certain approximate Formul\ae.]{ON CERTAIN APPROXIMATE FORMUL\AE.}
+
+\paragraph{104.} We shall now investigate certain approximate formul\ae\
+which are often useful in calculating spherical triangles when the
+radius of the sphere is large compared with the lengths of the
+sides of the triangles.
+
+\paragraph{105.} \textit{Given two sides and the included angle of a spherical
+triangle, to find the angle between the chords of these sides.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/085fc}
+\end{figure}
+
+Let $AB$, $AC$ be the two sides of the triangle $ABC$; let $O$ be
+the centre of the sphere. Describe a sphere round $A$ as a centre,
+and suppose it to meet $AO$, $AB$, $AC$ at $D$, $E$, $F$ respectively.
+Then the angle $EDF$ is the inclination of the planes $OAB$, $OAC$,
+and is therefore equal to $A$. From the spherical triangle $DEF$\\[1ex]
+${} \hfill \cos EF=\cos DE \cos DF + \sin DE \sin DF \cos A; \hfill$\\[1ex]
+$\rlap{and }\hfill
+  DE = \frac{1}{2} (\pi - c),\quad
+  DF = \frac{1}{2} (\pi - b);\hfill$\\[1ex]
+$\rlap{therefore }\hfill
+  \cos EF = \sin \frac{1}{2} b \sin \frac{1}{2} c
+          + \cos \frac{1}{2} b \cos \frac{1}{2} c \cos A. \hfill$
+
+If the sides of the triangle are small compared with the
+radius of the sphere, $EF$ will not differ much from $A$; suppose
+$EF=A - \theta$, then approximately
+\[
+  \cos EF = \cos A + \theta \sin A;
+\]
+%-----File: 086.png------------------------------------------------
+\begin{flalign*}
+&\text{and}&\sin\tfrac12b \sin\tfrac12c &= \sin^2\tfrac14(b + c) - \sin^2\tfrac14(b - c),&\phantom{and}\\[1.5ex]
+&&\cos\tfrac12b \cos\tfrac12c &= \cos^2\tfrac14(b + c) - \sin^2\tfrac14(b - c);
+\end{flalign*}
+therefore
+\begin{multline*}
+\cos A + \theta\sin A = \sin^2\tfrac14(b + c) - \sin^2\tfrac14(b - c)\\
+{}+ \Bigl\{1 - \sin^2\tfrac14(b + c) - \sin^2\tfrac14(b - c)\Bigr\}\cos A;
+\end{multline*}
+therefore
+\[
+  \theta\sin A = (1 - \cos A)\sin^2\tfrac14(b + c)
+                 - (1 + \cos A)\sin^2\tfrac14(b - c),
+\]
+therefore \hfill$%
+  \theta = \tan\tfrac12A \sin^2\tfrac14(b + c)
+         - \cot\tfrac12A \sin^2\tfrac14(b - c).
+\hfill\phantom{therefore}$
+
+This gives the \textit{circular measure} of $\theta$; the number of seconds in
+the angle is found by dividing the circular measure by the circular
+measure of one second, or approximately by the sine of one second
+(\textit{Plane Trigonometry}, Art.\ 123). If the lengths of the arcs corresponding
+to $a$ and $b$ respectively be $\alpha$ and $\beta$, and $r$ the radius of the
+sphere, we have $\dfrac{\alpha}{r}$ and $\dfrac{\beta}{r}$ as the circular measures of $a$ and $b$
+respectively; and the lengths of the sides of the chordal triangle
+are $2r\sin\dfrac{\alpha}{2r}$ and $2r\sin\dfrac{\beta}{2r}$ respectively. Thus when the sides of
+the spherical triangle and the radius of the sphere are known, we
+can calculate the angles and sides of the chordal triangle.
+
+\paragraph{106.} Legendre's Theorem. \textit{If the sides of a spherical triangle
+be small compared with the radius of the sphere, then each angle
+of the spherical triangle exceeds by one third of the spherical excess
+the corresponding angle of the plane triangle, the sides of
+which are of the same length as the arcs of the spherical triangle.}
+
+Let $A$, $B$, $C$ be the angles of the spherical triangle; $a$, $b$, $c$
+the sides; $r$ the radius of the sphere; $\alpha$, $\beta$, $\gamma$ the lengths of the
+arcs which form the sides, so that $\dfrac{\alpha}{r}$, $\dfrac{\beta}{r}$, $\dfrac{\gamma}{r}$ are the circular
+measures of $a$, $b$, $c$ respectively. Then
+%-----File: 087.png------------------------------------------------
+\begin{flalign*}
+&& \cos A &= \frac{\cos a - \cos b\cos c}{\sin b \sin c}\,; &&\\
+&\text{now }& \cos a
+&= 1 - \frac{\alpha^2}{2r^2} + \frac{\alpha^4}{24r^4} - \ldots, &\phantom{text}&\\
+&& \sin a &=\frac{\alpha}{r} - \frac{\alpha^3}{6r^3} +\ldots.
+\end{flalign*}
+
+Similar expressions hold for $\cos b$ and $\sin b$, and for $\cos c$
+and $\sin c$ respectively. Hence, if we neglect powers of the circular
+measure above the \textit{fourth}, we have
+\begin{align*}
+\cos A
+&=\frac{1 -\dfrac{\alpha^2}{2r^2} +\dfrac{\alpha^4}{24r^4}
+ - \left( 1 -\dfrac{\beta^2}{2r^2} +\dfrac{\beta^4}{24r^4} \right)
+   \left( 1 -\dfrac{\gamma^2}{2r^2} +\dfrac{\gamma^4}{24r^4} \right)}
+        {\dfrac{\beta\gamma}{r^2}
+          \left( 1 - \dfrac{\beta^2}{6r^2} \right)
+          \left( 1 - \dfrac{\gamma^2}{6r^2} \right)}
+\\[1.5ex]
+&=\frac{\dfrac{1}{2r^2}(\beta^2 + \gamma^2 - \alpha^2)
+        + \dfrac{1}{24r^4}
+          (\alpha^4 - \beta^4 - \gamma^4 - 6\beta^2\gamma^2)}
+        {\dfrac{\beta\gamma}{r^2}
+          \left( 1 - \dfrac{\beta^2 + \gamma^2}{6r^2} \right)}
+\\[1.5ex]
+&=\frac{1}{2\beta\gamma}
+  \left\{\beta^2 + \gamma^2 - \alpha^2
+        + \dfrac{1}{12r^2}
+          (\alpha^2 - \beta^2 - \gamma^2 - 6\beta^2\gamma^2) \right\}
+  \left\{1 + \dfrac{\beta^2 + \gamma^2}{6r^2} \right\}
+\\[2ex]
+&=\dfrac{\beta^2 + \gamma^2 - \alpha^2}{2\beta\gamma}
+ +\dfrac{\alpha^4 + \beta^4 + \gamma^4
+        - 2\alpha^2\beta^2 - 2\beta^2\gamma^2 - 2\gamma^2\alpha^2}
+        {24\beta\gamma r^2}\,.
+\end{align*}
+
+Now let $A'$, $B'$, $C'$ be the angles of the plane triangle whose
+sides are $\alpha$, $\beta$, $\gamma$ respectively; then
+\begin{flalign*}
+&& \cos A' &= \frac{\beta^2 + \gamma^2 - \alpha^2}{2 \beta \gamma}\,,\\
+&\text{thus}&
+  \cos A &= \cos A' - \frac{\beta \gamma \sin^2 A'}{6r^2}\,.
+&\phantom{thus}&
+\end{flalign*}
+
+Suppose $A = A' + \theta$; then
+\[
+  \cos A = \cos A' - \theta \sin A' \text{ approximately};
+\]
+$\text{therefore} \hfill\displaystyle
+  \theta = \frac{\beta \gamma \sin A'}{6r^2}=\frac{S}{3r^2}\,,\hfill \phantom{therefore}$\\[1ex]
+%-----File: 088.png------------------------------------------------
+where $S$ denotes the area of the plane triangle whose sides are
+$\alpha$, $\beta$, $\gamma$. Similarly
+\[
+B = B' + \frac{S}{3r^2} \text{ and } C = C' + \frac{S}{3r^2}\,;
+\]
+hence approximately
+\[
+A+B+C = A'+B'+C'+\frac{S}{r^2} = \pi + \frac{S}{r^2}\,;
+\]
+therefore $\dfrac{S}{r^2}$ is approximately equal to the spherical excess of the
+spherical triangle, and thus the theorem is established.
+
+It will be seen that in the above approximation the area of
+the spherical triangle is considered equal to the area of the plane
+triangle which can be formed with sides of the same length.
+
+\paragraph{107.} Legendre's Theorem may be used for the approximate
+solution of spherical triangles in the following manner.
+
+(1) Suppose the three sides of a spherical triangle known;
+then the values of $\alpha$, $\beta$, $\gamma$ are known, and by the formul\ae\ of
+Plane Trigonometry we can calculate $S$ and $A'$, $B'$, $C'$; then
+$A$, $B$, $C$ are known from the formul\ae.
+\[
+A = A' + \frac{S}{3r^2}, \quad
+B = B' + \frac{S}{3r^2}, \quad
+C = C' + \frac{S}{3r^2}.
+\]
+
+(2) Suppose two sides and the included angle of a spherical
+triangle known, for example $A$, $b$, $c$. Then
+\[
+S = \tfrac{1}{2}\beta\gamma\sin A' = \tfrac{1}{2}\beta\gamma\sin A \text{ approximately.}
+\]
+Then $A'$ is known from the formula $A'=A-\dfrac{S}{3r^2}$. Thus in the
+plane triangle two sides and the included angle are known;
+therefore its remaining parts can be calculated, and then those
+of the spherical triangle become known.
+%-----File: 089.png------------------------------------------------
+
+(3) Suppose two sides and the angle opposite to one of them
+in a spherical triangle known, for example $A$, $a$, $b$. Then
+\[
+\sin B' = \frac{\beta}{\alpha}\sin A' = \frac{\beta}{\alpha}\sin A \text{ approximately;}
+\]
+and $C'=\pi-A'-B' = \pi-A-B'$ approximately; then $S=\frac{1}{2}\alpha\beta\sin C'$.
+Hence $A'$ is known and the plane triangle can be solved, since two
+sides and the angle opposite to one of them are known.
+
+(4) Suppose two angles and the included side of a spherical
+triangle known, for example $A$, $B$, $c$.
+\[
+\text{Then } S=\frac{\gamma^2 \sin A' \sin B'}{2\sin(A'+B')}
+= \frac{\gamma^2\sin A \sin B}{2\sin(A+B)} \text{ nearly.}
+\]
+Hence in the plane triangle two angles and the included side are
+known.
+
+(5) Suppose two angles and the side opposite to one of them
+in a spherical triangle known, for example $A$, $B$, $a$. Then
+\begin{gather*}
+C'=\pi-A'-B'=\pi-A-B,\text{ approximately, and} \\[1ex]
+S=\frac{\alpha^2\sin B' \sin C'}{2\sin(B'+C')}\,,
+\end{gather*}
+which can be calculated, since $B'$ and $C'$ are approximately
+known.
+
+\paragraph{108.} The importance of Legendre's Theorem in the application
+of Spherical Trigonometry to the measurement of the Earth's
+surface has given rise to various developments of it which enable
+us to test the degree of exactness of the approximation. We shall
+finish the present Chapter with some of these developments, which
+will serve as exercises for the student. We have seen that approximately
+the spherical excess is equal to $\dfrac{S}{r^2}$, and we shall
+begin with investigating a closer approximate formula for the
+spherical excess.
+%-----File: 090.png------------------------------------------------
+
+\paragraph{109.} \textit{To find an approximate value of the spherical excess.}
+
+Let $E$ denote the spherical excess; then
+\[
+\sin\frac{1}{2}E = \frac{\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin C}
+                         {\cos\tfrac{1}{2}c}\,;
+\]
+therefore approximately
+\begin{flalign*}
+&& \sin\tfrac{1}{2}E
+&= \sin C \frac{\alpha\beta}{4r^2}
+   \left(1-\frac{\alpha^2}{24r^2} \right)
+   \left(1-\frac{\beta^2}{24r^2} \right)
+   \left(1-\frac{\gamma^2}{8r^2} \right)^{-1} &&
+\\[2ex]
+&&\phantom{therefore }
+&= \sin C \frac{\alpha\beta}{4r^2}
+   \left( 1 + \frac{3\gamma^2-\alpha^2-\beta^2}{24r^2} \right);
+&&\\[2ex]
+&\rlap{therefore}&
+  E &= \sin C \frac{\alpha\beta}{2r^2}
+   \left( 1+\frac{3\gamma^2-\alpha^2-\beta^2}{24r^2} \right), \tag{1}
+&&\\[2ex]
+&\rlap{and}&
+  \sin C
+&= \sin \left( C' + \tfrac{1}{3} E \vphantom{\frac{\beta^2}{r^2}} \right)
+ = \sin C' + \tfrac{1}{3} E\cos C' &&
+\\[2ex]
+&&\multispan{2}{$\hfill\displaystyle
+   = \sin C' + \frac{\sin C' \cos C'}{3} \frac{\alpha\beta}{2r^2}
+   = \sin C' \left( 1+\frac{\alpha^2+\beta^2 -\gamma^2}{12r^2} \right).
+\hfill$} \tag{2}
+\end{flalign*}
+
+From (1) and (2)
+\[
+  E = \sin C' \frac{\alpha\beta}{2r^2}
+      \left( 1 + \frac{\alpha^2+\beta^2+\gamma^2}{24r^2} \right).
+\]
+
+Hence to this order of approximation the area of the spherical
+triangle exceeds that of the plane triangle by the fraction
+$\dfrac{\alpha^2+\beta^2+\gamma^2}{24r^2}$ of the latter.
+
+\paragraph{110.} \textit{To find an approximate value of $\dfrac{\sin A}{\sin B}$.}
+\[
+  \frac{\operatorname{Sin} A}{\operatorname{Sin} B}
+= \frac{\sin a}{\sin b}\,;
+\]
+hence approximately
+$\displaystyle
+\frac{\sin A}{\sin B} =
+  \frac{\alpha\left(
+   1 - \dfrac{\alpha^2}{6r^2} + \dfrac{\alpha^4}{120r^4} \right)}
+       {\beta \left(
+   1 - \dfrac{\beta^2}{6r^2} + \dfrac{\beta^4}{120r^4} \right)}
+$
+%-----File: 091.png------------------------------------------------
+\begin{align*}
+&=\frac{\alpha}{\beta}
+  \left( 1 - \frac{\alpha^2}{6r^2} + \frac{\alpha^4}{120r^4}
+           + \frac{\beta^2}{6r^2} - \frac{\alpha^2\beta^2}{36r^4}
+           - \frac{\beta^4}{120r^4} + \frac{\beta^4}{36r^4} \right)
+\\[2ex]
+&=\frac{\alpha}{\beta}
+  \left\{1 + \frac{\beta^2-\alpha^2}{6r^2}
+            + \frac{\alpha^4-\beta^4}{120r^4}
+            + \frac{\beta^2 (\beta^2-\alpha^2)}{36r^4} \right\}
+\\[2ex]
+&=\frac{\alpha}{\beta}
+   \left\{1 + \frac{\beta^2 - \alpha^2}{6r^2}
+               \left( 1 + \frac{\beta^2}{6r^2}
+                   - \frac{\alpha^2 + \beta^2}{20r^2} \right) \right\}
+\\[2ex]
+&=\frac{\alpha}{\beta}
+  \left\{1 + \frac{\beta^2 - \alpha^2}{6r^2}
+            \left(1 + \frac{7\beta^2-3\alpha^2}{60r^2}\right)\right\}.
+\end{align*}
+
+\paragraph{111.} \textit{To express $\cot B-\cot A$ approximately.}
+\[
+  \operatorname{Cot}B-\cot A
+= \frac{1}{\sin B} (\cos B - \frac{\sin B}{\sin A}\cos A);
+\]
+hence, approximately, by Art.\ 110,
+\[
+  \cot B - \cot A
+= \frac{1}{\sin B}
+  (\cos B - \frac{\beta}{\alpha}\cos A
+          - \frac{\beta}{\alpha} \frac{\alpha^2-\beta^2}{6r^2}\cos A).
+\]
+
+Now we have shewn in Art.\ 106, that approximately
+\begin{flalign*}
+\multispan{6}{\hfil$
+  \cos A
+= \dfrac{\beta^2 + \gamma^2 - \alpha^2}{2\beta\gamma}
++ \dfrac{\alpha^4 + \beta^4 + \gamma^4
+        -2\alpha^2\beta^2 -2\beta^2\gamma^2 - 2\gamma^2\alpha^2}
+        {24\beta\gamma r^2},
+$\hfil}\\[1ex]
+&\text{therefore}&
+  \cos B - \frac{\beta}{\alpha}\cos A
+&= \frac{\alpha^2-\beta^2}{\alpha\gamma}\text{ approximately,}
+&\phantom{therefore}
+\\[1ex]
+&\text{and}&
+  \cot B - \cot A
+&=\frac{\alpha^2-\beta^2}{\alpha\gamma\sin B}
+ -\frac{\alpha^2-\beta^2}{\alpha\gamma\sin B}
+  \frac{\beta^2+\gamma^2-\alpha^2}{12r^2}
+\\[1ex]
+&&&=\frac{\alpha^2-\beta^2}{\alpha\gamma\sin B}
+    \left( 1 - \frac{\beta^2+\gamma^2-\alpha^2}{12r^2} \right).
+\end{flalign*}
+
+\paragraph{112.} The approximations in Arts.\ 109 and 110 are true so
+far as terms involving $r^4$; that in Art.\ 111 is true so far as
+terms involving $r^2$, and it will be seen that we are thus able
+to carry the approximations in the following Article so far as
+terms involving $r^4$.
+%-----File: 092.png------------------------------------------------
+
+\paragraph{113.} \textit{To find an approximate value of the error in the length
+of a side of a spherical triangle when calculated by Legendre's
+Theorem.}
+
+Suppose the side $\beta$ known and the side $\alpha$ required; let $3\mu$ denote
+the spherical excess which is adopted. Then the approximate
+value $\dfrac{\beta\sin(A-\mu)}{\sin(B-\mu)}$ is taken for the side of which $\alpha$ is the real
+value. Let $x=\alpha-\dfrac{\beta(A-\mu)}{\sin(B-\mu)}$; we have then to find $x$ approximately.
+Now approximately
+\begin{gather*}
+\frac{\sin (A-\mu)}{\sin(B-\mu)}=\dfrac{\sin A-\mu\cos A-\dfrac{\mu^2}{2}\sin A}
+{\sin B-\mu\cos B-\dfrac{\mu^2}{2}\sin B}\\[2ex]
+\begin{aligned}
+&=\frac{\sin A}{\sin B}\left(1-\mu\cot A-\frac{\mu^2}{2}\right)
+\left(1-\mu\cot B-\frac{\mu^2}{2}\right)^{-1}\\
+&=\frac{\sin A}{\sin B}\left\{1+\mu (\cot B-\cot A)+
+\mu^2\cot B(\cot B-\cot A)\right\}\\
+&=\frac{\sin A}{\sin B}+\frac{\mu\sin A}{\sin B}
+(\cot B-\cot A)(1+\mu\cot B).
+\end{aligned}
+\end{gather*}
+
+Also the following formul\ae\ are true so far as terms involving $r^2:$
+\begin{gather*}
+\frac{\sin A}{\sin B}=\frac{\alpha}{\beta}
+\left(1+\frac{\beta^2-\alpha^2}{6r^2}\right),\\[2ex]
+\cot B-\cot A=\frac{\alpha^2-\beta^2}{\alpha\gamma\sin B}
+\left(1-\frac{\beta^2+\gamma^2-\alpha^2}{12r^2}\right),\\[2ex]
+1+\mu\cot B=1+\frac{\alpha^2+\gamma^2-\beta^2}{12r^2}.
+\end{gather*}
+
+Hence, approximately,
+\[
+\frac{\sin A}{\sin B}(\cot B-\cot A)(1+\mu\cot B)=\frac{\alpha^2-\beta^2}
+{\beta\gamma\sin B}.
+\]
+%-----File: 093.png------------------------------------------------
+Therefore \hfil$
+  x = \alpha - \dfrac{\beta\sin A}{\sin B}
+    - \dfrac{\mu (\alpha^2-\beta^2) }{\gamma\sin B}
+$\hfil\phantom{Therefore}\\[1ex]
+\[
+= \frac{\alpha (\beta^2-\alpha^2) }{6}
+  \left\{\frac{6\mu}{\alpha\gamma\sin B}
+- \dfrac1{r^2} + \frac{3\alpha^2-7\beta^2}{60r^4} \right\},
+\text{ by Art.\ 110.}
+\]
+
+If we calculate $\mu$ from the formula $\mu=\dfrac{\alpha\gamma\sin B}{6r^2}$ we obtain
+\[
+x=\frac{\alpha(\beta^2-\alpha^2)(3\alpha^2-7\beta^2)}{360r^4}\,.
+\]
+
+If we calculate $\mu$ from an equation corresponding to (1) of
+Art.\ 109, we have
+\begin{flalign*}
+&&\mu &= \frac{\alpha\gamma\sin B}{6r^2}
+         \left( 1 + \frac{3\beta^2-\alpha^2-\gamma^2}{24r^2} \right);
+\\[1.5ex]
+&\text{therefore}& x
+&= \frac{\alpha (\beta^2-\alpha^2) (\alpha^2 + \beta^2 - 5\gamma^2)}
+        {720r^4}\,.
+&\phantom{therefore}
+\end{flalign*}
+
+\section*{\centering\normalsize MISCELLANEOUS EXAMPLES.}
+
+1. If the sides of a spherical triangle $AB$, $AC$ be produced to
+$B'$, $C'$, so that $BB'$, $CC'$ are the semi-supplements of $AB$, $AC$
+respectively, shew that the arc $B'C'$ will subtend an angle at the
+centre of the sphere equal to the angle between the chords of $AB$
+and $AC$.
+\medskip
+
+2. Deduce Legendre's Theorem from the formula
+\[
+\tan^2\frac{A}{2}
+= \frac{\sin\tfrac12(a+b-c) \sin\tfrac12(c+a-b) }
+       {\sin\tfrac12(b+c-a) \sin\tfrac12(a+b+c) }\,.
+\]
+
+3. Four points $A$, $B$, $C$, $D$ on the surface of a sphere are
+joined by arcs of great circles, and $E$, $F$ are the middle points
+of the arcs $AC$, $BD$: shew that
+\[
+\cos AB + \cos BC + \cos CD + \cos DA = 4 \cos AE \cos BF \cos FE.
+\]
+
+4. If a quadrilateral $ABCD$ be inscribed in a small circle on
+a sphere so that two opposite angles $A$ and $C$ may be at opposite
+extremities of a diameter, the sum of the cosines of the sides is
+constant.
+%-----File: 094.png------------------------------------------------
+\medskip
+
+5. In a spherical triangle if $A = B = 2C$, shew that
+\[
+\cos a\cos\frac{a}{2} = \cos\left( c+\frac{a}{2}\right).
+\]
+
+6. $ABC$ is a spherical triangle each of whose sides is a quadrant;
+$P$ is any point within the triangle: shew that
+\[
+\cos PA \cos PB \cos PC + \cot BPC \cot CPA \cot APB = 0,
+\]
+and \hfill$
+\tan ABP \tan BCP \tan CAP = 1.
+$\hfill\phantom{and}
+\medskip
+
+7. If $O$ be the middle point of an equilateral triangle $ABC$,
+and $P$ any point on the surface of the sphere, then
+\begin{gather*}
+\tfrac{1}{4} (\tan PO \tan OA)^2 (\cos PA + \cos PB + \cos PC)^2 = \\
+\cos^2 PA + \cos^2 PB + \cos^2 PC - \cos PA \cos PB - \cos PB \cos PC - \cos PC \cos PA.
+\end{gather*}
+
+8. If $ABC$ be a triangle having each side a quadrant, $O$ the
+pole of the inscribed circle, $P$ any point on the sphere, then
+\[
+(\cos PA + \cos PB + \cos PC)^2 = 3\cos^2 PO.
+\]
+
+9. From each of three points on the surface of a sphere arcs
+are drawn on the surface to three other points situated on a great
+circle of the sphere, and their cosines are $a$, $b$, $c$; $a'$, $b'$, $c'$; $a''$, $b''$, $c''$.
+Shew that $ab''c' + a'bc'' + a''b'c = ab'c'' + a'b''c + a''bc'$.
+\medskip
+
+10. From Arts.~110 and 111, shew that approximately
+\[
+\log\beta = \log\alpha + \log\sin B - \log\sin A + \frac{S}{3r^2}(\cot A-\cot B).
+\]
+
+11. By continuing the approximation in Art.~106 so as to
+include the terms involving $r^4$, shew that approximately
+\[
+\cos A = \cos A' - \frac{\beta\gamma\sin^2 A'}{6r^2}
+ + \frac{\beta\gamma(\alpha^2-3\beta^2-3\gamma^2)\sin^2 A'}{180r^4}\,.
+\]
+
+12. From the preceding result shew that if $A = A' + \theta$ then
+approximately
+\[
+\theta = \frac{\beta\gamma\sin A'}{6r^2}
+  \left( 1+\frac{7\beta^2 + 7\gamma^2 + \alpha^2}{120 r^2} \right)\,.
+\]
+%-----File: 095.png------------------------------------------------
+
+\chapter[Geodetical Operations.]{GEODETICAL OPERATIONS.}
+
+\paragraph{114.} One of the most important applications of Trigonometry,
+both Plane and Spherical, is to the determination of the
+figure and dimensions of the Earth itself, and of any portion of its
+surface. We shall give a brief outline of the subject, and for
+further information refer to Woodhouse's \textit{Trigonometry}, to the
+article \textit{Geodesy} in the \textit{English Cyclop\ae dia}, and to Airy's treatise
+on the \textit{Figure of the Earth} in the \textit{Encyclop\ae dia Metropolitana}.
+For practical knowledge of the details of the operations it will
+be necessary to study some of the published accounts of the great
+surveys which have been effected in different parts of the world,
+as for example, the \textit{Account of the measurement of two sections of
+the Meridional arc of India}, by Lieut.-Colonel Everest, 1847; or
+the \textit{Account of the Observations and Calculations of the Principal
+Triangulation in the Ordnance Survey of Great Britain
+and Ireland}, 1858.
+
+\paragraph{115.} An important part of any survey consists in the measurement
+of a horizontal line, which is called a \textit{base}. A level plain
+of a few miles in length is selected and a line is measured on it with
+every precaution to ensure accuracy. Rods of deal, and of metal,
+hollow tubes of glass, and steel chains, have been used in different
+surveys; the temperature is carefully observed during the operations,
+and allowance is made for the varying lengths of the rods
+or chains, which arise from variations in the temperature.
+
+\paragraph{116.} At various points of the country suitable stations are
+selected and signals erected; then by supposing lines to be drawn
+connecting the signals, the country is divided into a series of
+triangles. The angles of these triangles are observed, that is, the
+angles which any two signals subtend at a third. For example,
+suppose $A$ and $B$ to denote the extremities of the \textit{base}, and $C$ a
+%-----File: 096.png------------------------------------------------
+signal at a third point visible from $A$ and $B$; then in the triangle
+$ABC$ the angles $ABC$ and $BAC$ are observed, and then $AC$ and $BC$
+can be calculated. Again, let $D$ be a signal at a fourth point,
+such that it is visible from $C$ and $A$; then the angles $ACD$ and
+$CAD$ are observed, and as $AC$ is known, $CD$ and $AD$ can be
+calculated.
+
+\paragraph{117.} Besides the original \textit{base} other lines are measured in
+convenient parts of the country surveyed, and their measured lengths
+are compared with their lengths obtained by calculation through a
+series of triangles from the original base. The degree of closeness
+with which the measured length agrees with the calculated
+length is a test of the accuracy of the survey. During the progress
+of the Ordnance Survey of Great Britain and Ireland, several
+lines have been measured; the last two are, one near Lough
+Foyle in Ireland, which was measured in 1827 and 1828, and one
+on Salisbury Plain, which was measured in 1849. The line near
+Lough Foyle is nearly 8 miles long, and the line on Salisbury
+Plain is nearly 7 miles long; and the difference between the length
+of the line on Salisbury Plain as measured and as calculated from
+the Lough Foyle base is less than 5 inches (\textit{An Account of the
+Observations~\ldots} page~419).
+
+\paragraph{118.} There are different methods of effecting the calculations
+for determining the lengths of the sides of all the triangles in the
+survey. One method is to use the exact formul\ae\ of Spherical
+Trigonometry. The radius of the Earth may be considered known
+very approximately; let this radius be denoted by $r$, then if $\alpha$ be
+the length of any arc the circular measure of the angle which the
+arc subtends at the centre of the earth is $\dfrac{\alpha}{r}$. The formul\ae\ of
+Spherical Trigonometry gives expressions for the trigonometrical
+functions of $\dfrac{\alpha}{r}$, so that $\dfrac{\alpha}{r}$ may be found and then $\alpha$. Since in
+practice $\dfrac{\alpha}{r}$ is always very small, it becomes necessary to pay
+%-----File: 097.png------------------------------------------------
+attention to the methods of securing accuracy in calculations
+which involve the logarithmic trigonometrical functions of small
+angles (\textit{Plane Trigonometry}, Art.~205).
+
+Instead of the exact calculation of the triangles by Spherical
+Trigonometry, various methods of approximation have been proposed;
+only two of these methods however have been much used.
+One method of approximation consists in deducing from the angles
+of the spherical triangles the angles of the \textit{chordal triangles}, and
+then computing the latter triangles by Plane Trigonometry (see
+Art.\ 105). The other method of approximation consists in the
+use of Legendre's Theorem (see Art.~106).
+
+\paragraph{119.} The three methods which we have indicated were all
+used by Delambre in calculating the triangles in the French
+survey (\textit{Base du Syst\`eme M\'etrique}, Tome~\textsc{iii}.\ page~7). In the
+earlier operations of the Trigonometrical survey of Great Britain
+and Ireland, the triangles were calculated by the chord method;
+but this has been for many years discontinued, and in place of it
+Legendre's Theorem has been universally adopted (\textit{An Account
+of the Observations~\ldots} page~244). The triangles in the Indian
+Survey are stated by Lieut.-Colonel Everest to be computed on
+Legendre's Theorem. (\textit{An Account of the Measurement~\ldots} page
+\textsc{clviii}.)
+
+\paragraph{120.} If the three angles of a plane triangle be observed, the
+fact that their sum ought to be equal to two right angles affords a
+test of the accuracy with which the observations are made. We
+shall proceed to shew how a test of the accuracy of observations of
+the angles of a spherical triangle formed on the Earth's surface
+may be obtained by means of the \textit{spherical excess}.
+
+\paragraph{121.} \textit{The area of a spherical triangle formed on the Earth's
+surface being known in square feet, it is required to establish a rule
+for computing the spherical excess in seconds.}
+
+Let $n$ be the number of seconds in the spherical excess, $s$ the
+number of square feet in the area of the triangle, $r$ the number of
+%-----File: 098.png------------------------------------------------
+feet in the radius of the Earth. Then if $E$ be the circular measure
+of the spherical excess,
+\[
+s=Er^2,
+\]
+and $\hfill\displaystyle
+E=\frac{n\pi}{180\centerdot 60\centerdot 60}
+= \frac{n}{206265}\ \text{ approximately;}
+\hfill$\phantom{and }\\[1.5ex]
+therefore $\hfill\displaystyle
+s=\frac{nr^2}{206265}\,.
+\hfill$\phantom{therefore }\\[1ex]
+
+Now by actual measurement the mean length of a degree on
+the Earth's surface is found to be 365155 feet; thus
+\[
+\frac{\pi r}{180}=365155.
+\]
+
+With the value of $r$ obtained from this equation it is found by
+logarithmic calculation, that
+\[
+\log n = \log s - 9.326774.
+\]
+Hence $n$ is known when $s$ is known.
+
+This formula is called General Roy's rule, as it was used by
+him in the Trigonometrical survey of Great Britain and Ireland.
+Mr Davies, however, claims it for Mr Dalby. (See Hutton's
+\textit{Course of Mathematics}, by Davies, Vol.~\textsc{ii}.\ p.~47.)
+
+\paragraph{122.} In order to apply General Roy's rule, we must know
+the area of the spherical triangle. Now the area is not known
+\textit{exactly} unless the elements of the spherical triangle are known
+\textit{exactly}; but it is found that in such cases as occur in practice an
+approximate value of the area is sufficient. Suppose, for example,
+that we use the area of the \textit{plane triangle} considered in Legendre's
+Theorem, instead of the area of the \textit{Spherical Triangle itself;}
+then it appears from Art.\ 109, that the error is approximately
+denoted by the fraction $\dfrac{\alpha^2+\beta^2+\gamma^2}{24r^2}$ of the former area, and this
+fraction is less than $.0001$, if the sides do not exceed 100 miles
+in length. Or again, suppose we want to estimate the influence
+of errors in the angles on the calculation of the area; let the
+%-----File: 099.png------------------------------------------------
+circular measure of an error be $h$, so that instead of $\dfrac{\alpha\beta\sin C}{2}$
+we ought to use $\dfrac{\alpha\beta\sin(C+h)}{2}$; the error then bears to the area
+approximately the ratio expressed by $h\cot C$. Now in modern
+observations $h$ will not exceed the circular measure of a few
+seconds, so that, if $C$ be not very small, $h\cot C$ is practically insensible.
+
+\paragraph{123.} The following example was selected by Woodhouse from
+the triangles of the English survey, and has been adopted by other
+writers. The observed angles of a triangle being respectively
+$42^\circ\, 2'\, 32''$, $67^\circ\, 55'\, 39''$, $70^\circ\, 1'\, 48''$, the sum of the errors made
+in the observations is required, supposing the side opposite to the
+angle $A$ to be $27404.2$ feet. The area is calculated from the expression
+$\dfrac{a^2\sin B\sin C}{2\sin A}$, and by General Roy's rule it is found
+that $n=.23$. Now the sum of the observed angles is $180^\circ-1''$,
+and as it ought to have been $180^\circ +.23''$, it follows that the sum
+of the errors of the observations is $1''.23$. This total error may
+be distributed among the observed angles in such proportion as
+the opinion of the observer may suggest; one way is to increase
+each of the observed angles by one-third of $1''.23$, and take the
+angles thus corrected for the true angles.
+
+\paragraph{124.} An investigation has been made with respect to the
+form of a triangle, in which errors in the observations of the
+angles will exercise the least influence on the lengths of the sides,
+and although the reasoning is allowed to be vague it may be
+deserving of the attention of the student. Suppose the three
+angles of a triangle observed, and one side, as $a$, known, it is
+required to find the form of the triangle in order that the other
+sides may be least affected by errors in the observations. The
+spherical excess of the triangle may be supposed known with
+sufficient accuracy for practice, and if the sum of the observed
+angles does not exceed two right angles by the proper spherical
+excess, let these angles be altered by adding the same quantity to
+%-----File: 100.png------------------------------------------------
+each, so as to make their sum correct. Let $A$, $B$, $C$ be the angles
+thus furnished by observation and altered if necessary; and let
+$\delta A$, $\delta B$ and $\delta C$ denote the respective errors of $A$, $B$ and $C$. Then
+$\delta A + \delta B + \delta C = 0$, because by supposition the sum of $A$, $B$ and $C$
+is correct. Considering the triangle as approximately plane, the
+true value of the side $c$ is $\dfrac{a \sin (C + \delta C)}{\sin (A + \delta A)}$, that is, $\dfrac{a \sin (C + \delta C)}{\sin (A - \delta B - \delta C)}$.
+Now approximately
+\begin{align*}
+& \sin (C + \delta C) = \sin C + \delta C \cos C,\quad
+\text{(\textit{Plane Trig.}\ Chap.~\textsc{xii.})},
+\\
+& \sin (A - \delta B - \delta C) = \sin A - (\delta B + \delta C) \cos A.
+\end{align*}
+
+Hence approximately
+\begin{align*}
+c &= \frac{a \sin C}{\sin A}
+     \Bigl\{1 + \delta C \cot C \Bigr\}
+     \Bigl\{1 -(\delta B + \delta C) \cot A \Bigr\}^{-1}
+\\[1.5ex]
+  &= \frac{a \sin C}{\sin A}
+     \Bigl\{1 + \delta B \cot A + \delta C (\cot C + \cot A) \Bigr\};
+\end{align*}
+and $\cot C + \cot A = \dfrac{\sin (A + C)}{\sin A \sin C} = \dfrac{\sin B}{\sin A \sin C}$ approximately.\\[1ex]
+
+Hence the error of $c$ is approximately
+\[
+\frac{a \sin B}{\sin^2 A} \delta C + \frac{a \sin C \cos A}{\sin^2 A)} \delta B.
+\]
+
+Similarly the error of $b$ is approximately
+\[
+\frac{a \sin C}{\sin^2 A} \delta B + \frac{a \sin B \cos A}{\sin^2 A} \delta C.
+\]
+
+Now it is impossible to assign exactly the signs and magnitudes
+of the errors $\delta B$ and $\delta C$, so that the reasoning must be vague. It
+is obvious that to make the error small $\sin A$ must not be small.
+And as the sum of $\delta A$, $\delta B$ and $\delta C$ is zero, two of them must have
+the same sign, and the third the opposite sign; we may therefore
+consider that it is more probable than any two as $\delta B$ and $\delta C$ have
+different signs, than that they have the same sign.
+%-----File: 101.png------------------------------------------------
+
+If $\delta B$ and $\delta C$ have different signs the errors of $b$ and $c$ will
+be less when $\cos A$ is positive than when $\cos A$ is negative;
+$A$ therefore ought to be less than a right angle. And if $\delta B$ and
+$\delta C$ are probably not very different, $B$ and $C$ should be nearly
+equal. These conditions will be satisfied by a triangle differing
+not much from an equilateral triangle.
+
+If two angles only, $A$ and $B$, be observed, we obtain the same
+expressions as before for the errors in $b$ and $c$; but we have
+no reason for considering that $\delta B$ and $\delta C$ are of different signs
+rather than of the same sign. In this case then the supposition
+that $A$ is a right angle will probably make the errors smallest.
+
+\paragraph{125.} The preceding article is taken from the Treatise on
+Trigonometry in the \textit{Encyclop\ae dia Metropolitana}. The least
+satisfactory part is that in which it is considered that $\delta B$ and $\delta C$
+may be supposed nearly equal; for since $\delta A + \delta B + \delta C = 0$, if we
+suppose $\delta B$ and $\delta C$ nearly equal and of opposite signs, we do in
+effect suppose $\delta A = 0$ nearly; thus in observing three angles, we
+suppose that in one observation a certain error is made, in a
+second observation the same numerical error is made but with
+an opposite sign, and in the remaining observation no error is
+made.
+
+\paragraph{126.} We have hitherto proceeded on the supposition that the
+Earth is a sphere; it is however approximately a spheroid of small
+eccentricity. For the small corrections which must in consequence
+be introduced into the calculations we must refer to the works
+named in Art.\ 114. One of the results obtained is that the error
+caused by regarding the Earth as a sphere instead of a spheroid increases
+with the departure of the triangle from the well-conditioned
+or equilateral form (\textit{An Account of the Observations~\ldots} page~243).
+Under certain circumstances the spherical excess is the same on a
+spheroid as on a sphere (\textit{Figure of the Earth} in the \textit{Encyclop\ae dia
+Metropolitana}, pages 198 and 215).
+
+\paragraph{127.} In geodetical operations it is sometimes required to determine
+the horizontal angle between two points, which are at a
+%-----File: 102.png------------------------------------------------
+small angular distance from the horizon, the angle which the
+objects subtend being known, and also the angles of elevation
+or depression.
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/102fc}
+\end{figure}
+
+Suppose $OA$ and $OB$ the directions in which the two points
+are seen from $O$; and let the angle $AOB$ be observed. Let $OZ$ be
+the direction at right angles to the observer's horizon; describe
+a sphere round $O$ as a centre, and let vertical planes through $OA$
+and $OB$ meet the horizon at $OC$ and $OD$ respectively: then the
+angle $COD$ is required.
+
+Let $AOB = \theta$, $COD = \theta + x$, $AOC = h$, $BOD = k$;
+from the triangle $AZB$
+\[
+  \cos AZB=\frac{\cos \theta-\cos ZA \,\cos ZB}{\sin ZA \,\sin ZB}=
+  \frac{\cos \theta-\sin h \,\sin k}{\cos h \,\cos k}\,;
+\]
+and $\cos AZB=\cos COD=\cos (\theta + x)$; thus
+\[
+  \cos (\theta + x)=\frac{\cos \theta-\sin h \,\sin k}{\cos h \,\cos k}\,.
+\]
+
+This formula is exact; by approximation we obtain
+\[
+  \cos \theta-x \sin \theta=\frac{\cos \theta-hk}
+{1-\frac{1}{2}(h^2+k^2)} \,;
+\]
+%-----File: 103.png------------------------------------------------
+\begin{flalign*}
+&\text{therefore }&
+  x\sin \theta &= hk-\tfrac{1}{2}(h^2+k^2)\cos \theta, \text{ nearly},
+&\phantom{therefore}
+\\[1.5ex]
+&\text{and}&
+  x &= \frac{2hk - (h^2+k^2)
+                   (cos^2 \tfrac{1}{2}\theta-\sin^2 \tfrac{1}{2}\theta)}
+            {2 \sin \theta}
+\\[1.5ex]
+&&&= \tfrac{1}{4}(h + k)^2 \tan \tfrac{1}{2}\theta
+   - \tfrac{1}{4}(h - k)^2 \cot \tfrac{1}{2}\theta.
+\end{flalign*}
+
+This process, by which we find the angle $COD$ from the angle
+$AOB$, is called \textit{reducing an angle to the horizon}.
+
+\chapter[On small variations in the parts of a Spherical Triangle.]{ON SMALL VARIATIONS IN THE PARTS OF A SPHERICAL TRIANGLE.}
+\chaptermark{ON SMALL VARIATIONS.}
+
+\paragraph{128.} It is sometimes important to know what amount of
+error will be introduced into one of the calculated parts of a
+triangle by reason of any small error which may exist in the
+given parts. We will here consider an example.
+
+\paragraph{129.} \textit{A side and the opposite angle of a spherical triangle
+remain constant: determine the connexion between the small variations
+of any other pair of elements}.
+
+Suppose $C$ and $c$ to remain constant.
+
+(1) Required the connexion between the small variations of
+the other sides. We suppose $a$ and $b$ to denote the sides of one
+triangle which can be formed with $C$ and $c$ as fixed elements, and
+$a + \delta a$ and $b + \delta b$ to denote the sides of another such triangle;
+then we require the ratio of $\delta a$ to $\delta b$ when both are extremely
+small. We have
+\begin{flalign*}
+&& \cos c = &\cos a \,\cos b + \sin a \,\sin b \,\cos C,
+\\
+&\text{and}
+& \cos c = &\cos (a + \delta a) \cos (b + \delta b)
+ + \sin (a + \delta a) \sin (b + \delta b) \cos C;&\phantom{\text{and}}
+\\
+&\text{also}
+&& \cos (a + \delta a) = \cos a - \sin a \,\delta a, \text{nearly},
+\\
+&\text{and}
+&& \sin (a + \delta a) = \sin a + \cos a \,\delta a, \text{nearly},
+\end{flalign*}
+%-----File: 104.png------------------------------------------------
+with similar formul\ae\ for $\cos (b + \delta b)$ and $\sin (b + \delta b)$. (See \textit{Plane
+Trigonometry}, Chap.~\textsc{xii}.) Thus
+\begin{multline*}
+\cos c = (\cos a - \sin a\,\delta a) (\cos b - \sin b\,\delta b) \\
+{}     + (\sin a + \cos a\,\delta a) (\sin b + \cos b\,\delta b) \cos C.
+\end{multline*}
+
+Hence by subtraction, if we neglect the product $\delta a$, $\delta b$,
+\begin{multline*}
+0 = \delta a (\sin a\,\cos b - \cos a\,\sin b\,\cos C) \\
+{}+ \delta b (\sin b\,\cos a - \cos b\,\sin a\,\cos C);
+\end{multline*}
+this gives the ratio of $\delta a$ to $\delta b$ in terms of $a$, $b$, $C$. We may
+express the ratio more simply in terms of $A$ and $B$; for, dividing
+by $\sin a \sin b$, we get from Art.~44,
+\[
+  \frac{\delta a}{\sin a} \cot B\,\sin C
++ \frac{\delta b}{\sin b} \cot A\,\sin C = 0;
+\]
+therefore
+$\hfill
+\delta a \,\cos B + \delta b \,\cos A = 0.
+\hfill\phantom{therefore}$
+
+(2) Required the connexion between the small variations of
+the other angles. In this case we may by means of the polar
+triangle deduce from the result just found, that
+\[
+  \delta A \,\cos b + \delta B \,\cos a = 0;
+\]
+this may also be found independently as before.
+
+(3) Required the connexion between the small variations of
+a side and the opposite angle $(A,\ a)$.
+\begin{flalign*}
+&\text{\indent Here }&
+  \sin A \sin c &= \sin C \,\sin a,
+\\
+&\text{and }&
+  \sin (A + \delta A) \sin c &= \sin C \,\sin (a + \delta a);
+\\
+\intertext{hence by subtraction }
+&& \cos A \,\sin c \,\delta A &= \sin C \,\cos a \,\delta a,
+\\
+&\text{and therefore }&
+  \delta A \cot A &= \delta a \cot a.
+&\phantom{and therefore }&
+\end{flalign*}
+
+(4) Required the connexion between the small variations of
+a side and the adjacent angle $(a,\ B)$.
+%-----File: 105.png------------------------------------------------
+
+We have $\hfill
+  \cot C \sin B = \cot c \sin a - \cos B \cos a;
+\hfill\phantom{\indent We have}$\\[1ex]
+proceeding as before we obtain
+\[
+  \cot C \cos B \delta B
+= \cot c \cos a \delta a
++ \cos B \sin a \delta a
++ \cos a \sin B \delta B;
+\]
+therefore
+\[
+  (\cot C \cos B - \cos a \sin B) \delta B
+= (\cot c \cos a + \cos B \sin a) \delta a;
+\]
+therefore $\hfill\displaystyle
+-\frac{\cos A}{\sin C} \delta B = \frac{\cos b}{\sin c} \delta a;
+\hfill\phantom{therefore}$\\[2ex]
+therefore $\hfill\displaystyle
+\delta B \cos A = - \delta a \cot b \sin B.
+\hfill\phantom{therefore}$
+
+\paragraph{130.} Some more examples are proposed for solution at the
+end of this Chapter; as they involve no difficulty they are left for
+the exercise of the student.
+
+\section*{\centering\normalsize EXAMPLES.}
+
+1. In a spherical triangle, if $C$ and $c$ remain constant while
+$a$ and $b$ receive the small increments $\delta a$ and $\delta b$ respectively, shew
+that
+\[
+  \frac{\delta a}{\surd{(1 - n^2 \sin^2 a)}} +
+  \frac{\delta b}{\surd{(1 - n^2 \sin^2 b)}} = 0 \text{ where }
+n = \frac{\sin C}{\sin c}\,.
+\]
+
+2. If $C$ and $c$ remain constant, and a small change be made
+in $a$, find the consequent changes in the other parts of the triangle.
+Find also the change in the area.
+\medskip
+
+3. Supposing $A$ and $c$ to remain constant, prove the following
+equations, connecting the small variations of pairs of the other
+elements:
+\begin{gather*}
+  \sin C \delta b =  \sin a \delta B,\quad
+  \delta b \sin C = -\delta C \tan a,\quad
+  \delta a \tan C =  \delta B \sin a,
+\\
+  \delta a \tan C = -\delta C \tan a,\quad
+  \delta b \cos C =  \delta a,\quad
+  \delta B \cos a = -\delta C.
+\end{gather*}
+
+4. Supposing $b$ and $c$ to remain constant, prove the following
+equations connecting the small variations of pairs of the other
+elements:
+\begin{align*}
+& \delta B \tan C =  \delta C \tan B, &
+& \delta a \cot C = -\delta B \sin a,
+\\
+& \delta a = \delta A \sin c \sin B,  &
+& \delta A \sin B \cos C = -\delta B \sin A.
+\end{align*}
+%-----File: 106.png------------------------------------------------
+
+5. Supposing $B$ and $C$ to remain constant, prove the following
+equations connecting the small variations of pairs of the
+other elements:
+\begin{align*}
+  \delta b \tan c &= \delta c \tan b, &\quad
+  \delta A \cot c &= \delta b \sin A,
+\\
+  \delta A &= \delta a \sin b \sin C, &\quad
+  \delta a \sin B \cos c &= \delta b \sin A.
+\end{align*}
+
+6. If $A$ and $C$ are constant, and $b$ be increased by a small
+quantity, shew that $a$ will be increased or diminished according as
+$c$ is less or greater than a quadrant.
+
+\chapter[On the connexion of Formul\ae\ in Plane and Spherical Trigonometry.]{ON THE CONNEXION OF FORMUL\AE\ IN PLANE AND SPHERICAL TRIGONOMETRY.}
+\chaptermark{CONNEXION OF FORMUL\AE\ IN TRIGONOMETRY.}
+
+\paragraph{131.} The student must have perceived that many of the
+results obtained in \textit{Spherical} Trigonometry resemble others with
+which he is familiar in \textit{Plane} Trigonometry. We shall now pay
+some attention to this resemblance. We shall first shew how we
+may deduce formul\ae\ in Plane Trigonometry from formul\ae\ in
+Spherical Trigonometry; and we shall then investigate some
+theorems in Spherical Trigonometry which are interesting principally
+on account of their connexion with known results in Plane
+Geometry and Trigonometry.
+
+\paragraph{132.} \textit{From any formula in Spherical Trigonometry involving
+the elements of a triangle, one of them being a side, it is required
+to deduce the corresponding formula in Plane Trigonometry.}
+
+Let $\alpha$, $\beta$, $\gamma$ be the lengths of the sides of the triangle, $r$ the
+radius of the sphere, so that $\dfrac{\alpha}{r}$, $\dfrac{\beta}{r}$,
+$\dfrac{\gamma}{r}$ are the circular measures
+of the sides of the triangle; expand the functions of
+$\dfrac{\alpha}{r}$, $\dfrac{\beta}{r}$, $\dfrac{\gamma}{r}$
+which occur in any proposed formula in powers of
+$\dfrac{\alpha}{r}$, $\dfrac{\beta}{r}$, $\dfrac{\gamma}{r}$
+respectively; then if we suppose $r$ to become indefinitely great,
+%-----File: 107.png------------------------------------------------
+the limiting form of the proposed formula will be a relation in
+Plane Trigonometry.
+
+For example, in Art.~106, from the formula
+\[
+\cos A = \frac{\cos a - \cos b \cos c }{\sin b \sin c}
+\]
+we deduce
+\[
+\cos A = \frac{\beta^2 + \gamma^2 - \alpha^2}{2 \beta \gamma}
+  + \frac{\alpha^4 + \beta^4 + \gamma^4 - 2 \alpha^2 \beta^2 - 2 \beta^2 \gamma^2 - 2 \gamma^2 \alpha^2}{24 \beta \gamma r^2} + \ldots;
+\]
+now suppose $r$ to become infinite; then ultimately
+\[
+\cos A = \frac{\beta^2 + \gamma^2 - \alpha^2}{2 \beta \gamma}\,;
+\]
+and this is the expression for the cosine of the angle of a plane
+triangle in terms of the sides.
+
+Again, in Art.~110, from the formula
+\begin{flalign*}
+&&& \frac{\sin A}{\sin B} = \frac{\sin a}{\sin b} &&
+\\[1ex]
+&\text{we deduce }&&
+\frac{\sin A}{\sin B} = \frac{\alpha}{\beta} + \frac{\alpha (\beta^2 - \alpha^2)}{6 \beta r^2} + \ldots;
+\\
+\intertext{now suppose $r$ to become infinite; then ultimately}
+&&& \frac{\sin A}{\sin B} = \frac{\alpha}{\beta},
+\end{flalign*}
+that is, in a plane triangle the sides are as the sines of the opposite
+angles.
+
+\paragraph{133.} \textit{To find the equation to a small circle of the sphere.}
+
+The student can easily draw the required diagram.
+
+Let $O$ be the pole of a small circle, $S$ a fixed point on the
+sphere, $SX$ a fixed great circle of the sphere. Let $OS = \alpha$,
+$OSX = \beta$; then the position of $O$ is determined by means of these
+angular co-ordinates $\alpha$ and $\beta$. Let $P$ be any point on the circumference
+of the small circle, $PS = \theta$, $PSX = \phi$, so that $\theta$ and $\phi$ are
+%-----File: 108.png------------------------------------------------
+the angular co-ordinates of $P$. Let $OP = r$. Then from the
+triangle $OSP$
+\[
+\cos r = \cos \alpha \cos \theta + \sin \alpha \sin \theta \cos (\phi - \beta); \tag{1}
+\]
+this gives a relation between the angular co-ordinates of any point
+on the circumference of the circle.
+
+If the circle be a great circle then $r = \dfrac{\pi}{2}$; thus the equation
+becomes
+\[
+0 = \cos \alpha \cos \theta + \sin \alpha \sin \theta \cos (\phi - \beta). \tag{2}
+\]
+
+It will be observed that the angular co-ordinates here used are
+analogous to the \textit{latitude} and \textit{longitude} which serve to determine
+the positions of places on the Earth's surface; $\theta$ is the \textit{complement
+of the latitude} and $\phi$ is the \textit{longitude}.
+
+\paragraph{134.} Equation (1) of the preceding Article may be written
+thus:
+\begin{multline*}
+\cos r \left(\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2}\right)
+\\
+= \cos \alpha \left(\cos^2 \frac{\theta}{2}
+                  - \sin^2 \frac{\theta}{2}\right)
++ 2 \sin \alpha \sin \frac{\theta}{2} \cos \frac{\theta}{2}
+  \cos (\phi - \beta).
+\end{multline*}
+
+Divide by $\cos^2 \dfrac{\theta}{2}$ and rearrange; hence
+\[
+  \tan^2 \frac{\theta}{2} (\cos r + \cos \alpha)
+- 2 \tan \frac{\theta}{2} \sin \alpha \cos (\phi - \beta)
++ \cos r - \cos \alpha = 0.
+\]
+
+Let $\tan \dfrac{\theta_1}{2}$ and $\tan \dfrac{\theta_2}{2}$ denote the values of $\tan \dfrac{\theta}{2}$ found from
+this quadratic equation; then by \textit{Algebra}, Chapter \textsc{xxii.}
+\[
+  \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}
+= \frac{\cos r - \cos \alpha}{\cos r + \cos \alpha}
+= \tan \frac{\alpha + r}{2} \tan \frac{\alpha - r}{2}.
+\]
+
+Thus the value of the product $\tan \dfrac{\theta_{1}}{2} \tan \dfrac{\theta_{2}}{2}$ is \textit{independent} of $\phi$;
+this result corresponds to the well-known property of a circle in
+Plane Geometry which is demonstrated in Euclid \textsc{iii.}\ 36 \textit{Corollary}.
+
+\paragraph{135.} Let three arcs $OA$, $OB$, $OC$ meet at a point. From any
+point $P$ in $OB$ draw $PM$ perpendicular to $OA$, and $PN$ perpendicular
+to $OC$. The student can easily draw the required diagram.
+%-----File: 109.png------------------------------------------------
+
+Then, by Art.\ 65,
+\[
+\sin PM = \sin OP \sin AOB,\quad \sin PN = \sin OP \sin COB;
+\]
+therefore
+$\hfill\displaystyle
+\frac{\sin PM}{\sin PN} = \frac{\sin AOB}{\sin COB}.
+\hfill\phantom{therefore}$\\
+
+Thus the ratio of $\sin PM$ to $\sin PN$ is independent of the position
+of $P$ on the arc $OB$.
+
+\paragraph{136.} Conversely suppose that from any other point $p$ arcs $pm$
+and $pn$ are drawn perpendicular to $OA$ and $OC$ respectively; then if
+\[
+\frac{\sin pm}{\sin pn} = \frac{\sin PM}{\sin PN},
+\]
+it will follow that $p$ is on the same great circle as $O$ and $P$.
+
+\paragraph{137.} From two points $P_1$ and $P_2$ arcs are drawn perpendicular
+to a fixed arc; and from a point $P$ on the same great circle
+as $P_1$ and $P_2$ a perpendicular is drawn to the same fixed arc. Let
+$PP_1 = \theta_1$ and $PP_2 = \theta_2$; and let the perpendiculars drawn from $P$,
+$P_1$, and $P_2$ be denoted by $x$, $x_1$ and $x_2$. Then will
+\[
+\sin x
+= \frac{\sin \theta_2}{\sin(\theta_1 + \theta_2)} \sin x_1
++ \frac{\sin \theta_1}{\sin(\theta_1 + \theta_2)} \sin x_2.
+\]
+
+Let the arc $P_1 P_2$, produced if necessary, cut the fixed arc at a
+point $O$; let $\alpha$ denote the angle between the arcs. We will suppose
+that $P_1$ is between $O$ and $P_2$, and that $P$ is between $P_1$ and $P_2$.
+
+Then, by Art.~65,
+\begin{align*}
+\sin x_1 = \sin \alpha \sin OP_1 &= \sin \alpha \sin (OP - \theta_1) \\
+    &= \sin \alpha (\sin OP \cos \theta_1 - \cos OP \sin \theta_1);\\
+\sin x_2 = \sin \alpha \sin OP_2 &= \sin \alpha \sin (OP + \theta_2) \\
+    &= \sin \alpha (\sin OP \cos \theta_2 + \cos OP \sin \theta_2).
+\end{align*}
+Multiply the former by $\sin\theta_2$, and the latter by $\sin\theta_1$, and add;
+thus
+\begin{align*}
+\sin \theta_2 \sin x_1 + \sin \theta_1 \sin x_2 &= \sin (\theta_1 + \theta_2) \sin \alpha \sin OP\\
+     &= \sin (\theta_1 + \theta_2) \sin x.
+\end{align*}
+%-----File: 110.png------------------------------------------------
+
+The student should convince himself by examination that the
+result holds for all relative positions of $P$, $P_1$ and $P_2$, when due
+regard is paid to algebraical signs.
+
+\paragraph{138.} The principal use of Art.\ 137 is to determine whether
+three given points are on the same great circle; an illustration
+will be given in Art.\ 146.
+
+\paragraph{139.} \textit{The arcs drawn from the angles of a spherical triangle
+perpendicular to the opposite sides respectively meet at a point.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/110fc}
+\end{figure}
+
+Let $CF$ be perpendicular to $AB$. From $F$ suppose arcs drawn
+perpendicular to $CB$ and $CA$ respectively; denote the former by
+$\xi$ and the latter by $\eta$. Then, by Art.\ 135,
+\[
+\frac{\sin\xi}{\sin\eta} = \frac{\sin FCB}{\sin FCA}.
+\]
+
+But, by Art.\ 65,
+\[
+  \cos B = \cos CF \sin FCB, \quad \cos A = \cos CF \sin FCA;
+\]
+therefore $\hfill\displaystyle
+  \frac{\sin\xi}{\sin\eta}
+= \frac{\cos B}{\cos A}
+= \frac{\cos B \cos C}{\cos A \cos C}. \hfill\phantom{therefore}$\\[2ex]
+And if from \textit{any} point in $CF$ arcs are drawn perpendicular to
+$CB$ and $CA$ respectively, the ratio of the sine of the former perpendicular
+to the sine of the latter perpendicular is equal to $\dfrac{\sin\xi}{\sin\eta}$
+by Art.\ 135.
+%-----File: 111.png------------------------------------------------
+
+In like manner suppose $AD$ perpendicular to $BC$; then if from
+any point in $AD$ arcs are drawn perpendicular to $AC$ and $AB$
+respectively, the ratio of the sine of the former perpendicular to
+the sine of the latter perpendicular is equal to $\dfrac{\cos A \cos C}{\cos A \cos B}$.\\[1ex]
+
+Let $CF$ and $AD$ meet at $P$, and from $P$ let perpendiculars be
+drawn on the sides $a$, $b$, $c$ of the triangle; and denote these perpendiculars
+by $x$, $y$, $z$ respectively: then we have shewn that
+\begin{flalign*}
+&&\frac{\sin x}{\sin y} &= \frac{\cos B \cos C}{\cos A \cos C},\\[1ex]
+&\text{and that}& \frac{\sin y}{\sin z} &= \frac{\cos A \cos C}{\cos A \cos B};&\phantom{and that}\\
+\intertext{hence it follows that}
+&&\frac{\sin x}{\sin z} &= \frac{\cos B \cos C}{\cos B \cos A},
+\end{flalign*}
+and this shews that the point $P$ is on the arc drawn from $B$ perpendicular
+to $AC$.
+
+Thus the three perpendiculars meet at a point, and this point
+is determined by the relations
+\[
+\frac{\sin x}{\cos B \cos C} = \frac{\sin y}{\cos C \cos A} = \frac{\sin z}{\cos A \cos B}.
+\]
+
+\paragraph{140.} In the same manner it may be shewn that the arcs
+drawn from the angles of a spherical triangle to the middle points
+of the opposite sides meet at a point; and if from this point arcs
+$x$, $y$, $z$ are drawn perpendicular to the sides $a$, $b$, $c$ respectively,
+\[
+\frac{\sin x}{\sin B \sin C} = \frac{\sin y}{\sin C \sin A} = \frac{\sin z}{\sin A \sin B}.
+\]
+
+\paragraph{141.} It is known in Plane Geometry that a certain circle
+touches the inscribed and escribed circles of any triangle; this
+circle is called the \textit{Nine points circle}: see \textit{Appendix to Euclid},
+pages 317, 318, and \textit{Plane Trigonometry}, Chapter \textsc{xxiv}.
+%-----File: 112.png------------------------------------------------
+
+We shall now shew that a small circle can always be determined
+on the sphere to touch the inscribed and escribed circles of
+any spherical triangle.
+
+\paragraph{142.} Let $\alpha$ denote the distance from $A$ of the pole of the
+small circle inscribed within a spherical triangle $ABC$. Suppose
+that a small circle of angular radius $\rho$ touches this inscribed circle
+internally; let $\beta$ be the distance from $A$ of the pole of this touching
+circle; let $\gamma$ be the angle between arcs drawn from $A$ to the
+pole of the inscribed circle and the pole of the touching circle
+respectively. Then we must have
+\[
+\cos(\rho-r) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\cos\gamma. \tag{1}
+\]
+Suppose that this touching circle also touches externally the
+escribed circle of angular radius $r_1$; then if $\alpha_1$ denote the distance
+from $A$ of the pole of this escribed circle, we must have
+\[
+\cos(\rho+r_1) = \cos\alpha_1\cos\beta + \sin\alpha_1\sin\beta\cos\gamma. \tag{2}
+\]
+
+Similarly, if $\alpha_2$ and $\alpha_3$ denote the distances from $A$ of the poles
+of the other escribed circles, in order that the touching circle may
+touch these escribed circles externally, we must also have
+\begin{gather*}
+\cos(\rho+r_2) = \cos\alpha_2\cos\beta + \sin\alpha_2\sin\beta \cos\left(\frac{\pi}{2}-\gamma\right), \tag{3}\\[1.5ex]
+\cos(\rho+r_3) = \cos\alpha_3\cos\beta + \sin\alpha_3\sin\beta \cos\left(\frac{\pi}{2}+\gamma\right). \tag{4}
+\end{gather*}
+
+We shall shew that real values of $\rho$, $\beta$, and $\gamma$ can be found to
+satisfy these four equations.
+
+Eliminate $\cos \gamma$ from (1) and (2); thus
+\begin{multline*}
+\cos\rho(\cos r\sin\alpha_1-\cos r_1\sin\alpha) +
+\sin\rho(\sin r\sin\alpha_1+\sin r_1\sin\alpha) \\
+= \cos\beta (\cos\alpha \sin\alpha_1 - \cos\alpha_1 \sin\alpha). \tag{5}
+\end{multline*}
+
+Suppose that the inscribed circle touches $AB$ at the distance $m$
+from $A$, and that the escribed circle of angular radius $r_1$ touches
+$AB$ at the distance $m_1$ from $A$. Then, by Art.~65,
+%-----File: 113.png------------------------------------------------
+\[
+  \cot \alpha = \cot m \cos \frac{A}{2}, \quad
+  \cos \alpha = \cos r\cos m, \quad
+  \sin r = \sin \alpha \sin\frac{A}{2};
+\]
+therefore $\hfill\displaystyle
+  \frac{\cos r}{\sin \alpha}
+= \frac{\cot \alpha}{\cos m} =
+\frac{1}{\sin m}\cos\frac{A}{2}.
+\hfill\phantom{therefore }$\\
+
+Similarly we may connect $\alpha_1$ and $r_1$ with $m_1$. Thus we
+obtain from (5)
+\begin{multline*}
+\cos\rho \cos\frac{A}{2}
+\left(\frac{1}{\sin m}-\frac{1}{\sin m_1}\right) +
+2\sin\rho\sin\frac{A}{2}
+\\
+= \cos\beta\cos\frac{A}{2}
+\left(\cot m-\cot m_1\right);
+\end{multline*}
+therefore \ $
+   \cos\rho(\sin m_1-\sin m)
++ 2\sin\rho\sin m\sin m_1\tan\dfrac{A}{2} \hfill$ \\
+\rightline{$ = \cos\beta\sin(m_1-m)$.}
+
+But by Arts.~89 and 90 we have $m = s-a$, and $m_1 = s$; therefore
+by the aid of Art.~45 we obtain
+\[
+\tag{6}
+2\cos\rho \sin\frac{a}{2}\cos\frac{b+c}{2} +
+2n\sin\rho=\cos\beta\sin a ,
+\]
+where $n$ has the meaning assigned in Art.~46.
+
+In like manner if we eliminate $\sin \gamma$ between (3) and (4),
+putting $m_2$ for $s-c$, and $m_3$ for $s-b$, we obtain
+\begin{multline*}
+  \cos\rho(\sin m_2+\sin m_3)
+-2\sin\rho\sin m_2\sin m_3\cot\frac{A}{2}
+\\
+=\cos\beta\sin(m_2+m_3),
+\end{multline*}
+\begin{flalign*}
+\tag{7}
+\text{therefore}\quad&
+2\cos\rho\sin\frac{a}{2}\cos\frac{b-c}{2}-2n\sin\rho=\cos\beta\sin a. &&
+\end{flalign*}
+
+From (6) and (7) we get
+\[
+\tag{8}
+\tan\rho=
+\frac{\sin\dfrac{a}{2}\sin\dfrac{b}{2}\sin\dfrac{c}{2}}{n} =
+\frac{1}{2}\tan R,\ \text{by Art.~92}
+\]
+and $\hfill
+\cos\beta=
+\dfrac{\cos\dfrac{b}{2}\cos\dfrac{c}{2}\cos\rho}{\cos\dfrac{a}{2}}. \hfill\phantom{and } (9)$\\[2ex]
+%-----File: 114.png------------------------------------------------
+
+We may suppose that $\cos\dfrac{a}{2}$ is not less than $\cos\dfrac{b}{2}$ or $\cos\dfrac{c}{2}$, so
+that we are sure of a possible value of $\cos\beta$ from (9).
+
+It remains to shew that when $\rho$ and $\beta$ are thus determined, all
+the four fundamental equations are satisfied.
+
+It will be observed that, $\rho$ and $\beta$ being considered known,
+$\cos\gamma$ can be found from (1) or (2), and $\sin\gamma$ can be found from
+(3) or (4): we must therefore shew that (1) and (2) give the
+\textit{same} value for $\cos\gamma$, and that (3) and (4) give the \textit{same} value
+for $\sin\gamma$; and we must also shew that these values satisfy the
+condition $\cos^2\gamma+\sin^2\gamma=1$.
+
+From (1) we have
+\[
+\frac{\cos\rho\sin r}{\sin\alpha}\left(\cot r+\tan\rho-\cos m\cot r\frac{\cos\beta}{\cos\rho}\right)=\sin\beta\cos\gamma,
+\]
+that is,
+\begin{gather*}
+\frac{\cos\rho\sin\dfrac{A}{2}}{n}
+ \left\{\sin s+\sin\tfrac12a\sin\tfrac12b\sin\tfrac12c
+  - \frac{\cos(s-a)\sin s\cos\dfrac{b}{2}\cos\dfrac{c}{2}}{\cos\dfrac{a}{2}}\right\}\\
+=\sin\beta\cos\gamma;\\
+\intertext{this reduces to}
+\frac{\cos\rho\sin\dfrac{A}{2}}{n}
+ \left\{\cos\frac{a}{2}\sin\frac{b+c}{2}
+  - \frac{\sin(b+c)\cos\dfrac{b}{2}\cos\dfrac{c}{2}}{2\cos\dfrac{a}{2}} \right\}
+  =\sin\beta\cos\gamma:
+\end{gather*}
+and it will be found that (2) reduces to the same; so that (1) and
+(2) give the same value for $\cos\gamma$.
+
+In like manner it will be found that (3) and (4) agree in
+reducing to
+\[
+\frac{\cos\rho\cos\dfrac{A}{2}}{n}
+ \left\{\cos\dfrac{a}{2}\sin\frac{c-b}{2}
+  - \frac{\sin(c-b)\cos\dfrac{b}{2}\cos\dfrac{c}{2}}{2\cos\dfrac{a}{2}} \right\}
+  =\sin\beta\cos\gamma.
+\]
+%-----File: 115.png------------------------------------------------
+
+It only remains to shew that the condition $\cos^2\gamma+\sin^2\gamma=1$ is
+satisfied.
+\[
+\text{Put $k$ for $\frac{\cos\beta}{\cos\rho}$, that is for $\frac{\cos\dfrac b2 \cos\dfrac c2}{\cos\dfrac a2}$;}
+\]
+put $X$ for $\cot r\{1-k\cos(s-a)\}$, and $Y$ for $\cot r_1\{1-k\cos s\}$.
+
+Then (1) and (2) may be written respectively thus:
+\begin{align*}
+(X\cos\rho+\sin\rho)\sin\frac A2 &=\sin\beta\cos\gamma,\tag{10}\\
+(Y\cos\rho-\sin\rho)\sin\frac A2 &=\sin\beta\cos\gamma.\tag{11}
+\end{align*}
+From (10) and (11) by addition
+\begin{flalign*}
+&&\multispan{2}{\hfil$(X+Y)\sin\dfrac A2\cos\rho = 2\sin\beta\cos\gamma$;\hfil}&&\\
+&\text{therefore}&
+  4\sin^2\beta\cos^2\gamma&=(X^2+Y^2+2XY)\sin^2\frac A2\cos^2\rho.
+&\phantom{therefore}\tag{12}
+\end{flalign*}
+
+But from (10) and (11) by subtraction
+\begin{flalign*}
+&&\multispan{2}{\hfil$(X-Y)\cos\rho=-2\sin\rho$;\hfil}&&\\
+&\text{therefore}&
+  (X^2+Y^2)\cos^2\rho&=4\sin^2\rho+2XY\cos^2\rho.&\phantom{therefore}
+\end{flalign*}
+
+Substitute in (12) and we obtain
+\[
+\sin^2\beta\cos^2\gamma=(\sin^2\rho+XY\cos^2\rho)\sin^2\frac A2.\tag{13}
+\]
+
+Again, put
+\[
+\text{$X_1$ for $\cot r_2\{1-k\cos(s-c)\}$, and $Y_1$ for $\cot r_3\{1-k\cos(s-b)\}$.}
+\]
+
+Then (3) and (4) may be written respectively thus:
+\begin{align*}
+(X_1\cos\rho-\sin\rho)\cos\frac A2 &= \sin\beta\sin\gamma,\tag{14}\\
+(Y_1\cos\rho-\sin\rho)\cos\frac A2 &= -\sin\beta\sin\gamma.\tag{15}
+\end{align*}
+
+From (14) and (15) by subtraction
+\[
+(X_1-Y_1)\cos\frac A2\cos\rho = 2\sin\beta\sin\gamma,
+\]
+%-----File: 116.png------------------------------------------------
+and from (14) and (15) by addition,
+\[
+  (X_1+Y_1)\cos\rho = 2\sin\rho,
+\]
+whence
+\[
+  \sin^2 \beta \sin^2 \gamma = (\sin^2 \rho - X_1 Y_1 \cos^2\rho) \cos^2 \frac{A}{2}. \tag{16}
+\]
+
+Hence from (13) and (16) it follows that we have to establish
+the relation
+\[
+  \sin^2 \beta = \sin^2\rho + \left( XY\sin^2\frac{A}{2} - X_1 Y_1 \cos^2 \frac{A}{2} \right) \cos^2\rho.
+\]
+
+But $\sin^2\beta = 1-\cos^2\beta = \sin^2\rho + \cos^2 \rho - k^2\cos^2\rho$,
+so that the relation
+reduces to
+\[
+  1-k^2 = XY\sin^2\frac{A}{2} - X_1 Y_1 \cos^2 \frac{A}{2}.
+\]
+
+Now
+\begin{align*}
+XY\sin^2\frac{A}{2}
+&= \frac{\cot r \cot r_1 \{1-k\cos s\} \{1-k\cos(s-a)\} \sin(s-b)\sin(s-c)}{\sin b \sin c} \\[1.5ex]
+&= \frac{\{1-k\cos s\} \{1-k\cos(s-a) \} }{\sin b \sin c }.
+\end{align*}
+
+Similarly \ $
+X_1 Y_1 \cos^2 \dfrac{A}{2} = \dfrac{\{1-k\cos(s-b) \} \{1-k\cos(s-c) \} }{\sin b \sin c}$.\\[2ex]
+
+Subtract the latter from the former; then we obtain
+\begin{gather*}
+\frac{k}{\sin b \sin c} \{\cos(s-b) + \cos(s-c) -\cos s -\cos(s-a) \} \\[1.5ex]
+{}+ \frac{k^2}{\sin b \sin c} \{\cos s \cos(s-a) - \cos(s-b)\cos(s-c) \},
+\end{gather*}
+that is $\hfill
+  \dfrac{2k\cos\dfrac{a}{2} }{\sin b\sin c}
+  \left\{\cos\dfrac{b-c}{2} -\cos\dfrac{b+c}{2} \right\}
+\hfill\phantom{that is}$\\[1ex]
+\[
+{}+\frac{k^2}{\sin b \sin c}
+   \left\{\cos\frac{b+c+a}{2} \cos\frac{b+c-a}{2}
+         - \cos\frac{a+c-b}{2} \cos\frac{a+b-c}{2} \right\}
+\]
+that is\hfill
+$\displaystyle \dfrac{4\sin\dfrac{b}{2} \sin\dfrac{c}{2}
+          \cos\dfrac{b}{2} \cos\dfrac{c}{2} }
+       {\sin b \sin c}
++ \dfrac{k^2}{\sin b\sin c}
+  \left\{\sin^2\dfrac{c-b}{2}
+        - \sin^2\dfrac{c+b}{2} \right\},$\hfill\phantom{\text{that is}}\\[1ex]
+that is $1-k^2$; which was to be shewn.
+%-----File: 117.png------------------------------------------------
+
+\paragraph{143.} Thus the existence of a circle which touches the inscribed
+and escribed circles of any spherical triangle has been
+established.
+
+The distance of the pole of this touching circle from the
+angles $B$ and $C$ of the triangle will of course be determined by
+formul\ae\ corresponding to (9); and thus it follows that
+\[
+\frac{\cos\dfrac{a}{2} \cos\dfrac{c}{2} \cos\rho }{\cos\dfrac{b}{2} } \text{ and }
+\frac{\cos\dfrac{a}{2} \cos\dfrac{b}{2} \cos\rho }{\cos\dfrac{c}{2} },
+\]
+must both be less than unity.
+
+\paragraph{144.} Since the circle which has been determined touches
+the inscribed circle internally and touches the escribed circles
+externally, it is obvious that it must meet all the sides of the
+spherical triangle. We will now determine the position of the
+points of meeting.
+
+Suppose the touching circle intersects the side $AB$ at points
+distant $\lambda$ and $\mu$ respectively from $A$.
+
+Then by Art.~134 we have
+\[
+\tan\frac{\lambda}{2} \tan\frac{\mu}{2} = \frac{\cos\rho - \cos\beta}{\cos\rho + \cos\beta}
+= \frac{\cos\dfrac{a}{2} - \cos\dfrac{b}{2} \cos\dfrac{c}{2} }{\cos\dfrac{a}{2} + \cos\dfrac{b}{2} \cos\dfrac{c}{2} }. \tag{1}
+\]
+
+In the same way we must have by symmetry
+\[
+\tan\frac{c - \lambda}{2} \tan\frac{c - \mu}{2}
+= \frac{\cos\dfrac{b}{2} - \cos\dfrac{a}{2} \cos\dfrac{c}{2} }{\cos\dfrac{b}{2} + \cos\dfrac{a}{2} \cos\dfrac{c}{2}}. \tag{2}
+\]
+
+From (2), when we substitute the value of $\tan\dfrac{\lambda}{2} \tan\dfrac{\mu}{2}$ given
+by (1), we obtain
+\begin{gather*}
+\tan\frac{\lambda}{2} + \tan\frac{\mu}{2} =
+\frac{\cos^2 \dfrac{a}{2} - \cos^2 \dfrac{b}{2} \cos^2 \dfrac{c}{2} + \cos^2 \dfrac{b}{2} \sin^2 \dfrac{c}{2} }{\cos \dfrac{b}{2} \sin \dfrac{c}{2} \left(\cos \dfrac{a}{2} + \cos \dfrac{b}{2} \cos \dfrac{c}{2}\right) }\\
+%-----File: 118.png------------------------------------------------
+= \frac{\cos \dfrac{a}{2} - \cos \dfrac{b}{2} \cos \dfrac{c}{2} }
+       {\cos \dfrac{b}{2}   \sin \dfrac{c}{2} }
++ \frac{\cos \dfrac{b}{2}   \sin \dfrac{c}{2} }
+       {\cos \dfrac{a}{2} + \cos \dfrac{b}{2} \cos \dfrac{c}{2} }. \tag{3}
+\end{gather*}
+
+From (1) and (3) we see that we may put
+\begin{align*}
+  \tan\frac{\lambda}{2}
+&= \frac{\cos \dfrac{a}{2} - \cos \dfrac{b}{2} \cos \dfrac{c}{2} }
+        {\cos \dfrac{b}{2}   \sin \dfrac{c}{2} }. \tag{4}
+\\[2ex]
+  \tan\frac{\mu}{2}
+&= \frac{\cos \dfrac{b}{2}   \sin \dfrac{c}{2} }
+        {\cos \dfrac{a}{2} + \cos \dfrac{b}{2} \cos \dfrac{c}{2} }. \tag{5}
+\end{align*}
+
+Similar formul\ae\ of course hold for the points of intersection of
+the touching circle with the other sides.
+
+\paragraph{145.} Let $z$ denote the perpendicular from the pole of the
+touching circle on $AB$; then
+\begin{align*}
+\sin z &= \sin \beta \sin \left(\frac{A}{2} + \gamma\right) \\[1.5ex]
+  &= \sin \beta \left(\sin \frac{A}{2} \cos \gamma + \cos \frac{A}{2} \sin \gamma\right).
+\end{align*}
+
+But from (2) and (3) of Art.~142 we have
+\[
+\sin \beta \cos \gamma = \frac{\cos \rho \sin \dfrac{A}{2} }{n}
+  \left(Z - \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2} \right),
+\]
+where\hfill
+$\displaystyle Z = \sin (s - a) - \cos s \sin (s - a) \cos \frac{b}{2} \cos \frac{c}{2} \sec \frac{a}{2},$\hfill\phantom{\text{where}}
+\begin{flalign*}
+&\text{and }&&
+\sin \beta \sin \gamma = \frac{\cos \rho \cos \dfrac{A}{2} }{n}
+  \left(Z_1 - \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}\right),\\[2ex]
+&\text{where}&&
+Z_1 = \sin (s - b) - \cos (s - c) \sin (s - b) \cos \frac{b}{2} \cos \frac{c}{2} \sec \frac{a}{2}.
+&\phantom{where}&
+\end{flalign*}
+%-----File: 119.png------------------------------------------------
+Therefore
+\[
+\sin z = \frac{\cos \rho}{n} \left\{Z \sin^2 \frac{A}{2} +
+Z_1 \cos^2 \frac{A}{2} -
+\sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}\right\}.
+\]
+
+Now $Z \sin^2 \dfrac{A}{2}$
+\[
+= \frac{\sin (s - a) \sin (s - b) \sin (s - c)}{\sin b \sin c}
+\left\{1 - \cos s \cos \frac{b}{2} \cos \frac{c}{2} \sec \frac{a}{2}\right\},
+\]
+and $Z_1 \cos^2 \dfrac{A}{2}$
+\[
+= \frac{\sin s \sin (s - a) \sin (s - b)}{\sin b \sin c}
+  \left\{1 - \cos (s - c) \cos \frac{b}{2} \cos \frac{c}{2}
+  \sec \frac{a}{2}\right\}.
+\]
+
+Therefore $\hfill\displaystyle
+Z \sin^2 \frac{A}{2} + Z_1 \cos^2 \frac{A}{2}
+\hfill\phantom{\indent Therefore}$\\[1ex]
+is equal to the product of
+\[
+\frac{\sin (s - a) \sin (s - b)}{\sin b \sin c}
+\]
+into
+\begin{align*}
+\multispan{2}{\hfill$\displaystyle
+  \sin (s - c) + \sin s - \cos \frac{b}{2} \cos \frac{c}{2} \sec \frac{a}{2}
+  \left\{\sin (s - c) \cos s + \cos (s - c) \sin s
+\vphantom{\frac b2} \right\} $\hfill}\\[2ex]
+& = \frac{\sin (s - a) \sin (s - b)}{\sin b \sin c}
+  \left\{2 \sin \frac{a + b}{2} \cos \frac{c}{2} -
+  \cos \frac{b}{2} \cos \frac{c}{2} \sec \frac{a}{2} \sin (2s - c)\right\} \\[2ex]
+& = \frac{\sin (s - a) \sin (s - b)}{2 \sin b \sin \dfrac{c}{2}}
+  \left\{2 \sin \frac{a + b}{2} -
+  \sin (a + b) \cos \frac{b}{2} \sec \frac{a}{2}\right\} \\[2ex]
+& = \frac{\sin (s - a) \sin (s - b) \sin \dfrac{a + b}{2}}
+  {\sin b \sin \dfrac{c}{2}}
+  \left\{1 - \frac{\cos \dfrac{a + b}{2} \cos \dfrac{b}{2}}{\cos \dfrac{a}{2}}
+  \right\} \\[2ex]
+& = \frac{\sin (s - a) \sin (s - b) \sin^2 \dfrac{a + b}{2} \sin \dfrac{b}{2}}
+  {\sin b \sin \dfrac{c}{2} \cos \dfrac{a}{2}} =
+  \frac{\sin (s - a) \sin (s - b) \sin^2 \dfrac{a + b}{2}}
+  {2 \cos \dfrac{a}{2} \cos \dfrac{b}{2} \sin \dfrac{c}{2}}.
+\end{align*}
+%-----File: 120.png------------------------------------------------
+
+Therefore
+\begin{gather*}
+\sin z = \frac{\cos\rho}{n} \sin\frac{a}{2} \sin\frac{b}{2} \sin\frac{c}{2} \left\{\frac{2 \sin^2\dfrac{a + b}{2}\sin(s - a)\sin(s - b)}{\sin^2\dfrac{c}{2}\sin a\sin b}- 1\right\}\\[2ex]
+= \frac{\cos\rho}{n} \sin\frac{a}{2} \sin\frac{b}{2} \sin\frac{c}{2} \left\{2 \cos^2\frac{A - B}{2} - 1\right\}\text{; by (2) of Art.\ 54.}
+\end{gather*}
+
+Thus $\hfill\displaystyle
+  \sin z = \frac{\cos\rho}{n} \sin\frac{a}{2} \sin\frac{b}{2} \sin\frac{c}{2} \cos(A - B) \hfill\phantom{\indent Thus}$
+\[
+  = \sin\rho \cos(A - B).
+\]
+
+Similar expressions hold for the perpendiculars from the pole
+of the touching circle on the other sides of the spherical triangle.
+
+\paragraph{146.} Let $P$ denote the point determined in Art.\ 139; $G$ the
+point determined in Art.\ 140, and $N$ the pole of the touching
+circle. We shall now shew that $P$, $G$, and $N$ are on a great
+circle.
+
+Let $x$, $y$, $z$ denote the perpendiculars from $N$ on the sides
+$a$, $b$, $c$ respectively of the spherical triangle; let $x_1$, $y_1$, $z_1$ denote
+the perpendiculars from $P$; and $x_2$, $y_2$, $z_2$ the perpendiculars from
+$G$. Then by Arts.\ 145,
+139, and 140 we have
+\begin{align*}
+\frac{\sin x}{\cos(B - C)} &= \frac{\sin y}{\cos(C - A)} = \frac{\sin z}{\cos(A - B)},\\[1.5ex]
+\frac{\sin x_1}{\cos B \cos C} &= \frac{\sin y_1}{\cos C \cos A} = \frac{\sin z_1}{\cos A \cos B},\\[1.5ex]
+\frac{\sin x_2}{\sin B \sin C} &= \frac{\sin y_2}{\sin C \sin A} = \frac{\sin z_2}{\sin A \sin B}.
+\end{align*}
+
+Hence it follows that
+\begin{align*}
+\sin x &= t_1 \sin x_1 + t_2 \sin x_2,\\
+\sin y &= t_1 \sin y_1 + t_2 \sin y_2,\\
+\sin z &= t_1 \sin z_1 + t_2 \sin z_2,
+\end{align*}
+%-----File: 121.png------------------------------------------------
+where $t_1$ and $t_2$ are certain quantities the values of which are not
+required for our purpose.
+
+Therefore by Art.\ 137 a certain point \textit{in the same great circle}
+as $P$ and $G$ is at the perpendicular distances $x$, $y$, $z$ from the sides
+$a$, $b$, $c$ respectively of the spherical triangle: and hence this point
+must be the point $N$.
+
+\paragraph{147.} The resemblance of the results which have been obtained
+to those which are known respecting the Nine points circle in
+Plane Geometry will be easily seen.
+
+The result $\tan\rho = \dfrac12 \tan R$ corresponds to the fact that the
+radius of the Nine points circle is half the radius of the circumscribing
+circle of the triangle.
+
+From equation (4) of Art.\ 144 by supposing the radius of the
+sphere to become infinite we obtain $\lambda = \dfrac{b^2 + c^2 - a^2}{2c}$: this corresponds
+to the fact that the Nine points circle passes through the feet of
+the perpendiculars from the angles of a triangle on the opposite
+sides.
+
+From equation (5) of Art.\ 144 by supposing the radius of the
+sphere to become infinite we obtain $\mu = \dfrac{c}{2}$: this corresponds to the
+fact that the Nine points circle passes through the middle points
+of the sides of a triangle.
+
+From Art.\ 145 by supposing the radius of the sphere to become
+infinite we obtain $z = \dfrac12 R \cos(A - B)$: this is a known
+property of the Nine points circle.
+
+In Plane Geometry the points which correspond to the $P$, $G$,
+and $N$ of Art.\ 146 are on a straight line.
+
+\paragraph{148.} The results which have been demonstrated with respect
+to the circle which touches the inscribed and escribed circles of a
+spherical triangle are mainly due to Dr Hart and Dr Salmon.
+See the \textit{Quarterly Journal of Mathematics}, Vol.~\textsc{vi}. page~67.
+%-----File: 122.png------------------------------------------------
+
+\section*{\centering\normalsize EXAMPLES.}
+
+1. From the formula $\sin\dfrac{a}{2}=\Surd{\left\{\dfrac{-\cos S\cos(S-A)}{\sin B\sin C}\right\}}$ deduce the expression for the area of a plane triangle, namely
+$\dfrac{a^2\sin B\sin C}{2\sin A}$, when the radius of the sphere is indefinitely increased.
+\medskip
+
+2. Two triangles $ABC$, $abc$, spherical or plane, equal in all
+respects, differ slightly in position: shew that
+\[
+\cos ABb\cos BCc\cos CAa+\cos ACc\cos CBb\cos BAa=0.
+\]
+
+3. Deduce formul\ae\ in Plane Trigonometry from Napier's
+Analogies.
+\medskip
+
+4. Deduce formul\ae\ in Plane Trigonometry from Delambre's
+Analogies.
+\medskip
+
+5. From the formula
+$\cos\dfrac{c}{2}\cos\dfrac{A+B}{2}
+=\sin\dfrac{C}{2}\cos\dfrac{a+b}{2}$ deduce
+the area of a plane triangle in terms of the sides and one of the
+angles.
+\medskip
+
+6. What result is obtained from Example 7 to Chapter VI.,
+by supposing the radius of the sphere infinite?
+\medskip
+
+7. From the angle $C$ of a spherical triangle a perpendicular is
+drawn to the arc which joins the middle points of the sides $a$ and
+$b$: shew that this perpendicular makes an angle $S-B$ with the
+side $a$, and an angle $S-A$ with the side $b$.
+\medskip
+
+8. From each angle of a spherical triangle a perpendicular is
+drawn to the arc which joins the middle points of the adjacent
+sides. Shew that these perpendiculars meet at a point; and that
+%-----File: 123.png------------------------------------------------
+if $x$, $y$, $z$ are the perpendiculars from this point on the sides $a$, $b$, $c$
+respectively,
+\[
+  \frac{\sin x}{\sin(S-B)\sin(S-C)}
+= \frac{\sin y}{\sin(S-C)\sin(S-A)}
+= \frac{\sin z}{\sin(S-A)\sin(S-B)}.
+\]
+
+9. Through each angle of a spherical triangle an arc is drawn
+so as to make the same angle with one side which the perpendicular
+on the base makes with the other side. Shew that these
+arcs meet at a point; and that if $x$, $y$, $z$ are the perpendiculars
+from this point on the sides $a$, $b$, $c$ respectively,
+\[
+\frac{\sin x}{\cos A}=\frac{\sin y}{\cos B}=\frac{\sin z}{\cos C}.
+\]
+
+10. Shew that the points determined in Examples 8 and 9,
+and the point $N$ of Art.\ 146 are on a great circle.
+
+State the corresponding theorem in Plane Geometry.
+\medskip
+
+11. If one angle of a spherical triangle remains constant while
+the adjacent sides are increased, shew that the area and the sum
+of the angles are increased.
+\medskip
+
+12. If the arcs bisecting two angles of a spherical triangle and
+terminated at the opposite sides are equal, the bisected angles will
+be equal provided their sum be less than $180^\circ$.
+
+[Let $BOD$ and $COE$ denote these two arcs which are given
+equal. If the angles $B$ and $C$ are not equal suppose $B$ the greater.
+Then $CD$ is greater than $BE$ by Art.\ 58. And as the angle $OBC$
+is greater than the angle $OCB$, therefore $OC$ is greater than $OB$;
+therefore $OD$ is greater than $OE$. Hence the angle $ODC$ is
+greater than the angle $OEB$, by Example 11. Then construct
+a spherical triangle $BCF$ on the other side of $BC$, equal to $CBE$.
+Since the angle $ODC$ is greater than the angle $OEB$, the angle
+$FDC$ is greater than the angle $DFC$; therefore $CD$ is less than
+$CF$, so that $CD$ is less than $BE$. See the corresponding problem
+in Plane Geometry in the \textit{Appendix to Euclid}, page 317.]
+%-----File: 124.png------------------------------------------------
+
+\chapter[Polyhedrons.]{POLYHEDRONS.}
+
+\paragraph{149.} A polyhedron is a solid bounded by any number of
+plane rectilineal figures which are called its faces. A polyhedron
+is said to be \textit{regular} when its faces are similar and equal regular
+polygons, and its solid angles equal to one another.
+
+\paragraph{150.} \textit{If $\mathrm{S}$ be the number of solid angles in any polyhedron,
+$\mathrm{F}$ the number of its faces, $\mathrm{E}$ the number of its edges, then
+$\mathrm{S+F=E+2}$.}
+
+Take any point within the polyhedron as centre, and describe
+a sphere of radius $r$, and draw straight lines from the centre
+to each of the angular points of the polyhedron; let the points
+at which these straight lines meet the surface of the sphere be
+joined by arcs of great circles, so that the surface of the sphere is
+divided into as many polygons as the polyhedron has faces.
+
+Let $s$ denote the sum of the angles of any one of these polygons,
+$m$ the number of its sides; then the area of the polygon is
+$r^2\{s-(m-2)\pi\}$ by Art.~99. The sum of the areas of all the
+polygons is the surface of the sphere, that is, $4\pi r^2$. Hence since
+the number of the polygons is $F$, we obtain
+\[
+\textstyle
+4\pi = \sum s - \pi\sum m + 2F\pi.
+\]
+
+Now $\sum s$ denotes the sum of all the angles of the polygons, and
+is therefore equal to $2\pi \times$ the number of solid angles, that is, to
+$2\pi S$; and $\sum m$ is equal to the number of all the sides of all the
+polygons, that is, to $2E$, since every edge gives rise to an arc
+which is common to two polygons. Therefore
+\[
+4\pi = 2\pi S-2\pi E+2F\pi;
+\]
+therefore $\hfill
+S+F=E+2.
+\hfill\phantom{therefore }$
+%-----File: 125.png------------------------------------------------
+
+\paragraph{151.} \textit{There can be only five regular polyhedrons.}
+
+Let $m$ be the number of sides in each face of a regular polyhedron,
+$n$ the number of plane angles in each solid angle; then
+the entire number of plane angles is expressed by $mF$, or by $nS$,
+or by $2E$; thus
+\[
+  mF=nS=2E, \text{ and } S+F=E+2;
+\]
+from these equations we obtain
+\[
+  S = \frac{4 m}{2(m+n)-mn}, \quad
+  E = \frac{2mn}{2(m+n)-mn}, \quad
+  F = \frac{4 n}{2(m+n)-mn}.
+\]
+
+These expressions must be positive integers, we must therefore
+have $2(m+n)$ greater than $mn$; therefore
+\[
+  \frac{1}{m}+\frac{1}{n} \text{ must be greater than } \frac{1}{2};
+\]
+but $n$ cannot be less than 3, so that $\dfrac{1}{n}$ cannot be greater than $\dfrac{1}{3}$,
+and therefore $\dfrac{1}{m}$ must be greater than $\dfrac{1}{6}$; and as $m$ must be an
+integer and cannot be less than 3, the only admissible values of $m$
+are 3, 4, 5. It will be found on trial that the only values of $m$
+and $n$ which satisfy all the necessary conditions are the following:
+each regular polyhedron derives its name from the number of its
+plane faces.
+
+\begin{center}
+\begin{tabular}{|c|c|c|c|c|l|}
+\hline
+$m$&$n$&$S$&$E$&$F$&\multicolumn{1}{c|}{Name of regular Polyhedron.}\\[1ex]
+\hline
+3&3&4&6&4&Tetrahedron or regular Pyramid.\\[1ex]
+4&3&8&12&6&Hexahedron or Cube.\\[1ex]
+3&4&6&12&8&Octahedron.\\[1ex]
+5&3&20&30&12&Dodecahedron.\\[1ex]
+3&5&12&30&20&Icosahedron.\\[1ex]
+\hline
+\end{tabular}
+\end{center}
+
+It will be seen that the demonstration establishes something
+more than the enunciation states; for it is not assumed that the
+faces are equilateral and equiangular and all equal. It is in fact
+%-----File: 126.png------------------------------------------------
+demonstrated that, \textit{there cannot be more than five solids each of
+which has all its faces with the same number of sides, and all its
+solid angles formed with the same number of plane angles.}
+
+\paragraph{152.} \textit{The sum of all the plane angles which form the solid
+angles of any polyhedron is $2(S-2)\pi$.}
+
+For if $m$ denote the number of sides in any face of the polyhedron,
+the sum of the interior angles of that face is $(m-2)\pi$
+by Euclid I.~32, Cor.~1. Hence the sum of all the interior angles
+of all the faces is $\sum(m-2)\pi$, that is $\sum m\pi-2F\pi$, that is
+$2(E-F)\pi$, that is $2(S-2)\pi$.
+
+\paragraph{153.} \textit{To find the inclination of two adjacent faces of a regular
+polyhedron.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/126fc}
+\end{figure}
+
+Let $AB$ be the edge common to the two adjacent faces, $C$ and
+$D$ the centres of the faces; bisect $AB$ at $E$, and join $CE$ and $DE$;
+$CE$ and $DE$ will be perpendicular to $AB$, and the angle $CED$ is
+the angle of inclination of the two adjacent faces; we shall denote
+it by $I$. In the plane containing $CE$ and $DE$ draw $CO$ and $DO$
+at right angles to $CE$ and $DE$ respectively, and meeting at $O$;
+about $O$ as centre describe a sphere meeting $OA$, $OC$, $OE$ at $a$, $c$, $e$
+respectively, so that $cae$ forms a spherical triangle. Since $AB$ is
+perpendicular to $CE$ and $DE$, it is perpendicular to the plane
+$CED$, therefore the plane $AOB$ which contains $AB$ is perpendicular
+to the plane $CED$; hence the angle $cea$ of the spherical triangle is
+a right angle. Let $m$ be the number of sides in each face of the
+polyhedron, $n$ the number of the plane angles which form each solid
+%-----File: 127.png------------------------------------------------
+angle. Then the angle $ace=ACE=\dfrac{2\pi}{2m}=\dfrac{\pi}{m}$; and the angle $cae$
+is half one of the $n$ equal angles formed on the sphere round $a$,
+that is, $cae=\dfrac{2\pi}{2n}=\dfrac{\pi}{n}$. From the right-angled triangle $cae$
+\[
+\cos{cae}=\cos{cOe}\sin{ace},
+\]
+\begin{flalign*}
+&\text{that is }&
+  \cos{\frac{\pi}{n}}
+&=\cos{\left( \frac{\pi}{2} - \frac{I}{2} \right)}
+  \sin{\frac{\pi}{m}}; &&
+\\[1ex]
+&\text{therefore }&
+  \sin{\frac{I}{2}}
+&=\frac{\cos \dfrac{\pi}{n}}
+       {\sin{\dfrac{\pi}{m}}}. &\phantom{therefore }&
+\end{flalign*}
+
+\paragraph{154.} \textit{To find the radii of the inscribed and circumscribed
+spheres of a regular polyhedron.}
+
+Let the edge $AB=a$, let $OC=r$ and $OA=R$, so that $r$ is
+the radius of the inscribed sphere, and $R$ is the radius of the
+circumscribed sphere. Then
+\begin{flalign*}
+&& CE &= AE\cot{ACE} = \frac{a}{2}\cot{\frac{\pi}{m}}, \\[1.5ex]
+&& r  &= CE\tan{CEO} = CE\tan{\frac{I}{2}}
+       =\frac{a}{2}\cot{\frac{\pi}{m}}\tan{\frac{I}{2}};
+\\[1.5ex]
+&\text{also }&
+   r  &= R\cos{aOc} = R\cot{eca}\cot{eac}
+       = R\cot{\frac{\pi}{m}} \cot{\frac{\pi}{n}};
+\\[1.5ex]
+&\text{therefore }&
+   R  &= r\tan{\frac{\pi}{m}} \tan{\frac{\pi}{n}}
+       = \frac{a}{2} \tan{\frac{I}{2}} \tan{\frac{\pi}{n}}.
+&\phantom{therefore }&
+\end{flalign*}
+
+\paragraph{155.} \textit{To find the surface and volume of a regular polyhedron.}
+
+The area of one face of the polyhedron is $\dfrac{ma^2}{4}\cot{\dfrac{\pi}{m}}$, and
+therefore the surface of the polyhedron is $\dfrac{mFa^2}{4}\cot{\dfrac{\pi}{m}}$.\\[.5ex]
+
+Also the volume of the pyramid which has one face of the
+polyhedron for base and $O$ for vertex is $\dfrac{r}{3}\centerdot\dfrac{ma^2}{4}\cot{\dfrac{\pi}{m}}$, and
+therefore the volume of the polyhedron is $\dfrac{mFra^2}{12}\cot{\dfrac{\pi}{m}}$.\\[.5ex]
+%-----File: 128.png------------------------------------------------
+
+\paragraph{156.} \textit{To find the volume of a parallelepiped in terms of its
+edges and their inclinations to one another.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/128fc}
+\end{figure}
+
+Let the edges be $OA=a$, $OB=b$, $OC=c$; let the inclinations
+be $BOC=a$, $COA=\beta$, $AOB=\gamma$. Draw $CE$ perpendicular to the
+plane $AOB$ meeting it at $E$. Describe a sphere with $O$ as a
+centre, meeting $OA$, $OB$, $OC$, $OE$ at $a$, $b$, $c$, $e$ respectively.
+
+The volume of the parallelepiped is equal to the product of its
+base and altitude $=ab\sin{\gamma}\centerdot CE=abc\sin{\gamma}\sin{cOe}$. The spherical
+triangle $cae$ is right-angled at $e$; thus
+\[
+\sin{cOe}=\sin{cOa}\sin{cae}=\sin{\beta}\sin{cab},
+\]
+and from the spherical triangle $cab$
+\[
+\sin{cab}
+= \frac{\surd{( 1 - \cos^2{\alpha} - \cos^2{\beta} - \cos^2{\gamma}
+              + 2\cos{\alpha}\cos{\beta}\cos{\gamma} )} }
+       {\sin{\beta}\sin{\gamma} };
+\]
+therefore the volume of the parallelepiped
+\[
+= abc\surd{( 1 - \cos^2{\alpha} - \cos^2{\beta} - \cos^2{\gamma}
+           + 2\cos{\alpha} \cos{\beta} \cos{\gamma} )}.
+\]
+
+\paragraph{157.} \textit{To find the diagonal of a parallelepiped in terms of the
+three edges which it meets and their inclinations to one another.}
+
+Let the edges be $OA=a$, $OB=b$, $OC=c$; let the inclinations
+be $BOC=\alpha$, $COA=\beta$, $AOB=\gamma$. Let $OD$ be the diagonal required,
+and let $OE$ be the diagonal of the face $OAB$. Then
+\begin{align*}
+OD^2 &= OE^2+ED^2+2OE\centerdot ED\cos{COE}\\
+&=a^2+b^2+2ab\cos{\gamma}+c^2+2cOE\cos{COE}.
+\end{align*}
+%-----File: 129.png------------------------------------------------
+
+Describe a sphere with $O$ as centre meeting $OA$, $OB$, $OC$, $OE$
+at $a$, $b$, $c$, $e$ respectively; then (see Example~14, Chap.~\textsc{iv.})
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/129fc}
+\end{figure}
+\begin{align*}
+\cos{cOe} &= \frac{\cos{cOb}\sin{aOe}+\cos{cOa}\sin{bOe}}{\sin{aOb}}\\[1ex]
+&= \frac{\cos{\alpha}\sin{aOe}+\cos{\beta}\sin{bOe}}{\sin{\gamma}};
+\end{align*}
+therefore\\[-1ex]
+\[
+  OD^2 = a^2+b^2+c^2+2ab\cos{\gamma}
+       + \frac{2cOE}{\sin{\gamma}}
+        (\cos{\alpha}\sin{aOe} + \cos{\beta}\sin{bOe}),
+\]
+\begin{flalign*}
+&\text{and }&
+  OE\sin{aOe} &= b\sin{\gamma}, &\qquad
+  OE\sin{bOe} &= a\sin{\gamma}; &&
+\\
+&\text{therefore }& \multispan{4}{\hfill$
+ OD^2 = a^2+b^2+c^2
+       + 2ab\cos{\gamma} + 2bc\cos{\alpha} + 2ca\cos{\beta}$.
+\hfill} &\phantom{therefore}&
+\end{flalign*}
+
+\paragraph{158.} \textit{To find the volume of a tetrahedron.}
+
+A tetrahedron is one-sixth of a parallelepiped which has the
+same altitude and its base double that of the tetrahedron; thus if
+the edges and their inclinations are given we can take one-sixth
+of the expression for the volume in Art.~156. The volume of a
+tetrahedron may also be expressed in terms of its six edges; for
+in the figure of Art.~156 let $BC=a'$, $CA=b'$, $AB=c'$; then
+\[
+  \cos{\alpha}=\frac{b^2+c^2-a'^2}{2bc}, \quad
+  \cos{\beta} =\frac{c^2+a^2-b'^2}{2ca}, \quad
+  \cos{\gamma}=\frac{a^2+b^2-c'^2}{2ab},
+\]
+and if these values are substituted for $\cos{\alpha}$, $\cos{\beta}$, and $\cos{\gamma}$ in
+the expression obtained in Art.~156, the volume of the tetrahedron
+will be expressed in terms of its six edges.
+
+The following result will be obtained, in which $V$ denotes the
+volume of the tetrahedron,
+%-----File: 130.png------------------------------------------------
+\begin{align*}
+144&\hspace{3pt}V^2 = -a'^2 b'^2 c'^2\\
+& + a^2a'^2 (b'^2+c'^2-a'^2)
+    + b^2b'^2 (c'^2+a'^2-b'^2)
+    + c^2c'^2 (a'^2+b'^2-c'^2)\\
+& - a'^2 (a^2-b^2) (a^2-c^2)
+    - b'^2 (b^2-c^2) (b^2-a^2)
+    - c'^2 (c^2-a^2) (c^2-b^2).
+\end{align*}
+
+Thus for a \textit{regular} tetrahedron we have $144\hspace{3pt}V^2=2a^6$.
+
+\paragraph{159.} If the vertex of a tetrahedron be supposed to be situated
+at any point in the plane of its base, the volume vanishes;
+hence if we equate to zero the expression on the right-hand side
+of the equation just given, we obtain a relation which must hold
+among the six straight lines which join four points taken arbitrarily
+in a plane.
+
+Or we may adopt Carnot's method, in which this relation is
+established independently, and the expression for the volume of a
+tetrahedron is deduced from it; this we shall now shew, and we
+shall add some other investigations which are also given by
+Carnot.
+
+It will be convenient to alter the notation hitherto used, by
+interchanging the accented and unaccented letters.
+
+\paragraph{160.} \textit{To find the relation holding among the six straight lines
+which join four points taken arbitrarily in a plane.}
+
+Let $A$, $B$, $C$, $D$ be the four points. Let $AB=c$, $BC=a$,
+$CA=b$; also let $DA=a'$, $DB=b'$, $DC=c'$.
+
+If $D$ falls \textit{within} the triangle $ABC$, the sum of the angles
+$ADB$, $BDC$, $CDA$ is equal to four right angles; so that
+\[
+\cos{ADB}=\cos{(BDC+CDA)}.
+\]
+
+Hence by ordinary transformations we deduce
+\[
+1=\cos^2{ADB}+\cos^2{BDC}+\cos^2{CDA}-2\cos{ADB}\cos{BDC}\cos{CDA}.
+\]
+
+If $D$ falls \textit{without} the triangle $ABC$, one of the three angles
+at $D$ is equal to the sum of the other two, and the result just
+given still holds.\\[1ex]
+
+Now $\cos{ADB}=\dfrac{a'^2+b'^2-c^2}{2a'b'}$, and the other cosines may be
+expressed in a similar manner; substitute these values in the
+%-----File: 131.png------------------------------------------------
+above result, and we obtain the required relation, which after
+reduction may be exhibited thus,
+\begin{align*}
+&  0=-a^2b^2c^2\\
++ &a'^2a^2(b^2+c^2-a^2)
+   + b'^2b^2(c^2+a^2-b^2)
+   + c'^2c^2(a^2+b^2-c^2)\\
+- &a^2(a'^2-b'^2)(a'^2-c'^2)
+   - b^2(b'^2-c'^2)(b'^2-a'^2)
+   - c^2(c'^2-a'^2)(c'^2-b'^2).
+\end{align*}
+
+\paragraph{161.} \textit{To express the volume of a tetrahedron in terms of its
+six edges.}
+
+Let $a$, $b$, $c$ be the lengths of the sides of a triangle $ABC$
+forming one face of the tetrahedron, which we may call its base;
+let $a'$, $b'$, $c'$ be the lengths of the straight lines which join $A$, $B$, $C$
+respectively to the vertex of the tetrahedron. Let $p$ be the length
+of the perpendicular from the vertex on the base; then the lengths
+of the straight lines drawn from the foot of the perpendicular to
+$A$, $B$, $C$ respectively are $\surd{(a'^2-p^2)}$, $\surd{(b'^2-p^2)}$, $\surd{(c'^2-p^2)}$. Hence
+the relation given in Art.~160 will hold if we put $\surd{(a'^2-p^2)}$ instead
+of $a'$, $\surd{(b'^2-p^2)}$ instead of $b'$, and $\surd{(c'^2-p^2)}$ instead of $c'$.
+We shall thus obtain
+\begin{align*}
+&p^2(2a^2b^2+2b^2c^2+2c^2a^2 -a^4-b^4-c^4) = -a^2b^2c^2\\
+& + a'^2a^2(b^2+c^2-a^2)
+   + b'^2b^2(c^2+a^2-b^2)
+   + c'^2c^2(a^2+b^2-c^2)\\
+& - a^2(a'^2-b'^2)(a'^2-c'^2)
+   - b^2(b'^2-c'^2)(b'^2-a'^2)
+   - c^2(c'^2-a'^2)(c'^2-b'^2).
+\end{align*}
+
+The coefficient of $p^2$ in this equation is sixteen times the
+square of the area of the triangle $ABC$; so that the left-hand
+member is $144\,V^2$, where $V$ denotes the volume of the tetrahedron.
+Hence the required expression is obtained.
+
+
+\paragraph{162.} \textit{To find the relation holding among the six arcs of great
+circles which join four points taken arbitrarily on the surface of a
+sphere.}
+
+Let $A$, $B$, $C$, $D$ be the four points. Let $AB=\gamma$, $BC=\alpha$,
+$CA=\beta$; let $DA=\alpha'$, $DB=\beta'$, $DC=\gamma'$.
+
+As in Art.\ 160 we have
+\[
+1=\cos^2{ADB}+\cos^2{BDC}+\cos^2{CDA}-2\cos{ADB}\cos{BDC}\cos{CDA}.
+\]
+%-----File: 132.png------------------------------------------------
+
+Now $\cos{ADB}=\dfrac{\cos{\gamma}-\cos{\alpha'}\cos{\beta'}} {\sin{\alpha'}\sin{\beta}'}$, and the other cosines
+may be expressed in a similar manner; substitute these values in
+the above result, and we obtain the required relation, which after
+reduction may be exhibited thus,
+\begin{align*}
+1 &= \cos^2{\alpha} +\cos^2{\beta} +\cos^2{\gamma}
+   + \cos^2{\alpha'}+\cos^2{\beta'}+\cos^2{\gamma'}
+\\
+&{} - \cos^2{\alpha}\cos^2{\alpha'}
+    - \cos^2{\beta} \cos^2{\beta'}
+    - \cos^2{\gamma}\cos^2{\gamma'}
+\\
+&{} - 2( \cos{\alpha}\cos{\beta} \cos{\gamma}
+    + \cos{\alpha}\cos{\beta'}\cos{\gamma'}
+\\
+&{} + \cos{\beta} \cos{\alpha'}\cos{\gamma'}
+    + \cos{\gamma}\cos{\alpha'}\cos{\beta'} )
+\\
+&{} + 2( \cos{\alpha}\cos{\beta}\cos{\alpha'}\cos{\beta'}
+    + \cos{\beta}\cos{\gamma}\cos{\beta'}\cos{\gamma'}
+\\
+&{} + \cos{\gamma}\cos{\alpha}\cos{\gamma'}\cos{\alpha'} ).
+\end{align*}
+
+\paragraph{163.} \textit{To find the radius of the sphere circumscribing a tetrahedron.}
+
+Denote the edges of the tetrahedron as in Art.~161. Let the
+sphere be supposed to be circumscribed about the tetrahedron,
+and draw on the sphere the six arcs of great circles joining the
+angular points of the tetrahedron. Then the relation given in
+Art.~162 holds among the cosines of these six arcs.
+
+Let $r$ denote the radius of the sphere. Then
+\[
+  \cos{\alpha}=1-2\sin^2{\frac{\alpha}{2}}
+= 1-2\left( \frac{a}{2r} \right)^2 = 1 - \frac{a^2}{2r^2};
+\]
+and the other cosines may be expressed in a similar manner.
+Substitute these values in the result of Art.~162, and we obtain,
+after reduction, with the aid of Art.~161,
+\begin{flalign*}
+&\rlap{$4\times144\,V^2r^2=$}& \\
+&&& 2a^2b^2a'^2b'^2
+  + 2b^2c^2b'^2c'^2
+  + 2c^2a^2c'^2a'^2 - a^4a'^4 - b^4b'^4 - c^4c'^4. &&
+\end{flalign*}
+The right-hand member may also be put into factors, as we see
+by recollecting the mode in which the expression for the area of
+a triangle is put into factors. Let $aa'+bb'+cc'=2\sigma$; then
+\[
+36\hspace{3pt}V^2r^2=\sigma(\sigma-aa')(\sigma-bb')(\sigma-cc').
+\]
+%-----File: 133.png------------------------------------------------
+
+\section*{\centering\normalsize EXAMPLES.}
+
+1. If $I$ denote the inclination of two adjacent faces of a
+regular polyhedron, shew that $\cos I=\tfrac{1}{3}$ in the tetrahedron, $=0$
+in the cube, $=-\tfrac{1}{3}$ in the octahedron, $=-\tfrac{1}{5}\surd{5}$ in the dodecahedron,
+and $=-\tfrac{1}{3}\surd{5}$ in the icosahedron.
+\medskip
+
+2. With the notation of Art.\ 153, shew that the radius of
+the sphere which touches one face of a regular polyhedron and all
+the adjacent faces produced is $\tfrac{1}{2}a\cot{\dfrac{\pi}{m}}\cot{\tfrac{1}{2}}I$.
+\medskip
+
+3. A sphere touches one face of a regular tetrahedron and
+the other three faces produced: find its radius.
+\medskip
+
+4. If $a$ and $b$ are the radii of the spheres inscribed in and
+described about a regular tetrahedron, shew that $b=3a$.
+\medskip
+
+5. If $a$ is the radius of a sphere inscribed in a regular tetrahedron,
+and $R$ the radius of the sphere which touches the edges,
+shew that $R^2=3a^2$.
+\medskip
+
+6. If $a$ is the radius of a sphere inscribed in a regular tetrahedron,
+and $R'$ the radius of the sphere which touches one face and
+the others produced, shew that $R'=2a$.
+\medskip
+
+7. If a cube and an octahedron be described about a given
+sphere, the sphere described about these polyhedrons will be the
+same; and conversely.
+\medskip
+
+8. If a dodecahedron and an icosahedron be described about
+a given sphere, the sphere described about these polyhedrons will
+be the same; and conversely.
+\medskip
+
+9. A regular tetrahedron and a regular octahedron are inscribed
+in the same sphere: compare the radii of the spheres
+which can be inscribed in the two solids.
+\medskip
+
+10. The sum of the squares of the four diagonals of a parallelepiped
+is equal to four times the sum of the squares of the
+edges.
+%-----File: 134.png------------------------------------------------
+\medskip
+
+11. If with all the angular points of any parallelepiped as
+centres equal spheres be described, the sum of the intercepted
+portions of the parallelepiped will be equal in volume to one of
+the spheres.
+\medskip
+
+12. A regular octahedron is inscribed in a cube so that the
+corners of the octahedron are at the centres of the faces of the
+cube: shew that the volume of the cube is six times that of the
+octahedron.
+\medskip
+
+13. It is not possible to fill any given space with a number
+of regular polyhedrons of the same kind, except cubes; but this
+may be done by means of tetrahedrons and octahedrons which
+have equal faces, by using twice as many of the former as of
+the latter.
+\medskip
+
+14. A spherical triangle is formed on the surface of a sphere
+of radius $\rho$; its angular points are joined, forming thus a pyramid
+with the straight lines joining them with the centre: shew that
+the volume of the pyramid is
+\[
+  \tfrac{1}{3}\rho^3\surd{(\tan r\tan{r_1}\tan{r_2}\tan{r_3})},
+\]
+where $r$, $r_1$, $r_2$, $r_3$ are the radii of the inscribed and escribed circles
+of the triangle.
+\medskip
+
+15. The angular points of a regular tetrahedron inscribed
+in a sphere of radius $r$ being taken as poles, four equal small
+circles of the sphere are described, so that each circle touches
+the other three. Shew that the area of the surface bounded by
+each circle is $2\pi r^2\left( 1 - \dfrac{1}{\surd{3}} \right)$.
+\medskip
+
+16. If $O$ be any point within a spherical triangle $ABC$, the
+product of the sines of any two sides and the sine of the included
+angle
+\begin{multline*}
+=\sin{AO}\sin{BO}\sin{CO} \biggl\{\cot{AO}\sin{BOC} \\
++\cot{BO}\sin{COA}+\cot{CO}\sin{AOB} \biggr\}.
+\end{multline*}
+
+%-----File: 135.png------------------------------------------------
+
+\chapter[Arcs drawn to fixed points on the Surface of a Sphere.]{ARCS DRAWN TO FIXED POINTS ON THE SURFACE OF A SPHERE.}
+\chaptermark{ARCS DRAWN TO FIXED POINTS.}
+
+\paragraph{164.} In the present Chapter we shall demonstrate various
+propositions relating to the arcs drawn from any point on the
+surface of a sphere to certain fixed points on the surface.
+
+\paragraph{165.} $ABC$ is a spherical triangle having all its sides quadrants,
+and therefore all its angles right angles; $T$ is any point
+on the surface of the sphere: to shew that
+\[
+\cos^2{TA}+\cos^2{TB}+\cos^2{TC}=1.
+\]
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/135fc}
+\end{figure}
+
+By Art.~37 we have
+\begin{align*}
+\cos{TA} &= \cos{AB}\cos{TB}+\sin{AB}\sin{TB}\cos{TBA}\\
+&= \sin{TB}\cos{TBA}.
+\end{align*}
+
+Similarly\hfil
+$\displaystyle \cos{TC}=\sin{TB}\cos{TBC}=\sin{TB}\sin{TBA}$.
+\hfil\phantom{\indent Similarly}\\
+Square and add; thus
+\[
+\cos^2{TA}+\cos^2{TC}=\sin^2{TB}=1-\cos^2{TB};
+\]
+therefore $\hfill
+\cos^2{TA}+\cos^2{TB}+\cos^2{TC}=1.
+\hfill\phantom{therefore }$
+%-----File: 136.png------------------------------------------------
+
+\paragraph{166.} $ABC$ is a spherical triangle having all its sides quadrants,
+and therefore all its angles right angles; $T$ and $U$ are any
+points on the surface of the sphere: to shew that
+\[
+\cos TU = \cos TA \cos UA + \cos TB \cos UB + \cos TC \cos UC.
+\]
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/136fc}
+\end{figure}
+
+By Art.~37 we have
+\begin{flalign*}
+\multispan{6}{\hfill$
+  \cos TU = \cos TA \cos UA + \sin TA \sin UA \cos TAU,$
+\hfill}
+\\
+&\text{and }&
+  \cos TAU & = \cos (BAU - BAT)                        &&\\
+&&         & = \cos BAU \cos BAT + \sin BAU \sin BAT   &&\\
+&&         & = \cos BAU \cos BAT + \cos CAU \cos CAT;  &&
+\\
+&\text{therefore }& \multispan{2}{$
+  \cos TU = \cos TA \cos UA $\hfill}
+&\phantom{therefore}&
+\\
+\multispan{4}{$\phantom{therefore}
+  {}+ \sin TA \sin UA (\cos BAU \cos BAT + \cos CAU \cos CAT)
+$;\hfill} &&
+\end{flalign*}
+\begin{flalign*}
+&\text{and }&
+    \cos TB & = \sin TA \cos BAT, &\phantom{and }&\\
+&&  \cos UB & = \sin UA \cos BAU, &&\\
+&&  \cos TC & = \sin TA \cos CAT, &&\\
+&&  \cos UC & = \sin UA \cos CAU; &&
+\end{flalign*}
+therefore\\[-3ex]
+\[
+\cos TU = \cos TA \cos UA + \cos TB \cos UB + \cos TC \cos UC.
+\]
+
+\paragraph{167.} We leave to the student the exercise of shewing that
+the formul\ae\ of the two preceding Articles are perfectly general for
+all positions of $T$ and $U$, outside or inside the triangle $ABC$: the
+%-----File: 137.png------------------------------------------------
+demonstrations will remain essentially the same for all modifications
+of the diagrams. The formul\ae\ are of constant application
+in Analytical Geometry of three dimensions, and are demonstrated
+in works on that subject; we have given them here as they may
+be of service in Spherical Trigonometry, and will in fact now be
+used in obtaining some important results.
+
+\paragraph{168.} Let there be any number of fixed points on the surface
+of a sphere; denote them by $H_1$, $H_2$, $H_3$,\ldots. Let $T$ be any point
+on the surface of a sphere. We shall now investigate an expression
+for the sum of the cosines of the arcs which join $T$ with
+the fixed points.
+
+Denote the sum by $\Sigma$; so that
+\[
+  \Sigma = \cos TH_1 + \cos TH_2 + \cos TH_3 + \ldots.
+\]
+
+Take on the surface of the sphere a fixed spherical triangle
+$ABC$, having all its sides quadrants, and therefore all its angles
+right angles.
+
+Let $\lambda$, $\mu$, $\nu$ be the cosines of the arcs which join $T$ with
+$A$, $B$, $C$ respectively; let $l_1$, $m_1$, $n_1$ be the cosines of the arcs which
+join $H_1$ with $A$, $B$, $C$ respectively; and let a similar notation be
+used with respect to $H_2$, $H_3$,\ldots.
+
+Then, by Art.\ 166,
+\begin{align*}
+\Sigma & =l_1\lambda + m_1\mu + n_1\nu + l_2\lambda + m_2\mu + n_2\nu + \ldots\\
+           & = P \lambda + Q \mu + R \nu;
+\end{align*}
+where $P$ stands for $l_1 + l_2 + l_3 + \ldots$, with corresponding meanings
+for $Q$ and $R$.
+
+\paragraph{169.} It will be seen that $P$ is the value which $\Sigma$ takes when
+$T$ coincides with $A$, that $Q$ is the value which $\Sigma$ takes when $T$
+coincides with $B$, and that $R$ is the value which $\Sigma$ takes when $T$
+coincides with $C$. Hence the result expresses the general value
+of $\Sigma$ in terms of the cosines of the arcs which join $T$ to the fixed
+points $A$, $B$, $C$, and the particular values of $\Sigma$ which correspond
+to these three points.
+%-----File: 138.png------------------------------------------------
+
+\paragraph{170.} We shall now transform the result of Art.~168.
+\begin{flalign*}
+&\text{\indent Let }&&
+G=\surd{(P^2+Q^2+R^2)};
+&\phantom{\indent Let }&
+\end{flalign*}
+and let $\alpha$, $\beta$, $\gamma$ be three arcs determined by the equations
+\[
+\cos \alpha = \frac{P}{G},\ \cos \beta \frac{Q}{G},\ \cos\gamma \frac{R}{G};
+\]
+then\hfill
+$\Sigma=G(\lambda \cos \alpha + \mu \cos \beta + \nu \cos \gamma)$.
+\hfill\phantom{\text{then}}
+
+Since $\cos^2\alpha +\cos^2\beta+\cos^2\gamma = 1$, it is obvious that there will be
+some point on the surface of the sphere, such that $\alpha$, $\beta$, $\gamma$ are the
+arcs which join it to $A$, $B$, $C$ respectively; denote this point by
+$U$: then, by Art.~166,
+\[
+\cos TU= \lambda \cos \alpha + \mu \cos \beta + \nu \cos \gamma;
+\]
+and finally
+\[
+\Sigma=G \cos TU.
+\]
+
+Thus, whatever may be the position of $T$, the sum of the cosines
+of the arcs which join $T$ to the fixed points varies as the cosine
+of the single arc which joins $T$ to a certain fixed point $U$.
+
+We might take $G$ either positive or negative; it will be
+convenient to suppose it positive.
+
+\paragraph{171.} A sphere is described about a regular polyhedron;
+from any point on the surface of the sphere arcs are drawn to the
+solid angles of the polyhedron: to shew that the sum of the cosines
+of these arcs is zero.
+
+From the preceding Article we see that if $G$ is not zero
+there is \textit{one} position of $T$ which gives to $\Sigma$ its greatest positive
+value, namely, when $T$ coincides with $U$. But by the symmetry
+of a regular polyhedron there must always be \textit{more than one} position
+of $T$ which gives the same value to $\Sigma$. For instance, if we
+take a regular tetrahedron, as there are four faces there will at
+least be \textit{three other} positions of $T$ symmetrical with any assigned
+position.
+
+Hence $G$ must be zero; and thus the \textit{sum of the cosines of the
+arcs which join $\mathrm{T}$ to the solid angles of the regular polyhedron is
+zero for all positions of $\mathrm{T}$.}
+%-----File: 139.png------------------------------------------------
+
+\paragraph{172.} Since $G=0$, it follows that $P$, $Q$, $R$ must each be zero;
+these indeed are particular cases of the general result of Art.\ 171.
+See Art.\ 169.
+
+\paragraph{173.} The result obtained in Art.\ 171 may be shewn to hold
+also in some other cases. Suppose, for instance, that a rectangular
+parallelepiped is inscribed in a sphere; then the sum of the
+cosines of the arcs drawn from any point on the surface of the
+sphere to the solid angles of the parallelepiped is zero. For here
+it is obvious that there must always be at least \textit{one other} position
+of $T$ symmetrical with any assigned position. Hence by the
+argument of Art.\ 171 we must have $G = 0$.
+
+\paragraph{174.} Let there be any number of fixed points on the surface
+of a sphere; denote them by $H_1$, $H_2$, $H_3,\dots$ Let $T$ be any point on
+the surface of the sphere. We shall now investigate a remarkable
+expression for the sum of the squares of the cosines of the arcs
+which join $T$ with the fixed points.
+
+Denote the sum by $\Sigma$; so that
+\[
+\Sigma = \cos^2 TH_1 + \cos^2 TH_2 + \cos^2 TH_3 + \dots.
+\]
+
+Take on the surface of the sphere a fixed spherical triangle
+$ABC$, having all its sides quadrants, and therefore all its angles
+right angles.
+
+Let $\lambda$, $\mu$, $\nu$ be the cosines of the arcs which join $T$ with $A$, $B$, $C$
+respectively; let $l_1$, $m_1$, $n_1$ be the cosines of the angles which join
+$H_1$ with $A$, $B$, $C$ respectively; and let a similar notation be used
+with respect to $H_2$, $H_3$,\dots.
+
+Then, by Art.\ 166,
+\begin{gather*}
+\Sigma = (l_1\lambda + m_1\mu + n_1\nu)^2 + (l_2\lambda + m_2\mu + n_2\nu)^2 + \dots.\\
+\intertext{Expand each square, and rearrange the terms; thus}
+\Sigma = P\lambda^2 + Q\mu^2 + R\nu^2 + 2p\mu\nu + 2q\nu\lambda + 2r\lambda \mu,\\
+\text{where $P$ stands for $l_1^2 + l_2^2 + l_3^2 + \dots$,}\\
+\text{and $p$ stands for $m_1n_1 + m_2n_2 + m_3n_3 + \dots$,}
+\end{gather*}
+with corresponding meanings for $Q$ and $q$, and for $R$ and $r$.
+%-----File: 140.png------------------------------------------------
+
+We shall now shew that there is some position of the triangle
+$ABC$ for which $p$, $q$, and $r$ will vanish; so that we shall then
+have
+\[
+\Sigma=P\lambda^2+Q\mu^2+R\nu^2.
+\]
+
+Since $\Sigma$ is always a finite positive quantity there must be some
+position, or some positions, of $T$ for which $\Sigma$ has the largest value
+which it can receive. Suppose that $A$ has this position, or one
+of these positions if there are more than one. When $T$ is at $A$
+we have $\mu$ and $\nu$ each zero, and $\lambda$ equal to unity, so that $\Sigma$ is then
+equal to $P$.
+
+Hence, whatever be the position of $T$,
+$P$ is never less than $P\lambda^2+Q\mu^2+R\nu^2+2p\mu\nu+2q\nu\lambda+2r\lambda\mu$,
+that is, by Art.~165,
+\begin{gather*}
+P(\lambda^2+\mu^2+\nu^2) \text{ is never less than}
+\\
+P\lambda^2+Q\mu^2+R\nu^2+2p\mu\nu+2q\nu\lambda+2r\lambda\mu;
+\end{gather*}
+therefore\\[-2ex]
+\[
+(P-Q)\mu^2+(P-R)\nu^2\text{ is never less than }2p\mu\nu+2q\nu\lambda+2r\lambda\mu.
+\]
+
+Now suppose $\nu = 0$; then $T$ is situated on the great circle of
+which $AB$ is a quadrant, and whatever be the position of $T$ we
+have
+\begin{flalign*}
+&& (P-Q)\mu^2\text{ not less than } 2r\lambda\mu, &&
+\\
+&\text{and therefore }&
+\multispan{2}{\hfill$
+P-Q\text{ not less than } \dfrac{2r\lambda}{\mu}.
+$\hfill} &\phantom{and therefore }&
+\end{flalign*}
+
+But now $\dfrac{\lambda}{\mu}$ is equal to $\dfrac{\cos TA}{\cos TB}$; this is numerically equal to
+$\tan TB$, and so may be made numerically as great as we please,
+positive or negative, by giving a suitable position to $T$. Thus
+$P-Q$ must in some cases be less than $\dfrac{2r\lambda}{\mu}$ if $r$ have any value different
+from zero.
+
+Therefore $r$ must $= 0$.
+
+In like manner we can shew that $q$ must $= 0$.
+%-----File: 141.png------------------------------------------------
+
+Hence with the specified position for $A$ we arrive at the result
+that whatever may be the position of $T$
+\[
+\Sigma = P\lambda^2+Q\mu^2+R\nu^2+2p\mu\,\nu.
+\]
+
+Let us now suppose that the position of $B$ is so taken that
+when $T$ coincides with $B$ the value of $\Sigma$ is as large as it can be
+for any point in the great circle of which $A$ is the pole. When $T$
+is at $B$ we have $\lambda$ and $\nu$ each zero, and $\mu$ equal to unity, so that
+$\Sigma$ is then equal to $Q$. For any point in the great circle of which
+$A$ is the pole $\lambda$ is zero; and therefore for any such point
+\begin{gather*}
+Q \text{ is not less than }Q\mu^2+R\nu^2+2p\mu\,\nu,\\
+\intertext{that is, by Art.\ 165,}
+Q(\mu^2+\nu^2) \text{ is not less than }Q\mu^2+R\nu^2+2p\mu\,\nu;
+\end{gather*}
+therefore $Q-R$ is not less than $\dfrac{2p\mu}{\nu}$.
+
+Hence by the same reasoning as before we must have $p = 0$.
+
+Therefore we see that there must be some position of the
+triangle $ABC$, such that for every position of $T$
+\[
+\Sigma = P\lambda^2+Q\mu^2+R\nu^2.
+\]
+
+\paragraph{175.} The remarks of Art.\ 169 are applicable to the result
+just obtained.
+
+\paragraph{176.} In the final result of Art.\ 174 we may shew that $R$ is
+the least value which $\Sigma$ can receive. For, by Art.\ 165,
+\begin{align*}
+\Sigma &= P\lambda^2 + Q\mu^2 + R(1-\lambda^2-\mu^2)\\
+       &= R +(P-R)\lambda^2 + (Q-R)\mu^2;
+\end{align*}
+and by supposition neither $P-R$ nor $Q-R$ is negative, so that
+$\Sigma$ cannot be less than $R$.
+
+\paragraph{177.} A sphere is described about a regular polyhedron; from
+any point on the surface of the sphere arcs are drawn to the
+solid angles of the polyhedron: it is required to find the sum of
+the squares of the cosines of these arcs.
+%-----File: 142.png------------------------------------------------
+
+With the notation of Art.\ 174 we have
+\[
+\Sigma = P\lambda^2 + Q\mu^2 + R\nu^2.
+\]
+
+We shall shew that in the present case $P$, $Q$, and $R$ must \textit{all
+be equal}. For if they are not, one of them must be greater than
+each of the others, or one of them must be less than each of the
+others.
+
+If possible let the former be the case; suppose that $P$ is
+greater than $Q$, and greater than $R$.
+\begin{flalign*}
+&\text{\indent Now }& \Sigma &= P(1 - \mu^2 - \nu^2) + Q\mu^2 + R\nu^2&\phantom{\indent Now}\\
+&&& = P - (P - Q)\mu^2 - (P - R)\nu^2;
+\end{flalign*}
+this shews that $\Sigma$ is \textit{always less than} $P$ except when $\mu = 0$ and
+$\nu = 0$: that is $\Sigma$ \textit{is always less than} $P$ except when $T$ is at $A$, or
+at the point of the surface which is diametrically opposite to $A$.
+But by the symmetry of a regular polyhedron there must always
+be more than \textit{two} positions of $T$ which give the same value to $\Sigma$.
+For instance if we take a regular tetrahedron, as there are \textit{four}
+faces there will be at least \textit{three other} positions of $T$ symmetrical
+with any assigned position. Hence $P$ cannot be greater than $Q$
+and greater than $R$.
+
+In the same way we can shew that one of the three $P$, $Q$,
+and $R$, cannot be less than each of the others.
+
+Therefore $P = Q = R$; and therefore by Art.\ 165 for \textit{every
+position} of $T$ we have $\Sigma = P$.
+
+Since $P = Q = R$ each of them $= \frac{1}{3}(P + Q + R)$
+\begin{align*}
+&= \frac13 \{{l_1}^2 + {m_1}^2 + {n_1}^2 + {l_2}^2 + {m_2}^2 + {n_2}^2 + \dots\}\\[1.5ex]
+&= \frac{S}{3}, \text{ by Art.\ 165},
+\end{align*}
+where $S$ is the number of the solid angles of the regular polyhedron.
+%-----File: 143.png------------------------------------------------
+
+\textit{Thus the sum of the squares of the cosines of the arcs which
+join any point on the surface of the sphere to the solid angles of
+the regular polyhedron is one third of the number of the solid
+angles.}
+
+\paragraph{178.} Since $P=Q=R$ in the preceding Article, it will follow
+that when the fixed points of Art.\ 174 are the solid angles of
+a regular polyhedron, then for \textit{any} position of the spherical triangle
+$ABC$ we shall have $p=0$, $q=0$, $r=0$.
+
+For taking any position for the spherical triangle $ABC$ we
+have
+\[
+\Sigma = P\lambda^2 + Q\mu^2 + R\nu^2 + 2p\mu\nu + 2q\nu\lambda + 2r\lambda\mu;
+\]
+then at $A$ we have $\mu = 0$ and $\nu = 0$, so that $P$ is then the value
+of $\Sigma$; similarly $Q$ and $R$ are the values of $\Sigma$ at $B$ and $C$ respectively.
+But by Art.\ 177 we have the same value for $\Sigma$ whatever
+be the position of $T$; thus
+\begin{flalign*}
+&\phantom{therefore}&&&\phantom{therefore}&\\[-\baselineskip]
+&&\multispan{2}{\hfil$P = P(\lambda^2 + \mu^2 + \nu^2) + 2p\mu\nu + 2q\nu\lambda + 2r\lambda\mu$;\hfil}&\\
+&\text{therefore}&\multispan{2}{\hfil$0 = 2p\mu\nu + 2q\nu\lambda + 2r\lambda\mu$.\hfil}
+\end{flalign*}
+
+This holds then for every position of $T$. Suppose $T$ is at \textit{any
+point} of the great circle of which $A$ is the pole; then $\lambda = 0$: thus
+we get $p\mu\nu = 0$, and therefore $p=0$. Similarly $q=0$, and $r=0$.
+
+\paragraph{179.} Let there be any number of fixed points on the surface
+of a sphere; denote them by $H_1$, $H_2$, $H_3,\dots$; from any two points
+$T$ and $U$ on the surface of the sphere arcs are drawn to the fixed
+points: it is required to find the sum of the products of the corresponding
+cosines, that is
+\[
+\cos TH_1 \cos UH_1 + \cos TH_2 \cos UH_2 + \cos TH_3 \cos UH_3 + \dots.
+\]
+
+Let the notation be the same as in Art.\ 174; and let $\lambda'$, $\mu'$, $\nu'$
+be the cosines of the arcs which join $U$ with $A$, $B$, $C$ respectively.
+Then by Art.\ 166,
+\begin{gather*}
+\cos TH_1 \cos UH_1 = (\lambda l_1 + \mu m_1 + \nu n_1)(\lambda'l_1 + \mu'm_1 + \nu'n_1) =\\
+\lambda\lambda'{l_1}^2 + \mu\mu'{m_1}^2 + \nu\nu'{n_1}^2 + (\lambda\mu' + \mu\lambda')l_1m_1 + (\mu\nu' + \nu\mu')m_1n_1 + (\nu\lambda' + \lambda\nu')n_1 l_1.
+\end{gather*}
+%-----File: 144.png------------------------------------------------
+
+Similar results hold for $\cos TH_2 \cos UH_2$, $\cos TH_3 \cos UH_3$,\ldots.
+Hence, with the notation of Art.~174, the required sum is
+\[
+\lambda\lambda'P + \mu\mu'Q + \nu\nu'R + (\mu\nu' + \nu\mu')p + (\nu\lambda' + \lambda\nu')q + (\lambda\mu' + \mu\lambda')r.
+\]
+
+Now by properly choosing the position of the triangle $ABC$
+we have $p$, $q$, and $r$ each zero as in Art.~174; and thus the
+required sum becomes
+\[
+\lambda\lambda'P + \mu\mu'Q + \nu\nu'R.
+\]
+
+\paragraph{180.} The result obtained in Art.~174 may be considered
+as a particular case of that just given; namely the case in which
+the points $T$ and $U$ coincide.
+
+\paragraph{181.} A sphere is described about a regular polyhedron; from
+any two points on the surface of the sphere arcs are drawn to
+the solid angles of the polyhedron: it is required to find the sum
+of the products of the corresponding cosines.
+
+With the notation of Art.~179 we see that the sum is
+\[
+\lambda\lambda'P + \mu\mu'Q + \nu\nu'R.
+\]
+
+And here $P = Q = R = \dfrac{S}{3}$, by Art.~177.
+
+Thus the sum $= \dfrac{S}{3} (\lambda\lambda' + \mu\mu' + \nu\nu') = \dfrac{S}{3} \cos TU$.
+
+\textit{Thus the sum of the products of the cosines is equal to the
+product of the cosine $\mathrm{TU}$ into a third of the number of the solid
+angles of the regular polyhedron.}
+
+\paragraph{182.} The result obtained in Art.~177 may be considered as
+a particular case of that just given; namely, the case in which
+the points $T$ and $U$ coincide.
+
+\paragraph{183.} If $TU$ is a quadrant then $\cos TU$ is zero, and the
+sum of the products of the cosines in Art.~181 is zero. The
+results $p = 0$, $q = 0$, $r = 0$, are easily seen to be all special examples
+of this particular case.
+%-----File: 145.png------------------------------------------------
+
+\chapter[Miscellaneous Propositions.]{MISCELLANEOUS PROPOSITIONS.}
+
+\paragraph{184.} \textit{To find the locus of the vertex of a spherical triangle of
+given base and area.}
+
+Let $AB$ be the given base, $= c$ suppose, $AC = \theta$, $BAC = \phi$.
+Since the area is given the spherical excess is known; denote
+it by $E$; then by Art.~103,
+\begin{flalign*}
+&&
+  \cot \tfrac{1}{2}E
+= \cot \tfrac{1}{2}\theta \cot \tfrac{1}{2}c \operatorname{cosec} \phi
++ \cot \phi; &&
+\\[1.5ex]
+&\text{therefore }&\multispan{2}{$\hfill
+  \sin(\phi - \tfrac{1}{2}E)
+= \cot \tfrac{1}{2} \theta \cot \tfrac{1}{2}c \sin \tfrac{1}{2}E;
+\hfill$} &\phantom{therefore }&
+\\[1.5ex]
+&\text{therefore }&\multispan{2}{$\hfill
+  2\cot \tfrac{1}{2}c \sin \tfrac{1}{2}E \cos^2 \dfrac{\theta}{2}
+= \sin \theta \sin (\phi - \tfrac{1}{2}E);
+\hfill$} &\phantom{therefore }&
+\end{flalign*}
+therefore
+\[
+  \cos \theta \cot \tfrac{1}{2}c \sin \tfrac{1}{2} E
++ \sin \theta \cos \left(\phi - \tfrac{1}{2}E + \dfrac{\pi}{2}\right)
+= -\cot \tfrac{1}{2} c \sin \tfrac{1}{2} E.
+\]
+
+Comparing this with equation (1) of Art.~133, we see that the
+required locus is a circle. If we call $\alpha, \beta$ the angular co-ordinates
+of its pole, we have
+\begin{align*}
+  \tan \alpha &= \frac{1}{\cot\frac{1}{2}c\sin\frac{1}{2}E}
+= \frac{\tan\frac{1}{2}c}{\sin \frac{1}{2}E} ,
+\\
+  \beta &= \tfrac{1}{2} E - \frac{\pi}{2}.
+\end{align*}
+
+It may be presumed from symmetry that the pole of this
+circle is in the great circle which bisects $AB$ at right angles;
+and this presumption is easily verified. For the equation to
+that great circle is
+\[
+0 = \cos \theta \cos \left(\frac{\pi}{2} - \frac{c}{2}\right)
+  + \sin \theta \sin \left(\frac{\pi}{2} - \frac{c}{2}\right)
+    \cos (\phi - \pi)
+\]
+and the values $\theta = \alpha$, $\phi = \beta$ satisfy this equation.
+%-----File: 146.png------------------------------------------------
+
+\paragraph{185.} \textit{To find the angular distance between the poles of the
+inscribed and circumscribed circles of a triangle.}
+
+Let $P$ denote the pole of the inscribed circle, and $Q$ the pole
+of the circumscribed circle of a triangle $ABC$; then $PAB = \frac{1}{2}A$,
+by Art.\ 89, and $QAB = S-C$, by Art.\ 92; hence
+\begin{gather*}
+\cos PAQ = cos\tfrac{1}{2}(B-C);
+\\
+\text{and } \cos PQ = \cos PA \cos QA + \sin PA \sin QA \cos \tfrac{1}{2}(B-C).
+\end{gather*}
+
+Now, by Art.\ 62 (see the figure of Art.\ 89),
+\begin{align*}
+\cos PA &= \cos PE \cos AE = \cos r \cos(s-a), \\
+\sin PA &= \frac{\sin PE}{\sin PAE} = \frac{\sin r}{\sin\tfrac{1}{2}A};
+\end{align*}
+thus
+\[
+  \cos PQ
+= \cos R \cos r \cos(s-a)
++ \sin R \sin r \cos \frac{1}{2}(B-C)
+ \operatorname{cosec} \tfrac{1}{2}A.
+\]
+
+Therefore, by Art.\ 54
+\[
+\cos PQ
+= \cos R \cos r \cos(s-a)
++ \sin R \sin r \sin \tfrac{1}{2}(b+c)
+ \operatorname{cosec} \tfrac{1}{2}a,
+\]
+\begin{flalign*}
+&\text{therefore }&&
+  \frac{\cos PQ}{\cos R \sin r}
+= \cot r \cos(s-a)
++ \tan R \sin \frac{1}{2}(b+c) \operatorname{cosec} \frac{1}{2}a. &\phantom{therefore }&
+\\[.5ex]
+&\text{Now }&&
+  \cot r = \frac{\sin s}{n},\quad
+  \tan R = \frac{2 \sin \frac{1}{2} a \sin \tfrac{1}{2} b \sin \tfrac{1}{2} c}{n} , &&
+\end{flalign*}
+\begin{flalign*}
+&\text{therefore }&
+   \frac{\cos PQ}{\cos R \sin r}
+&= \frac{1}{n}
+   \Bigl\{\sin s \cos (s - a)
+         + 2\sin\tfrac{1}{2}(b+c) \sin\tfrac{1}{2}b \sin\tfrac{1}{2}c \Bigr\}
+&\phantom{therefore }&
+\\[1ex]
+&&&= \dfrac{1}{2n} (\sin a + \sin b + \sin c).
+\end{flalign*}
+\begin{flalign*}
+&\text{Hence }&
+  \left(\dfrac{\cos PQ }{\cos R \sin r}\right)^2 - 1
+&= \dfrac{1}{4n^2} (\sin a + \sin b + \sin c)^2 - 1 &&
+\\
+&&&= (\cot r + \tan R)^2 \text{ (by Art.\ 94)}; &&
+\\
+&\text{therefore }&
+  \cos^2 PQ &= \cos^2 R \sin^2 r + \cos^2 (R-r),
+&\phantom{therefore }&
+\\
+&\text{and }&
+  \sin^2 PQ &= \sin^2 (R-r) - \cos^2 R \sin^2 r. &&
+\end{flalign*}
+%-----File: 147.png------------------------------------------------
+
+\paragraph{186.} \textit{To find the angular distance between the pole of the
+circumscribed circle and the pole of one of the escribed circles of
+a triangle.}
+
+Let $Q$ denote the pole of the circumscribed circle, and $Q_1$ the
+pole of the escribed circle opposite to the angle $A$. Then it may
+be shewn that $QBQ_1 = \frac{1}{2} \pi + \frac{1}{2} (C-A)$, and
+\begin{align*}
+\cos QQ_1
+&= \cos R \cos r_1 \cos (s-c)
+ - \sin R \sin r_1 \sin \tfrac{1}{2}(C-A) \sec \tfrac{1}{2}B \\
+&= \cos R \cos r_1 \cos (s-c)
+ - \sin R \sin r_1 \sin \tfrac{1}{2}(c-a) \operatorname{cosec} \frac{1}{2}b.
+\end{align*}
+Therefore
+\[
+  \frac{\cos QQ_1}{\sin r_1 \cos R}
+= \cot r_1 \cos (s - c)
+- \tan R \sin \tfrac{1}{2} (c-a) \operatorname{cosec} \tfrac{1}{2} b;
+\]
+by reducing as in the preceding Article, the right-hand member of
+the last equation becomes
+\[
+\dfrac{1}{2n} (\sin b + \sin c - \sin a);
+\]
+\begin{flalign*}
+&\text{hence }&
+   \left(\frac{\cos QQ_1}{\cos R \sin r_1}\right)^2 - 1
+&= (\tan R - \cot r_1)^2, \text{ (Art.\ 94)};
+&\phantom{therefore }&
+\\
+&\text{therefore }&
+  \cos^2 QQ_1 &= \cos^2 R \sin^2 r_1 + \cos^2 (R + r_1), &&
+\\
+&\text{and }&
+  \sin^2 QQ_1 &= \sin^2 (R + r_1) - \cos^2 R \sin^2 r_1.
+\end{flalign*}
+
+\paragraph{187.} \textit{The arc which passes through the middle points of the
+sides of any triangle upon a given base will meet the base produced
+at a fixed point, the distance of which from the middle point of the
+base is a quadrant.}
+
+Let $ABC$ be any triangle, $E$ the middle point of $AC$, and $F$
+the middle point of $AB$; let the arc which joins $E$ and $F$ when
+produced meet $BC$ produced at $Q$. Then
+\[
+  \frac{\sin BQ}{\sin BF} = \frac{\sin BFQ}{\sin BQF},\qquad
+  \frac{\sin AQ}{\sin AF} = \frac{\sin AFQ}{\sin AQF};
+\]
+\begin{flalign*}
+&\text{therefore }&&
+  \frac{\sin BQ}{\sin AQ} = \frac{\sin AQF}{\sin BQF},
+&\phantom{therefore }&
+\end{flalign*}
+%-----File: 148.png------------------------------------------------
+similarly
+$\hfill\displaystyle
+\frac{\sin CQ}{\sin AQ} = \frac{\sin AQF}{\sin CQF};
+\hfill\phantom{similarly }$\\[2ex]
+therefore
+$\hfill
+\sin BQ = \sin CQ;\qquad \text{therefore } BQ + CQ = \pi.
+\hfill\phantom{therefore }$
+
+Hence if $D$ be the middle point of $BC$
+\[
+DQ = \tfrac{1}{2} (BQ + CQ) = \tfrac{1}{2} \pi.
+\]
+
+\paragraph{188.} \textit{If three arcs be drawn from the angles of a spherical
+triangle through any point to meet the opposite sides, the products
+of the sines of the alternate segments of the sides are equal.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/148fc}
+\end{figure}
+
+Let $P$ be any point, and let arcs be drawn from the angles
+$A$, $B$, $C$ passing through $P$ and meeting the opposite sides at
+$D$, $E$, $F$. Then
+\[
+  \frac{\sin BD}{\sin BP} = \frac{\sin BPD}{\sin BDP},\qquad
+  \frac{\sin CD}{\sin CP} = \frac{\sin CPD}{\sin CDP},
+\]
+therefore
+$\hfill\displaystyle
+  \frac{\sin BD }{\sin CD }
+= \frac{\sin BPD}{\sin CPD}\,
+  \frac{\sin BP }{\sin CP }.
+\hfill\phantom{therefore }$
+
+Similar expressions may be found for $\dfrac{\sin CE}{\sin AE}$ and $\dfrac{\sin AF}{\sin BF}$;
+and hence it follows obviously that
+\[
+  \dfrac{\sin BD}{\sin CD}\,
+  \dfrac{\sin CE}{\sin AE}\,
+  \dfrac{\sin AF}{\sin BF} = 1;
+\]
+therefore
+$\hfill
+\sin BD \sin CE \sin AF= \sin CD \sin AE \sin BF.
+\hfill\phantom{therefore }$
+%-----File: 149.png------------------------------------------------
+
+\paragraph{189.} Conversely, when the points $D$, $E$, $F$ in the sides of a
+spherical triangle are such that the relation given in the preceding
+Article holds, the arcs which join these points with the opposite
+angles respectively \textit{pass through a common point}. Hence the
+following propositions may be established: the perpendiculars
+from the angles of a spherical triangle on the opposite sides
+meet at a point; the arcs which bisect the angles of a spherical
+triangle meet at a point; the arcs which join the angles of a
+spherical triangle with the middle points of the opposite sides
+meet at a point; the arcs which join the angles of a spherical
+triangle with the points where the inscribed circle touches the
+opposite sides respectively meet at a point.
+
+Another mode of establishing such propositions has been
+exemplified in Arts.\ 139 and 140.
+
+\paragraph{190.} \textit{If $\mathrm{AB}$ and $\mathrm{A'B'}$ be any two equal arcs
+$\mathrm{AA'}$ and $\mathrm{AA'}$ and $\mathrm{BB'}$ be bisected at right angles by arcs meeting at $\mathrm{P}$,
+then $\mathrm{AB}$ and $\mathrm{A'B'}$ subtend equal angles at $\mathrm{P}$.}
+\begin{figure}[htp]
+\centering
+\includegraphics[width=5.0cm]{images/149fc}
+\end{figure}
+
+For $PA = PA'$ and $PB = PB'$; hence the sides of the triangle
+$PAB$ are respectively equal to those of $PA'B'$; therefore the angle
+$APB =$ the angle $A'PB'$.
+
+This simple proposition has an important application to the
+motion of a rigid body of which one point is fixed. For conceive
+a sphere capable of motion round its centre which is fixed; then it
+appears from this proposition that any two fixed points on the
+%-----File: 150.png------------------------------------------------
+sphere, as $A$ and $B$, can be brought into any other positions, as
+$A'$ and $B'$, by rotation round an axis passing through the centre of
+the sphere and a certain point $P$. Hence it may be inferred that
+any change of position in a rigid body, of which one point is fixed,
+may be effected by rotation round some axis through the fixed
+point.
+\nopagebreak\par
+\rightline{(De Morgan's \textit{Differential and Integral Calculus}, page 489.)}
+
+\paragraph{191.} Let $P$ denote any point within any \textit{plane} angle $AOB$,
+and from $P$ draw perpendiculars on the straight lines $OA$ and
+$OB$; then it is evident that these perpendiculars include an angle
+which is the \textit{supplement} of the angle $AOB$. The corresponding
+fact with respect to a \textit{solid} angle is worthy of notice. Let there
+be a solid angle formed by three plane angles, meeting at a point
+$O$. From any point $P$ within the solid angle, draw perpendiculars
+$PL$, $PM$, $PN$ on the three planes which form the solid angle;
+then the spherical triangle which corresponds to the three planes
+$LPM$, $MPN$, $NPL$ is the \textit{polar triangle} of the spherical triangle
+which corresponds to the solid angle at $O$. This remark is due to
+Professor De Morgan.
+
+\paragraph{192.} Suppose three straight lines to meet at a point and form
+a solid angle; let $\alpha$, $\beta$, and $\gamma$ denote the angles contained by these
+three straight lines taken in pairs: then it has been proposed to
+call the expression $\surd(1-\cos^2 \alpha-\cos^2 \beta-\cos^2 \gamma+2\cos \alpha \cos \beta \cos \gamma)$,
+the \textit{sine of the solid angle}. See Baltzer's \textit{Theorie\ldots der Determinanten},
+2nd edition, page~177. Adopting this definition it is easy
+to shew that the sine of a solid angle lies between zero and unity.
+
+We know that the area of a plane triangle is half the product
+of two sides into the sine of the included angle: by Art.\ 156 we
+have the following analogous proposition; the volume of a tetrahedron
+is one sixth of the product of three edges into the sine of
+the solid angle which they form.
+
+Again, we know in mechanics that if three forces acting at a
+point are in equilibrium, each force is as the sine of the angle
+between the directions of the other two: the following proposition
+is analogous; if four forces acting at a point are in equilibrium
+%-----File: 151.png------------------------------------------------
+each force is as the sine of the solid angle formed by the directions
+of the other three. See \textit{Statics}, Chapter~II\@.
+
+\paragraph{193.} Let a sphere be described about a regular polyhedron;
+let perpendiculars be drawn from the centre of the sphere on the
+faces of the polyhedron, and produced to meet the surface of
+the sphere: then it is obvious from symmetry that the points of
+intersection must be the angular points of another regular polyhedron.
+
+This may be verified. It will be found on examination that if
+$S$ be the number of solid angles, and $F$ the number of faces of one
+regular polyhedron, then another regular polyhedron exists which
+has $S$ faces and $F$ solid angles. See Art.~151.
+
+\paragraph{194.} \textit{Polyhedrons.} The result in Art.~150 was first obtained
+by Euler; the demonstration which is there given is due to
+Legendre. The demonstration shews that the result is true in
+many cases in which the polyhedron has \textit{re-entrant} solid angles;
+for all that is necessary for the demonstration is, that it shall be
+possible to take a point within the polyhedron as the centre of a
+sphere, so that the polygons, formed as in Art.~150, shall not have
+any coincident portions. The result, however, is generally true,
+even in cases in which the condition required by the demonstration
+of Art.~150 is not satisfied. We shall accordingly give
+another demonstration, and shall then deduce some important
+consequences from the result. We begin with a theorem which
+is due to Cauchy.
+
+\paragraph{195.} \textit{Let there be any network of rectilineal figures, not necessarily
+in one plane, but not forming a closed surface; let $\mathrm E$ be the
+number of edges, $\mathrm F$ the number of figures, and $\mathrm S$ the number of
+corner points: then $\mathrm{F + S = E + 1 }$.}
+
+This theorem is obviously true in the case of a single plane
+figure; for then $F = 1$, and $S = E$. It can be shewn to be generally
+true by induction. For assume the theorem to be true for
+a network of $F$ figures; and suppose that a rectilineal figure of
+$n$ sides is added to this network, so that the network and the
+additional figure have $m$ sides coincident, and therefore $m + 1$
+%-----File: 152.png------------------------------------------------
+corner points coincident. And with respect to the new network
+which is thus formed, let $E'$, $F'$, $S'$ denote the same things as
+$E$, $F$, $S$ with respect to the old network. Then
+\[
+E' = E + n-m,\quad F' = F+1,\quad S' = S + n-(m + 1);
+\]
+therefore
+$\hfill
+F' + S'- E' = F+S-E.
+\hfill\phantom{therefore }$
+
+But $F+S=E+ 1$, by hypothesis; therefore $F' + S'=E' + 1$.
+
+\paragraph{196.} To demonstrate Euler's theorem we suppose one face of
+a polyhedron removed, and we thus obtain a network of rectilineal
+figures to which Cauchy's theorem is applicable. Thus
+\begin{flalign*}
+&                &  F-1+S &= E + 1; &&\\
+&\text{therefore }& F + S &= E + 2. &\phantom{therefore }&
+\end{flalign*}
+
+\paragraph{197.} \textit{In any polyhedron the number of faces with an odd
+number of sides is even, and the number of solid angles formed
+with an odd number of plane angles is even.}
+
+Let $a$, $b$, $c$, $d$,\ldots\ldots denote respectively the numbers of faces
+which are triangles, quadrilaterals, pentagons, hexagons,\ldots\ldots.
+Let $\alpha$, $\beta$, $\gamma$, $\delta$,\ldots\ldots denote respectively the numbers of the solid
+angles which are formed with three, four, five, six,\ldots\ldots plane
+angles.
+
+Then, each edge belongs to \textit{two} faces, and terminates at \textit{two}
+solid angles; therefore
+\begin{align*}
+  & 2E=3a+4b+5c+6d+ \ldots\ldots, \\
+  & 2E= 3 \alpha + 4 \beta + 5 \gamma + 6 \delta + \ldots\ldots.
+\end{align*}
+
+From these relations it follows that $a + c + e + \ldots\ldots$, and
+$\alpha + \gamma + \epsilon + \ldots\ldots$ are \textit{even} numbers.
+
+\paragraph{198.} With the notation of the preceding Article we have
+\begin{align*}
+  & F=a+b+c+d+ \ldots\ldots, \\
+  & S= \alpha + \beta + \gamma + \delta + \ldots\ldots.
+\end{align*}
+
+From these combined with the former relations we obtain
+\begin{align*}
+  & 2E-3F = b + 2c + 3d + \ldots\ldots, \\
+  & 2E-3S = \beta + 2 \gamma + 3 \delta + \ldots\ldots.
+\end{align*}
+
+Thus $2E$ cannot be less than $3F$, or less than $3S$.
+%-----File: 153.png------------------------------------------------
+
+\paragraph{199.} From the expressions for $E$, $F$, and $S$, given in the
+two preceding Articles, combined with the result $2F + 2S=4 + 2E$,
+we obtain
+\begin{gather*}
+  2 (a + b + c + d + \ldots)
++ 2 (\alpha + \beta + \gamma + \delta + \ldots)
+= 4 + 3a + 4b + 5c + 6d + \ldots,
+\\
+  2 (a + b + c + d + \ldots)
++ 2 (\alpha + \beta + \gamma + \delta + \ldots)
+= 4 + 3 \alpha + 4 \beta + 5 \gamma + 6 \delta + \ldots,
+\end{gather*}
+therefore $\hfill
+  2 (\alpha + \beta + \gamma + \delta + \ldots)
+- (a + 2b + 3c + 4d + \ldots)
+= 4$, \hfill(1)
+\\
+\phantom{therefore} $\hfill
+  2 (a + b + c + d + \ldots)
+- (\alpha + 2 \beta + 3 \gamma + 4 \delta + \ldots)
+= 4$.\hfill(2)
+
+Therefore, by addition
+\[
+a + \alpha - (c + \gamma) - 2 (d + \delta) - 3 (e + \epsilon) - \ldots\ldots = 8.
+\]
+
+\textit{Thus the number of triangular faces together with the number
+of solid angles formed with three plane angles cannot be less
+than eight.}
+
+Again, from (1) and (2), by eliminating $\alpha$, we obtain
+\[
+3a + 2b + c - e - 2f - \ldots\ldots -2\beta - 4\gamma - \ldots\ldots = 12,
+\]
+so that $3a + 2b + c$ cannot be less than 12. From this result
+various inferences can be drawn; thus for example, \textit{a solid cannot
+be formed which shall have no triangular, quadrilateral, or pentagonal
+faces.}
+
+In like manner, we can shew that $3 \alpha + 2 \beta + \gamma$ cannot be less
+than 12.
+
+\paragraph{200.} Poinsot has shewn that in addition to the five well-known
+\textit{regular polyhedrons}, four other solids exist which are
+perfectly symmetrical in shape, and which might therefore also be
+called \textit{regular}. We may give an idea of the nature of Poinsot's
+results by referring to the case of a polygon. Suppose five points
+$A$, $B$, $C$, $D$, $E$, placed in succession at equal distances round the
+circumference of a circle. If we draw a straight line from each
+point to the next point, we form an ordinary regular pentagon.
+Suppose however we join the points by straight lines in the following
+order, $A$ to $C$, $C$ to $E$, $E$ to $B$, $B$ to $D$, $D$ to $A$; we thus
+form a star-shaped symmetrical figure, which might be considered
+a regular pentagon.
+%-----File: 154.png------------------------------------------------
+
+It appears that, in a like manner, four, and only four, new
+regular solids can be formed. To such solids, the faces of which
+intersect and cross, Euler's theorem does not apply.
+
+\paragraph{201.} Let us return to Art.~195, and suppose $e$ the number
+of edges \textit{in} the bounding contour, and $e'$ the number of edges
+\textit{within} it; also suppose $s$ the number of corners \textit{in} the bounding
+contour, and $s'$ the number \textit{within} it. Then
+\begin{flalign*}
+\multispan{6}{$\hfill E = e+e'; \ S=s+s'; \hfill$} \\
+&\text{therefore }& 1 + e + e' &= s + s' + F.
+&\phantom{therefore }&\\[1ex]
+&\text{\indent But }&       e  &= s;  &&\\
+&\text{therefore }&     1 + e' &= s' + F. &&
+\end{flalign*}
+
+We can now demonstrate an extension of Euler's theorem,
+which has been given by Cauchy.
+
+\paragraph{202.} \textit{Let a polyhedron be decomposed into any number of
+polyhedrons at pleasure; let $\mathrm P$ be the number thus formed, $\mathrm S$ the
+number of solid angles, $\mathrm F$ the number of faces, $\mathrm E$ the number of
+edges: then $\mathrm{S + F = E + P + 1 }$.}
+
+For suppose all the polyhedrons united, by starting with one
+and adding one at a time. Let $e$, $f$, $s$ be respectively the numbers
+of edges, faces, and solid angles in the first; let $e'$, $f'$, $s'$ be
+respectively the numbers of edges, faces, and solid angles in the
+second which are not common to it and the first; let $e''$, $f''$, $s''$
+be respectively the numbers of edges, faces, and solid angles in
+the third which are not common to it and the first or second;
+and so on. Then we have the following results, namely, the first
+by Art.~196, and the others by Art.~201;
+\begin{align*}
+    s + f &= e + 2,\\
+  s' + f' &= e' + 1,\\
+s'' + f'' &= e'' + 1,\\
+\multispan{2}{\dotfill}
+\end{align*}
+
+By addition, since
+$s + s' + s'' + \ldots = S$,
+$f + f' + f'' + \ldots = F$, and
+$e + e' + e'' + \ldots = E$, we obtain
+\[
+  S + F = E + P + 1.
+\]
+%-----File: 155.png------------------------------------------------
+
+\paragraph{203.} The following references will be useful to those who
+study the theory of polyhedrons. Euler, \textit{Novi Commentarii
+Acad\-emi{\ae}\ldots Pe\-tro\-pol\-i\-tan{\ae}}, Vol.~\textsc{iv.}\ 1758; Legendre, \textit{G{\'e}om{\'e}trie};
+Poinsot, \textit{Journal de l'\'Ecole Polytechnique}, Cahier \textsc{x}; Cauchy,
+\textit{Journal de l'\'Ecole Polytechnique}, Cahier \textsc{xvi}; Poinsot and Bertrand,
+\textit{Comptes Rendus\ldots de l'Acad\-\'emie des Sciences}, Vol.~\textsc{xlvi}; Catalan,
+\textit{Th{\'e}\-or\-{\`e}mes et Prob\-l{\`e}mes de G{\'e}o\-m{\'e}\-trie El\-{\'e}\-men\-taire}; Kirkman, \textit{Philosophical
+Transactions} for 1856 and subsequent years; Listing,
+\textit{Abhandlungen der K{\"o}nig\-lichen Ge\-sell\-schaft\ldots zu G{\"o}tt\-ingen}, Vol.~\textsc{x}.
+
+\section*{\centering\normalsize MISCELLANEOUS EXAMPLES.}
+
+1. Find the locus of the vertices of all right-angled spherical
+triangles having the same hypotenuse; and from the equation
+obtained, prove that the locus is a circle when the radius of the
+sphere is infinite.
+\medskip
+
+2. $AB$ is an arc of a great circle on the surface of a sphere, $C$
+its middle point: shew that the locus of the point $P$, such that
+the angle $APC =$ the angle $BPC$, consists of two great circles at
+right angles to one another. Explain this when the triangle
+becomes plane.
+\medskip
+
+3. On a given arc of a sphere, spherical triangles of equal
+area are described: shew that the locus of the angular point
+opposite to the given arc is defined by the equation
+\begin{multline*}
+  \tan^{-1} \left\{\frac{\tan (\alpha + \phi)}{\sin \theta}\right\}
++ \tan^{-1} \left\{\frac{\tan (\alpha - \phi)}{\sin \theta}\right\} \\
+{}
++ \tan^{-1} \left\{\frac{\tan \theta}{\sin (\alpha + \phi)}\right\}
++ \tan^{-1} \left\{\frac{\tan \theta}{\sin (\alpha - \phi)}\right\}
+= \beta,
+\end{multline*}
+where $2\alpha$ is the length of the given arc, $\theta$ the arc of the great
+circle drawn from any point $P$ in the locus perpendicular to the
+given arc, $\phi$ the inclination of the great circle on which $\theta$ is
+%-----File: 156.png------------------------------------------------
+measured to the great circle bisecting the given arc at right
+angles, and $\beta$ a constant.
+\medskip
+
+4. In any spherical triangle
+\[
+  \tan c = \frac{\cot A \cot a + \cot B \cot b}
+                {\cot a \cot b - \cos A \cos B}.
+\]
+
+5. If $\theta$, $\phi$, $\psi$ denote the distances from the angles $A$, $B$, $C$
+respectively of the point of intersection of arcs bisecting the
+angles of the spherical triangle $ABC,$ shew that
+\[
+  \cos \theta \sin(b - c)
++ \cos \phi   \sin(c - a)
++ \cos \psi   \sin(a - b) = 0.
+\]
+
+6. If $A'$, $B'$, $C'$ be the poles of the sides $BC$, $CA$, $AB$ of a
+spherical triangle $ABC$, shew that the great circles $AA'$, $BB'$, $CC'$
+meet at a point $P$, such that
+\[
+  \cos PA \cos BC = \cos PB \cos CA = \cos PC \cos AB.
+\]
+
+7. If $O$ be the point of intersection of arcs $AD$, $BE$, $CF$
+drawn from the angles of a triangle perpendicular to the opposite
+sides and meeting them at $D$, $E$, $F$ respectively, shew that
+\[
+  \frac{\tan AD}{\tan OD}, \qquad
+  \frac{\tan BE}{\tan OE}, \qquad
+  \frac{\tan CF}{\tan OF}
+\]
+are respectively equal to
+\[
+  1+\frac{\cos A}{\cos B \cos C}, \quad
+  1+\frac{\cos B}{\cos A \cos C}, \quad
+  1+\frac{\cos C}{\cos A \cos B}.
+\]
+
+8. If $p$, $q$, $r$ be the arcs of great circles drawn from the
+angles of a triangle perpendicular to the opposite sides, $(\alpha, \alpha')$,
+$(\beta, \beta')$, $(\gamma, \gamma')$ the segments into which these arcs are divided,
+shew that
+\begin{flalign*}
+&&&  \tan \alpha \tan \alpha'
+   = \tan \beta  \tan \beta'
+   = \tan \gamma \tan \gamma'; &&
+\\[2ex]
+&\text{and }&\multispan{2}{\hfill$\displaystyle
+  \frac{\cos p}{\cos \alpha \cos \alpha'}
+= \frac{\cos q}{\cos \beta  \cos \beta' }
+= \frac{\cos r}{\cos \gamma \cos \gamma'}.
+$\hfill} &\phantom{and }&
+\end{flalign*}
+%-----File: 157.png------------------------------------------------
+
+9. In a spherical triangle if arcs be drawn from the angles to
+the middle points of the opposite sides, and if $\alpha$, $\alpha'$ be the two
+parts of the one which bisects the side $a$,
+shew that
+\[
+\frac{\sin \alpha}{\sin \alpha'} = 2 \cos \frac{a}{2}.
+\]
+
+10. The arc of a great circle bisecting the sides $AB$, $AC$ of a
+spherical triangle cuts $BC$ produced at $Q$: shew that
+\[
+\cos AQ \sin \frac{a}{2} = \sin \frac{c - b}{2} \sin \frac{c + b}{2}.
+\]
+
+11. If $ABCD$ be a spherical quadrilateral, and the opposite
+sides $AB$, $CD$ when produced meet at $E$, and $AD$, $BC$ meet at $F$,
+the ratio of the sines of the arcs drawn from $E$ at right angles to
+the diagonals of the quadrilateral is the same as the ratio of those
+from $F$.
+\medskip
+
+12. If $ABCD$ be a spherical quadrilateral whose sides $AB$,
+$DC$ are produced to meet at $P$, and $AD$, $BC$ at $Q$, and whose
+diagonals $AC$, $BD$ intersect at $R$, then
+\[
+\sin AB \sin CD \cos P = \sin AD \sin BC \cos Q = \sin AC \sin BD \cos R.
+\]
+
+13. If $A'$ be the angle of the chordal triangle which corresponds
+to the angle $A$ of a spherical triangle, shew that
+\[
+\cos A' = \sin (S - A) \cos \frac{a}{2}.
+\]
+
+14. If the tangent of the radius of the circle described about a
+spherical triangle is equal to twice the tangent of the radius of the
+circle inscribed in the triangle, the triangle is equilateral.
+\medskip
+
+15. The arc $AP$ of a circle of the same radius as the sphere
+is equal to the greater of two sides of a spherical triangle, and
+the arc $AQ$ taken in the same direction is equal to the less; the
+sine $PM$ of $AP$ is divided at $E$, so that $\dfrac{EM}{PM} =$ the natural cosine
+%-----File: 158.png------------------------------------------------
+of the angle included by the two sides, and $EZ$ is drawn parallel
+to the tangent to the circle at $Q$. Shew that the remaining side
+of the spherical triangle is equal to the arc $QPZ$.
+\medskip
+
+16. If through any point $P$ within a spherical triangle $ABC$
+great circles be drawn from the angular points $A$, $B$, $C$ to meet
+the opposite sides at $a$, $b$, $c$ respectively, prove that
+\[
+\dfrac{\sin Pa \cos PA}{\sin Aa} +
+\dfrac{\sin Pb \cos PB}{\sin Bb} +
+\dfrac{\sin Pc \cos PC}{\sin Cc} = 1.
+\]
+
+17. $A$ and $B$ are two places on the Earth's surface on the
+same side of the equator, $A$ being further from the equator
+than $B$. If the bearing of $A$ from $B$ be more nearly due East
+than it is from any other place in the same latitude as $B$, find
+the bearing of $B$ from $A$.
+\medskip
+
+18. From the result given in example 18 of Chapter~V.\ infer
+the possibility of a regular dodecahedron.
+\medskip
+
+19. $A$ and $B$ are fixed points on the surface of a sphere, and
+$P$ is any point on the surface. If $a$ and $b$ are given constants,
+shew that a fixed point $S$ can always be found, in $AB$ or $AB$ produced,
+such that
+\[
+a \cos AP + b \cos BP = s \cos SP,
+\]
+where $s$ is a constant.
+\medskip
+
+20. $A$, $B$, $C$,\ldots are fixed points on the surface of a sphere;
+$a$, $b$, $c$,\ldots are given constants. If $P$ be a point on the surface of
+the sphere, such that
+\[
+a \cos AP + b \cos BP + c \cos CP + \ldots = \text{constant},
+\]
+shew that the locus of $P$ is a circle.
+%-----File: 159.png------------------------------------------------
+
+\chapter[Numerical Solution of Spherical Triangles.]{NUMERICAL SOLUTION OF SPHERICAL TRIANGLES.}
+
+\paragraph{204.} We shall give in this Chapter examples of the numerical
+solution of Spherical Triangles.
+
+We shall first take right-angled triangles, and then oblique-angled
+triangles.
+
+\section*{\centering\normalfont\large%
+\textit{Right-Angled Triangles.}}
+
+\paragraph{205.} Given $a = 37^\circ\, 48'\, 12''$, $b = 59^\circ\, 44'\, 16''$, $C = 90^\circ$.
+
+To find $c$ we have
+\begin{center}
+$\cos c = \cos a \cos b,$\\[.1ex]
+\begin{tabular}{rr@{}}
+$L \cos 37^\circ\, 48'\, 12'' =$ & $9.8976927$ \\
+$L \cos 59^\circ\, 44'\, 16'' =$ & $9.7023945$ \\
+\cline{2-2}
+$L \cos c + 10 =$ & $19.6000872$\\
+$c =$ & $66^\circ\, 32'\, 6''$.
+\end{tabular}
+\end{center}
+
+To find $A$ we have
+\begin{center}
+$\cot A = \cot a \sin b,$\\[.5ex]
+\begin{tabular}{rr@{}}
+$L \cot 37^\circ\, 48'\, 12'' =$ & $10.1102655$ \\
+$L \sin 59^\circ\, 44'\, 16'' =$ & $ 9.9363770$ \\
+\cline{2-2}
+$L \cot A + 10 =$ & $ 20.0466425$\\
+$A =$ & $41^\circ\, 55'\, 45''$.
+\end{tabular}
+\end{center}
+%-----File: 160.png------------------------------------------------
+
+To find $B$ we have
+\begin{center}
+$\cot B = \cot b \sin a,$\\[.1ex]
+\begin{tabular}{rr@{}}
+$L \cot 59^\circ\, 44'\, 16'' =$ & $ 9.7660175$\\
+$L \sin 37^\circ\, 48'\, 12'' =$ & $ 9.7874272$\\
+\cline{2-2}
+$L \cot B + 10 = $ & $19.5534447$\\
+$B =$ & $70^\circ\, 19'\, 15''$.
+\end{tabular}
+\end{center}
+
+\paragraph{206.}
+Given $A = 55^\circ\, 32'\, 45''$, $C = 90^\circ$, $c = 98^\circ\, 14'\, 24''$.
+
+To find $a$ we have
+\begin{center}
+$\sin a = \sin c \sin A,$\\[.5ex]
+\begin{tabular}{rr@{}}
+$L \sin 98^\circ\, 14'\, 24'' =$ & $ 9.9954932$ \\
+$L \sin 55^\circ\, 32'\, 45'' =$ & $ 9.9162323$ \\
+\cline{2-2}
+$L \sin a + 10                =$ & $19.9117255$ \\
+$a =$ & $54^\circ\, 41'\, 35''$.
+\end{tabular}
+\end{center}
+
+To find $B$ we have
+\[
+\cot B = \cos c \tan A.
+\]
+
+Here $\cos c$ \textit{is negative}; and therefore $\cot B$ will be negative,
+and $B$ greater than a right angle. The numerical value of $\cos c$
+is the same as that of $\cos 81^\circ\, 45'\, 36''$.
+\begin{center}
+\begin{tabular}{rr@{}}
+$L \cos 81^\circ\, 45'\, 36''$ = & $ 9.1563065$ \\
+$L \tan 55^\circ\, 32'\, 45''$ = & $10.1636102$ \\
+\cline{2-2}
+$L \cot (180^\circ - B) + 10$  = & $19.3199167$ \\
+$180^\circ - B =            $    & $ 78^\circ\, 12'\, 4''$ \\
+$            B =            $    & $101^\circ\, 47'\, 56''$.
+\end{tabular}
+\end{center}
+%-----File: 161.png------------------------------------------------
+
+To find $b$ we have
+\[
+\tan b = \tan c \cos A.
+\]
+
+Here $\tan c$ is \textit{negative}; and therefore $\tan b$ will be negative
+and $b$ greater than a quadrant.
+\begin{center}
+\begin{tabular}{rr@{}}
+$L \tan 81^\circ\, 45'\, 36'' =$ & $ 10.8391867 $ \\
+$L \cos 55^\circ\, 32'\, 45'' =$ & $  9.7526221 $ \\
+\cline{2-2}
+$L \tan (180^\circ- b) + 10   =$ & $ 20.5918088 $ \\
+$180^\circ - b                =$ & $ 75^\circ 38'\, 32''$ \\
+$b                            =$ & $104^\circ\, 21'\, 28''$.
+\end{tabular}
+\end{center}
+
+\paragraph{207.} Given $A = 46^\circ\, 15'\, 25''$, $C = 90^\circ$, $a = 42^\circ\, 18'\, 45''$.
+
+To find $c$ we have
+\begin{center}
+$\sin c=\dfrac{\sin a}{\sin A},$\\[1ex]
+$L \sin c = 10 + L \sin a - L \sin A,$ \\
+\begin{tabular}{rr@{}}
+$10 + L \sin 42^\circ\, 18'\, 45'' =$ & $19.8281272$\\
+$L \sin 46^\circ\, 15'\, 25''      =$ & $ 9.8588065$\\
+\cline{2-2}
+$L \sin c                          =$ & $9.9693207$\\
+$c = 68^\circ\, 42'\, 59'' $\text{ or }& $111^\circ\, 17'\, 1''$.
+\end{tabular}
+\end{center}
+
+To find $b$ we have
+\begin{center}
+$\sin b = \tan a \cot A,$ \\[1ex]
+\begin{tabular}{rr@{}}
+$L \tan 42^\circ\, 18'\, 45'' =$ & $9.9591983$\\
+$L \cot 46^\circ\, 15'\, 25'' =$ & $9.9809389$\\
+\cline{2-2}
+$L \sin b + 10                =$ & $19.9401372$\\
+$b = 60^\circ\, 36'\, 10'' $\text{ or }& $119^\circ\, 23'\, 50''$.
+\end{tabular}
+\end{center}
+%-----File: 162.png------------------------------------------------
+
+To find $B$ we have
+\begin{center}
+$\sin B = \dfrac{\cos A}{\cos a},$\\[1ex]
+$L\sin B = 10 + L\cos A - L\cos a,$\\
+\begin{tabular}{rr@{}}
+$10 + L\cos 46^\circ\, 15'\, 25'' =$ & $19.8397454$ \\
+$L\cos 42^\circ\, 18'\, 45''      =$ & $ 9.8689289$ \\
+\cline{2-2}
+$L\sin B                          =$ & $ 9.9708165$ \\
+$B = 69^\circ\, 13'\, 47''$\text{ or }& $ 110^\circ\, 46'\, 13''$.
+\end{tabular}
+\end{center}
+
+\section*{\centering\normalfont\large%
+\textit{Oblique-Angled Triangles.}}
+
+\paragraph{208.} Given $a = 70^\circ\, 14'\, 20''$, $b = 49^\circ\, 24'\, 10''$, $c = 38^\circ\, 46'\, 10''$.
+We shall use the formula given in Art.\ 45,
+\[
+\tan\tfrac12 A = \Surd{\left\{\frac{\sin(s - b)\sin(s - c)}{\sin s \sin(s - a)}\right\}}.
+\]
+\begin{flalign*}
+&\text{\indent Here}& s &= 79^\circ\, 12'\, 20'',&\phantom{\indent Here}\\
+&&s - a &= 8^\circ\, 58',\\
+&&s - b &= 29^\circ\, 48'\, 10'',\\
+&&s - c &= 40^\circ\, 26'\, 10''.
+\end{flalign*}
+\[
+\begin{array}{r@{\:}r}
+L\sin 29^\circ\, 48'\, 10'' = & \phantom{1}9.6963704\\[.3ex]
+L\sin 40^\circ\, 26'\, 10''= & \phantom{1}9.8119768 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  &                  19.5083472\\[1ex]
+L\sin 79^\circ\, 12'\, 20''= & \phantom{1}9.9922465\\[.3ex]
+L\sin 8^\circ\, 58' =& \phantom{1}9.1927342 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+&19.1849807\\[1ex]
+&19.5083472\\[.3ex]
+&19.1849807 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+\multicolumn{2}{r}{ 2\,\underline{)\: .3233665}} \\[.6ex]
+\multicolumn{2}{r}{L\tan\tfrac12 A - 10 = \phantom{19}.1616832}
+\end{array}
+\]
+\begin{align*}
+  \tfrac12 A&= \phantom{0}55^\circ\, 25'\, 38''\\
+  A &= 110^\circ\, 51'\, 16''.
+\end{align*}
+%-----File: 163.png------------------------------------------------
+
+Similarly to find $B$,
+\[
+\begin{array}{r@{\:}r}
+L \sin 8^\circ\, 58' = & \phantom{1} 9.1927342\\[.3ex]
+L \sin 40^\circ\, 26'\, 10'' = & \phantom{1} 9.8119768\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+&19.0047110\\[1ex]
+L \sin 79^\circ\, 12'\, 20'' = & \phantom{1} 9.9922465\\[.3ex]
+L \sin 29^\circ\, 48'\, 10'' = & \phantom{1} 9.6963704\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+&19.6886169\\[1ex]
+&19.0047110\\[.3ex]
+&19.6886169\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+\multicolumn{2}{r}{ 2\,\underline{)\: \overline{1}.3160941}}\\[.6ex]
+L \tan \tfrac{1}{2}B - 10 = & \phantom{1} \overline{1}.6580470\\[.3ex]
+L \tan \tfrac{1}{2}B = & \phantom{1} 9.6580470\\[.6ex]
+\multispan{2}{\hfil$\begin{aligned}
+\tfrac{1}{2}B &= 24^\circ\, 28'\, 2''\\
+B &= 48^\circ\, 56'\, 4''.
+\end{aligned}$\hfil}
+\end{array}
+\]
+Similarly to find $C$,
+\[
+\begin{array}{r@{\:}r}
+L \sin 8^\circ\, 58' = & \phantom{1} 9.1927342\\
+L \sin 29^\circ\, 48'\, 10'' = & \phantom{1} 9.6963704\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+&18.8891046\\[1ex]
+L \sin 79^\circ\, 12'\, 20'' = & \phantom{1} 9.9922465\\[.3ex]
+L \sin 40^\circ\, 26'\, 10'' = & \phantom{1} 9.8119768\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+&19.8042233\\[1ex]
+&18.8891046\\[.3ex]
+&19.8042233\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+\multicolumn{2}{r}{ 2\,\underline{)\: \overline{1}.0848813}} \\[.6ex]
+L \tan \tfrac{1}{2}C-10 = & \phantom{1} \overline{1}.5424406\\[.3ex]
+L \tan \tfrac{1}{2}C = & \phantom{1} 9.5424406\\[.6ex]
+\multispan{2}
+{\hfil$\begin{aligned}
+\tfrac{1}{2}C &= 19^\circ\, 13'\, 24''\\
+C &= 38^\circ\, 26'\, 48''.
+\end{aligned}$\hfil}
+\end{array}
+\]
+%-----File: 164.png------------------------------------------------
+
+\paragraph{209.} Given
+$a =  68^\circ\, 20'\, 25''$,
+$b =  52^\circ\, 18'\, 15''$,
+$C = 117^\circ\, 12'\, 20''$.
+
+By Art.\ 82,
+\begin{gather*}
+\begin{aligned}
+  \tan\tfrac12 (A + B)
+&= \dfrac{\cos\tfrac12 (a - b)}{\cos\tfrac12 (a + b)}\cot\tfrac12 C,\\
+  \tan\dfrac12 (A - B)
+&= \dfrac{\sin\tfrac12 (a - b)}{\sin\tfrac12 (a + b)}\cot\tfrac12 C.
+\end{aligned}\\
+  \tfrac12(a - b) =  8^\circ\,  1'\,  5'',\
+  \tfrac12(a + b) = 60^\circ\, 19'\, 20'',\
+  \tfrac12C       = 58^\circ\, 36'\, 10''.
+\\
+\begin{array}{r@{\:}r}
+  L\cos  8^\circ\,  1'\, 5''  =& \phantom{1} 9.9957335  \\[.3ex]
+  L\cot 58^\circ\, 36'\, 10'' =& \phantom{1} 9.7855690  \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                  &            19.7813025  \\[.6ex]
+  L\cos 60^\circ\, 19'\, 20'' =& \phantom{1} 9.6947120  \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac12 (A + B) =& 10.0865905  \\[.3ex]
+       \tfrac12 (A + B) =&\multicolumn{1}{@{\:}l}{ 50^\circ\, 40'\, 28''}\\[1ex]
+  L\sin  8^\circ\,  1'\,  5'' =& \phantom{1} 9.1445280  \\[.3ex]
+  L\cot 58^\circ\, 36'\, 10'' =& \phantom{1} 9.7855690  \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                  &            18.9300970  \\[.6ex]
+  L\sin 60^\circ\, 19'\, 20'' =& \phantom{1} 9.9389316  \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac12 (A - B) =& \phantom{1} 8.9911654\\[.3ex]
+\tfrac12 (A - B) =& \multicolumn{1}{@{\:}l}{ 5^\circ\, 35'\, 47''.}
+\end{array}
+\end{gather*}
+
+Therefore\quad\hfill
+$A = 56^\circ\, 16'\, 15''$, \quad
+$B = 45^\circ\,  4'\, 41''$. \hfill\phantom{\indent Therefore}
+
+If we proceed to find $c$ from the formula
+\[
+\sin c = \dfrac{\sin a \sin C}{\sin A},
+\]
+since $\sin C$ is greater than $\sin A$ we shall obtain two values for $c$
+both greater than $a$, and we shall not know which is the value to
+be taken.
+%-----File: 165.png------------------------------------------------
+
+We shall therefore determine $c$ from formula (1) of Art.~54,
+which is free from ambiguity,
+\begin{gather*}
+\cos \tfrac{1}{2} c = \dfrac{\cos \tfrac{1}{2} (a + b) \sin \tfrac{1}{2}C}{\cos\tfrac{1}{2}(A+B)}, \\
+\begin{array}{r@{\:}r@{}c}
+  L \cos 60^\circ\, 19'\, 20'' = & 9.6947120 \\[.3ex]
+  L \sin 58^\circ\, 36'\, 10'' = & 9.9312422 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                 & 19.6259542 \\[.6ex]
+  L \cos 50^\circ\, 40'\, 28'' = & 9.8019015 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L \cos \tfrac{1}{2} c        = & 9.8240527 \\[.3ex]
+  \tfrac{1}{2} c =
+&\multicolumn{2}{@{\:}l}{ 48^\circ\, 10'\, 22''} \\[.3ex]
+  c =
+&\multicolumn{2}{@{\:}l}{ 96^\circ\, 20'\, 44''.}
+\end{array}
+\end{gather*}
+
+Or we may adopt the second method of Art.~82. First, we
+determine $\theta$ from the formula $\tan \theta = \tan b \cos C$.
+
+Here $\cos C$ \textit{is negative}, and therefore $\tan \theta$ will be negative,
+and $\theta$ greater than a right angle. The numerical value of $\cos C$ is
+the same as that of $\cos 62^\circ\, 47'\, 40''$.
+\[
+\begin{array}{r@{\:}r@{}c}
+  L \tan 52^\circ\, 18'\, 15''       = & 10.1119488 \\[.3ex]
+  L \cos 62^\circ\, 47'\, 40''       = &  9.6600912 \\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L \tan (180^\circ\, - \theta) + 10 = & 19.7720400 \\[.3ex]
+  180^\circ\, - \theta =
+&\multicolumn{2}{@{\:}l}{ 30^\circ\, 36'\, 33'',} \\[.3ex]
+\text{therefore }\theta =
+&\multicolumn{2}{@{\:}l}{ 149^\circ\, 23'\, 27''.}
+\end{array}
+\]
+
+Next, we determine $c$ from the formula
+\[
+\cos c = \dfrac{\cos b \cos (a - \theta)}{\cos \theta}.
+\]
+
+Here $\cos \theta$ \textit{is negative}, and therefore $\cos c$ will be negative, and
+$c$ will be greater than a right angle. The numerical value of
+$\cos \theta$ is the same as that of $\cos (180^\circ - \theta)$, that is, of $\cos 30^\circ\, 36'\, 33''$;
+and the value of $\cos (a - \theta)$ is the same as that of $\cos (\theta - a)$, that
+is, of $\cos 81^\circ\, 3'\, 2''$.
+%-----File: 166.png------------------------------------------------
+\[
+\begin{array}{r@{\:}r}
+  L\cos 52^\circ\, 18'\, 15'' = & \phantom{1} 9.7863748 \\[.3ex]
+  L\cos 81^\circ\,  3'\,  2'' = & \phantom{1} 9.1919060\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                   &            18.9782808\\[.6ex]
+  L\cos 30^\circ\, 36'\, 33'' = & \phantom{1} 9.9348319\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\cos(180^\circ - c) = & \phantom{1} 9.0434489\\[.3ex]
+        180^\circ - c  =
+&\multicolumn{1}{@{\:}l}{ 83^\circ\, 39'\, 17''}\\[.3ex]
+  c =
+&\multicolumn{1}{@{\:}l}{ 96^\circ\, 20'\, 43''.}
+\end{array}
+\]
+
+Thus by taking only the nearest number of seconds in the
+tables the two methods give values of $c$ which differ by $1''$; if,
+however, we estimate fractions of a second both methods will
+agree in giving about $43\tfrac12$ as the number of seconds.
+
+\paragraph{210.} Given
+$a = 50^\circ\, 45'\, 20''$,
+$b = 69^\circ\, 12'\, 40''$,
+$A = 44^\circ\, 22'\, 10''$.
+\begin{flalign*}
+&\text{\indent By Art.\ 84,}& \sin B &= \dfrac{\sin b}{\sin a}\sin A,&\phantom{\indent By Art.\ 84,}
+\end{flalign*}
+\[
+\begin{array}{r@{\:}r@{}c}
+\phantom{\ 4,\ or\ 122\ 25\ 8.6.}
+  L\sin 69^\circ\, 12'\, 40'' = & \phantom{1} 9.9707626\\[.3ex]
+  L\sin 44^\circ\, 22'\, 10'' = & \phantom{1} 9.8446525\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                   &            19.8154151\\[.6ex]
+  L\sin 50^\circ\, 45'\, 20'' = & \phantom{1} 9.8889956\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\sin B                        = & \phantom{1} 9.9264195\\[.3ex]
+  B =
+&\multicolumn{2}{@{\:}l}{ 57^\circ\, 34'\, 51''.4,\text{ or }122^\circ\, 25'\, 8''.6.}
+\end{array}
+\]
+
+In this case there will be two solutions; see Art.\ 86. We
+will calculate $C$ and $c$ by Napier's analogies,
+\begin{align*}
+\tan \tfrac12 C &= \dfrac{\cos \tfrac12 (b - a)}{\cos \tfrac12 (b + a)}\cot \tfrac12 (B + A),\\[2ex]
+\tan \tfrac12 c &= \dfrac{\cos \tfrac12 (B + A)}{\cos \tfrac12 (B - A)}\tan \tfrac12 (b + a).
+\end{align*}
+
+First take the smaller value of $B$; thus
+\[
+  \tfrac12 (B + A) = 50^\circ\, 58'\, 30''.7,\
+  \tfrac12 (B - A) =  6^\circ\, 35'\, 20''.7,
+\]
+%-----File: 167.png------------------------------------------------
+\[
+\begin{array}{r@{\:}r@{}c}
+  L\cos  9^\circ\, 13'\, 40''   = & \phantom{1} 9.9943430  \\[.3ex]
+  L\cot 50^\circ\, 58'\, 30''.7 = & \phantom{1} 9.9087536\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                     &            19.9030966  \\[.6ex]
+  L\cos 59^\circ\, 59'          = & \phantom{1} 9.6991887\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac{1}{2} C                  = &            10.2039079\\[.3ex]
+       \tfrac{1}{2} C                  =
+&\multicolumn{2}{@{}l}{ 57^\circ\, 58'\, 55''.3}\\[.3ex]
+                C                  =
+&\multicolumn{2}{@{}l}{ 115^\circ\, 57'\, 50''.6.}\\[1ex]
+  L\cos 50^\circ\, 58'\, 30''.7 = & \phantom{1} 9.7991039  \\[.3ex]
+  L\tan 59^\circ\, 59'          = &            10.2382689\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                     &            20.0373728  \\[.6ex]
+  L\cos 6^\circ\, 36'\, 20''.7  = & \phantom{1} 9.9971072\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac{1}{2} c                  = &            10.0402656\\[.3ex]
+       \tfrac{1}{2} c                  =
+&\multicolumn{2}{@{}l}{ 47^\circ\, 39'\, 8''.2}\\[.3ex]
+                c                  =
+&\multicolumn{2}{@{}l}{ 95^\circ\, 18'\, 16''.4.}
+\end{array}
+\]
+Next take the larger value of $B$; thus
+\begin{gather*}
+  \tfrac{1}{2}(B + A) = 83^\circ\, 23'\, 39''.3,\quad
+  \tfrac{1}{2}(B - A) = 39^\circ\,  1'\, 29''.3.\\
+\begin{array}{r@{\:}r@{}c}
+  L\cos \phantom{0}9^\circ\, 13'\, 40''\phantom{.0} =
+& \phantom{1} 9.9943430\\[.3ex]
+  L\cot 83^\circ\, 23'\, 39''.3 = & \phantom{1} 9.0637297\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                     &            19.0580727  \\[.6ex]
+  L\cos 59^\circ\, 59'          = & \phantom{1} 9.6991887\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac{1}{2} C = & \phantom{1} 9.3588840\\[.3ex]
+       \tfrac{1}{2} C =
+&\multicolumn{2}{@{}l}{ 12^\circ\, 52'\, 15''.8}\\[.3ex]
+                C =
+&\multicolumn{2}{@{}l}{ 25^\circ\, 44'\, 31''.6.}\\[1ex]
+  L\cos 83^\circ\, 23'\, 39''.3 = & \phantom{1} 9.0608369  \\[.3ex]
+  L\tan 59^\circ\, 59'\,\phantom{00''.0} =   & 10.2382689\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+                                                & 19.2991058  \\[.6ex]
+  L\cos 39^\circ\, 1'\, 29''.3  = & \phantom{1} 9.8903494\\[-1.5ex]
+&\hrulefill\\[-.3ex]
+  L\tan\tfrac{1}{2} c = & \phantom{1} 9.4087564\\[.3ex]
+       \tfrac{1}{2} c =
+&\multicolumn{2}{@{}l}{ 14^\circ\, 22'\, 32''.6} \\[.3ex]
+                c =
+&\multicolumn{2}{@{}l}{ 28^\circ\, 45'\, 5''.2}
+\end{array}
+\end{gather*}
+%-----File: 168.png------------------------------------------------
+
+The student can obtain more examples, which can be easily
+verified, from those here worked out, by interchanging the given
+and required quantities, or by making use of the polar triangle.
+
+
+\section*{\centering\normalsize EXAMPLES.}
+\allowdisplaybreaks
+\begin{flalign*}
+\multispan{1}{\indent 1.\hfill\quad Given $b$}%
+& = 137^\circ\, 3'\, 48'',\
+A = 147^\circ\, 2'\, 54'',\
+C =  90^\circ.
+&&\\
+\textit{Results.}\quad c
+& =  47^\circ\, 57'\, 15'',\
+a = 156^\circ\, 10'\, 34'',\
+B = 113^\circ\, 28'.
+&&\\[1ex]
+\multispan{1}{\indent 2.\hfill Given $c$}%
+& = 61^\circ\,  4'\, 56'',\
+a = 40^\circ\, 31'\, 20'',\
+C = 90^\circ.
+&&\\
+\textit{Results.}\quad b
+& = 50^\circ\, 30'\, 29'',\
+B = 61^\circ\, 50'\, 28'',\
+A = 47^\circ\, 54'\, 21''.
+&&\\[1ex]
+\multispan{1}{\indent 3.\hfill Given $A$}%
+& = 36^\circ,\
+B = 60^\circ,\
+C = 90^\circ.
+&&\\
+\textit{Results.}\quad a
+& = 20^\circ\, 54'\, 18''.5,\
+b = 31^\circ\, 43'\,  3'',\
+c = 37^\circ\, 21'\, 38''.5.
+&&\\[1ex]
+\multispan{1}{\indent 4.\hfill Given $a$}%
+& = 59^\circ\, 28'\, 27'',\
+A = 66^\circ\,  7'\, 20'',\
+C = 90^\circ.
+&&\\
+\textit{Results.}\quad c
+& = 70^\circ\, 23'\, 42'',\
+b = 48^\circ\, 39'\, 16'',\
+B = 52^\circ\, 50'\, 20'',
+&&\\
+\indent\rlap{~or,}\phantom{\textit{Results.}\ {}}c
+& = 109^\circ\, 36'\, 18'',\
+b = 131^\circ\, 20'\, 44'',\
+B = 127^\circ\,  9'\, 40''.
+&&\\[1ex]
+\multispan{1}{\indent 5.\hfill Given $c$}%
+& =  90^\circ,\
+a = 138^\circ\,  4',\
+b = 109^\circ\, 41'.
+&&\\
+\textit{Results.}\quad C
+& = 113^\circ\, 28'\,  2'',\
+A = 142^\circ\, 11'\, 38'',\
+B = 120^\circ\, 15'\, 57''.
+&&\\[1ex]
+\multispan{1}{\indent 6.\hfill Given $c$}%
+& =  90^\circ,\
+A = 131^\circ\, 30',\
+B = 120^\circ\, 32'.
+&&\\
+\textit{Results.}\quad C
+& = 109^\circ\, 40'\, 20'',\
+a = 127^\circ\, 17'\, 51'',\
+b = 113^\circ\, 49'\, 31''.
+&&\\[1ex]
+\multispan{1}{\indent 7.\hfill Given $a$}%
+& = 76^\circ\, 35'\, 36'',\
+b = 50^\circ\, 10'\, 30'',\
+c = 40^\circ\,  0'\, 10''.
+&&\\
+\textit{Results.}\quad A
+& = 121^\circ\, 36'\, 20'',\
+B =  42^\circ\, 15'\, 13'',\
+C =  34^\circ\, 15'\,  3''.
+&&\\[1ex]
+\multispan{1}{\indent 8.\hfill\quad Given $A$}%
+&= 129^\circ\,  5'\, 28'',\
+B= 142^\circ\, 12'\, 42'',\
+C= 105^\circ\,  8'\, 10''.
+&&\\
+\textit{Results.}\quad a
+& = 135^\circ\, 49'\, 20'',\
+b = 144^\circ\, 37'\, 15'',\
+c =  60^\circ\,  4'\, 54''.  &&
+\end{flalign*}
+\end{spacing}
+\end{mainmatter}
+\vfill
+\begin{center}
+\noindent\hrulefill\nopagebreak\par
+\scriptsize
+CAMBRIDGE: PRINTED BY C.~J.\ CLAY, M.A.\ AND SON, AT THE UNIVERSITY PRESS.
+\end{center}
+%-----File: 169.png------------------------------------------------
+
+
+\chapter*{\centering\Large
+WORKS\\
+{\footnotesize BY}\\
+\textbf{I.\ TODHUNTER, D.Sc., F.R.S.}
+}
+
+\noindent\hangindent\parindent{\LARGE Natural Philosophy for Beginners. In two parts.}
+\linebreak
+Part~I\@. The Properties of Solid and Fluid Bodies. With numerous
+Examples. 18mo. 3\textit{s}.\ 6\textit{d}. Part~II\@. Sound, Light, and Heat. 3\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Euclid for Colleges and Schools. New Edition.}
+\linebreak
+18mo.\ cloth. 3\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to Exercises in Euclid. Crown 8vo.\ cloth.}
+\linebreak
+6\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Mensuration for Beginners. With numerous}
+\linebreak
+Examples. New Edition. 18mo. 2\textit{s}.\ 6\textit{d}. Key, 7\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Algebra for Beginners. With numerous Examples.}
+\linebreak
+New Edition. 18mo.\ cloth. 2\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to the Algebra for Beginners. New}
+\linebreak
+Edition. Crown 8vo.\ cloth. 6\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Trigonometry for Beginners. With numerous}
+\linebreak
+Examples. New Edition. 18mo.\ cloth. 2\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to the Trigonometry for Beginners.}
+\linebreak
+Crown 8vo.\ cloth. 8\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Mechanics for Beginners. With numerous Exam-}
+\linebreak
+ples. New Edition. 18mo.\ cloth. 4\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to the Mechanics for Beginners. Crown}
+\linebreak
+8vo.\ cloth. 6\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Algebra for the use of Colleges and Schools.}
+\linebreak
+With numerous Examples. New Edition. Crown 8vo.\ cloth. 7\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to the Algebra for the use of Colleges}
+\linebreak
+and Schools. New Edition. Crown 8vo.\ cloth. 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A Treatise on the Theory of Equations. New}
+\linebreak
+Edition. Crown 8vo.\ cloth. 7\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+%-----File: 170.png------------------------------------------------
+
+
+\noindent\hangindent\parindent{\LARGE Plane Trigonometry for Colleges and Schools.}
+\linebreak
+With numerous Examples. New Edition. Crown 8vo.\ cloth. 5\textit{s}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Key to the Plane Trigonometry for Colleges}
+\linebreak
+and Schools. Crown 8vo.\ cloth. 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A Treatise on Spherical Trigonometry for the}
+\linebreak
+use of Colleges and Schools. With numerous Examples. New
+Edition. Crown 8vo.\ cloth. 4\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A Treatise on Conic Sections. With numerous}
+\linebreak
+Examples. New Edition. Crown 8vo.\ cloth. 7\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A Treatise on the Differential Calculus. With}
+\linebreak
+numerous Examples. New Edition. Crown 8vo.\ cloth. 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A Treatise on the Integral Calculus. With}
+\linebreak
+numerous Examples. New Edition. Crown 8vo.\ cloth. 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE Examples of Analytical Geometry of Three}
+\linebreak
+Dimensions. New Edition. Crown 8vo.\ cloth. 4\textit{s}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE An Elementary Treatise on Laplace's, Lam\'e's,}
+\linebreak
+and Bessel's Functions. Crown 8vo.\ 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE A History of the Mathematical Theory of Probabil-}
+\linebreak
+ity, from the time of Pascal to that of Laplace. 8vo.\ cloth. 18\textit{s}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE The Conflict of Studies, and other Essays on}
+\linebreak
+Subjects connected with Education. 8vo.\ 10\textit{s}.\ 6\textit{d}.\\[.5ex]
+
+\noindent\hangindent\parindent{\LARGE William Whewell, D.D.\ Master of Trinity}
+\linebreak
+College, Cambridge. An account of his writings with selections
+from his literary and scientific correspondence. In Two Volumes.
+Demy 8vo.\ cloth. 25\textit{s}.\\[.5ex]
+
+
+\vfill
+\begin{center}
+\noindent\hrulefill\nopagebreak\par
+\small
+MACMILLAN AND CO\@. LONDON.
+\end{center}
+
+\newpage
+\pagestyle{empty}
+
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+\pagenumbering{gobble}
+\begin{verbatim}
+End of the Project Gutenberg EBook of Spherical Trigonometry, by I. Todhunter
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diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
diff --git a/README.md b/README.md
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index 0000000..2058a14
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #19770 (https://www.gutenberg.org/ebooks/19770)