Spherical Trigonometry, for the Use of Colleges and Schools
Miscellaneous Propositions
Excerpts
Miscellaneous Propositions
*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.*
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*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.*
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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.
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Let $P$ denote the pole of the inscribed circle, and $Q$ the pole of the circumscribed circle of a triangle $ABC$;
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then the spherical triangle which corresponds to the three planes $LPM$, $MPN$, $NPL$ is the *polar triangle* of the spherical triangle which corresponds to the solid angle at $O$.
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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 each force is as the sine of the solid angle formed by the directions of the other three.
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the volume of a tetrahedron is one sixth of the product of three edges into the sine of the solid angle which they form.
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The result, however, is generally true, even in cases in which the condition required by the demonstration of Art. 150 is not satisfied.
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We begin with a theorem which is due to Cauchy.
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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.
Equations
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\cos A' = \sin (S - A) \cos \frac{a}{2}The cosine of an angle of the chordal triangle equals the sine of S minus the corresponding angle times the cosine of half the side (example 13).
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\dfrac{\sin Pa \cos PA}{\sin Aa} + \dfrac{\sin Pb \cos PB}{\sin Bb} + \dfrac{\sin Pc \cos PC}{\sin Cc} = 1For a point P inside a spherical triangle with cevian great circles through P, the three weighted sine-cosine ratios sum to one (example 16).
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\cot \tfrac{1}{2}E = \cot \tfrac{1}{2}\theta \cot \tfrac{1}{2}c \operatorname{cosec} \phi + \cot \phiThe cotangent of half the spherical excess equals a combination of the cotangents of half the sides and the angle at A, for a triangle with given base and area.
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\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} EThe vertex of a spherical triangle of given base and area satisfies a linear trigonometric equation in theta and phi, which shows its locus is a circle.
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\beta = \tfrac{1}{2} E - \frac{\pi}{2}The angular co-ordinate beta of the pole of the locus circle equals half the spherical excess minus a right angle.
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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)The equation of the great circle that bisects the base AB at right angles, in the polar co-ordinates theta and phi.
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\cos PAQ = cos\tfrac{1}{2}(B-C)The angle between the arcs from A to the poles P and Q of the inscribed and circumscribed circles equals the cosine of half the difference of the angles B and C.
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\cos PQ = \cos PA \cos QA + \sin PA \sin QA \cos \tfrac{1}{2}(B-C)The spherical cosine rule applied to the triangle with vertices P, A, Q gives the cosine of the distance PQ between the two poles.
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\sin PA = \frac{\sin PE}{\sin PAE} = \frac{\sin r}{\sin\tfrac{1}{2}A}The sine of the distance PA from the vertex to the pole of the inscribed circle equals the sine of the inradius divided by the sine of half the angle A.
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\cos PQ = \cos R \cos r \cos(s-a) + \sin R \sin r \sin \tfrac{1}{2}(b+c) \operatorname{cosec} \tfrac{1}{2}aThe cosine of the distance between the two poles is expressed in terms of the inradius r, circumradius R, and the sides of the triangle.
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\cot r = \frac{\sin s}{n}The cotangent of the inradius equals the sine of the semi-perimeter divided by n, where n is a quantity defined from the sides.
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\tan R = \frac{2 \sin \frac{1}{2} a \sin \tfrac{1}{2} b \sin \tfrac{1}{2} c}{n}The tangent of the circumradius R equals twice the product of the sines of half the sides, divided by n.
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\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}aDividing the cosine formula for PQ by cos R sin r gives a form in cotangents and tangents of the radii.
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\cos^2 PQ = \cos^2 R \sin^2 r + \cos^2 (R-r)The squared cosine of the distance between the inscribed-circle and circumscribed-circle poles is given in terms of the circumradius R and inradius r.
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\sin^2 PQ = \sin^2 (R-r) - \cos^2 R \sin^2 rThe squared sine of the distance between the poles of the inscribed and circumscribed circles is given in terms of R and r.
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\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}BThe cosine of the distance between the circumscribed-circle pole Q and the escribed-circle pole Q1 is given by a spherical cosine relation.
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\cos^2 QQ_1 = \cos^2 R \sin^2 r_1 + \cos^2 (R + r_1)The squared cosine of the distance QQ1 between the circumscribed and escribed poles in terms of R and r1.
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\sin^2 QQ_1 = \sin^2 (R + r_1) - \cos^2 R \sin^2 r_1The squared sine of the distance QQ1 between the circumscribed and escribed poles in terms of R and r1.
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\sin BQ = \sin CQThe sines of the arcs from B and from C to the point Q, where the arc through the midpoints meets BC produced, are equal.
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BQ + CQ = \piThe arcs BQ and CQ together make a half great circle, so the fixed point Q lies a quadrant beyond the midpoint of the base.
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DQ = \tfrac{1}{2} (BQ + CQ) = \tfrac{1}{2} \piThe distance from the midpoint D of BC to the fixed point Q is a quadrant.
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\sin BD \sin CE \sin AF= \sin CD \sin AE \sin BFIf three arcs from the vertices of a spherical triangle pass through a common point, the products of the sines of alternate segments of the sides are equal.
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\dfrac{\sin BD}{\sin CD}\, \dfrac{\sin CE}{\sin AE}\, \dfrac{\sin AF}{\sin BF} = 1The product of the three sine ratios of the divided sides equals one (Ceva-type condition for concurrent arcs on a sphere).
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\surd(1-\cos^2 \alpha-\cos^2 \beta-\cos^2 \gamma+2\cos \alpha \cos \beta \cos \gamma)The sine of a solid angle formed by three plane angles alpha, beta, gamma is the square root of the stated expression, which lies between zero and one.
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2E=3a+4b+5c+6d+ \ldots\ldotsEach edge belongs to two faces, so twice the number of edges equals the sum of sides over all faces.
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2E= 3 \alpha + 4 \beta + 5 \gamma + 6 \delta + \ldots\ldotsEach edge terminates at two solid angles, so twice the number of edges equals the sum of plane-angle counts over all solid angles.
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F=a+b+c+d+ \ldots\ldotsThe total number of faces is the sum of the numbers of faces with each number of sides.
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S= \alpha + \beta + \gamma + \delta + \ldots\ldotsThe total number of solid angles is the sum of the numbers of solid angles with each number of plane angles.
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2E-3F = b + 2c + 3d + \ldots\ldotsTwice the number of edges minus three times the number of faces equals a sum with nonnegative coefficients, so 2E cannot be less than 3F.
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2E-3S = \beta + 2 \gamma + 3 \delta + \ldots\ldotsTwice the number of edges minus three times the number of solid angles equals a sum with nonnegative coefficients, so 2E cannot be less than 3S.
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2F + 2S=4 + 2ETwice the sum of faces and solid angles equals four plus twice the number of edges, which is Euler's relation rewritten.
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2 (\alpha + \beta + \gamma + \delta + \ldots) - (a + 2b + 3c + 4d + \ldots) = 4Combining the face and solid-angle counts gives an identity relating solid-angle counts to face-side counts.
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2 (a + b + c + d + \ldots) - (\alpha + 2 \beta + 3 \gamma + 4 \delta + \ldots) = 4The dual identity relating face counts to solid-angle counts holds for any polyhedron.
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a + \alpha - (c + \gamma) - 2 (d + \delta) - 3 (e + \epsilon) - \ldots\ldots = 8Adding the two identities shows the number of triangular faces plus triangular solid angles is at least eight.
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3a + 2b + c - e - 2f - \ldots\ldots -2\beta - 4\gamma - \ldots\ldots = 12Eliminating alpha gives a relation showing 3a + 2b + c cannot be less than 12.
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\mathrm{F + S = E + 1 }For a network of rectilineal figures not forming a closed surface, faces plus corner points equals edges plus one.
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F-1+S = E + 1Removing one face of a polyhedron yields a network to which Cauchy's theorem applies.
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F + S = E + 2For any polyhedron, faces plus solid angles equals edges plus two.
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1 + e + e' = s + s' + FSplitting edges, corners and faces into those on the bounding contour and those within it gives a relation for the network.
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1 + e' = s' + FSince the contour edges and corners are equal in number, the interior edges and corners satisfy 1 + e' = s' + F.
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\mathrm{S + F = E + P + 1 }If a polyhedron is decomposed into P polyhedrons, solid angles plus faces equals edges plus P plus one.
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\tan c = \frac{\cot A \cot a + \cot B \cot b} {\cot a \cot b - \cos A \cos B}The tangent of side c of a spherical triangle is given in terms of two angles and their opposite sides (example 4).
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\cos \theta \sin(b - c) + \cos \phi \sin(c - a) + \cos \psi \sin(a - b) = 0For the point where the angle bisectors of a spherical triangle meet, the stated weighted cosine sum vanishes (example 5).
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\cos PA \cos BC = \cos PB \cos CA = \cos PC \cos ABFor the point P where the great circles through the vertices and the poles of the opposite sides meet, the products of cosines are equal (example 6).
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\tan \alpha \tan \alpha' = \tan \beta \tan \beta' = \tan \gamma \tan \gamma'For the three cevian arcs perpendicular to the sides, the products of the tangents of their segments are equal (example 8).
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\frac{\cos p}{\cos \alpha \cos \alpha'} = \frac{\cos q}{\cos \beta \cos \beta' } = \frac{\cos r}{\cos \gamma \cos \gamma'}For the perpendicular arcs p, q, r of a spherical triangle, the ratios of their cosines to products of cosines of their segments are equal (example 8).
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\frac{\sin \alpha}{\sin \alpha'} = 2 \cos \frac{a}{2}For an arc from a vertex to the midpoint of the opposite side, the ratio of the sines of its two parts equals twice the cosine of half that side (example 9).
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\cos AQ \sin \frac{a}{2} = \sin \frac{c - b}{2} \sin \frac{c + b}{2}For the arc bisecting AB and AC meeting BC produced at Q, the cosine of AQ times the sine of half a equals a product of sines (example 10).
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\sin AB \sin CD \cos P = \sin AD \sin BC \cos Q = \sin AC \sin BD \cos RFor a spherical quadrilateral with the stated intersection points, the three products of sines and cosines are equal (example 12).
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a \cos AP + b \cos BP = s \cos SPFor fixed points A and B on a sphere and constants a and b, a fixed point S exists on AB such that the combined cosine relation holds for every P (example 19).
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a \cos AP + b \cos BP + c \cos CP + \ldots = \text{constant}For fixed points A, B, C, ... on a sphere with given constants, the locus of P satisfying a constant weighted cosine sum is a circle (example 20).
Problems
Exercise XV
Exercise XV, problem 1, p. 155
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.
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Exercise XV, problem 10, p. 155
The arc of a great circle bisecting the sides $AB$, $AC$ of a spherical triangle cuts $BC$ produced at $Q$: shew that AQ a2 = c - b2 c + b2.
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Exercise XV, problem 11, p. 155
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 $F$.
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Exercise XV, problem 12, p. 155
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 AB CD P = AD BC Q = AC BD R.
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Exercise XV, problem 13, p. 155
If $A'$ be the angle of the chordal triangle which corresponds to the angle $A$ of a spherical triangle, shew that A’ = (S - A) a2.
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Exercise XV, problem 14, p. 155
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.
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Exercise XV, problem 15, p. 155
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 $QPZ$.
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Exercise XV, problem 16, p. 155
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 Pa PAAa + Pb PBBb + Pc PCCc = 1.
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Exercise XV, problem 17, p. 155
$A$ 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 $A$.
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Exercise XV, problem 18, p. 155
From the result given in example 18 of Chapter V. infer the possibility of a regular dodecahedron.
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Exercise XV, problem 19, p. 155
$A$ 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 AP + b BP = s SP, where $s$ is a constant.
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Exercise XV, problem 2, p. 155
$AB$ 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.
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Exercise XV, problem 20, p. 155
$A$, $B$, $C$,…are fixed points on the surface of a sphere; $a$, $b$, $c$,…are given constants. If $P$ be a point on the surface of the sphere, such that a AP + b BP + c CP + …= constant, shew that the locus of $P$ is a circle.
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Exercise XV, problem 3, p. 155
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 multline* ^-1 (+ ) + ^-1 (- ) + ^-1 (+ ) + ^-1 (- ) = , 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.
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Exercise XV, problem 4, p. 155
In any spherical triangle c = A a + B b a b - A B.
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Exercise XV, problem 5, p. 155
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 (b - c) + (c - a) + (a - b) = 0.
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Exercise XV, problem 6, p. 155
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 PA BC = PB CA = PC AB.
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Exercise XV, problem 7, p. 155
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 ADOD, BEOE, CFOF are respectively equal to 1+AB C, 1+BA C, 1+CA B.
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Exercise XV, problem 8, p. 155
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 flalign* &&& ’ = ’ = ’; && [2ex] &and&2$\displaystyle \frac{\cos p}{\cos \alpha \cos \alpha'} = \frac{\cos q}{\cos \beta \cos \beta' } = \frac{\cos r}{\cos \gamma \cos \gamma'}. $ &and & flalign*
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Exercise XV, problem 9, p. 155
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 ’ = 2 a2.
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