Spherical Trigonometry

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.

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 5:

1. Tanr, tan r, tan r2 = tan r sin3s.

Tan R+cot r tan R1 + cotr1

tan R1 + cotr,

tan R ̧ + cot r2 = 1⁄2) (cot r + cotr, + cot r ̧ + cotr ̧).

3. Tan3 R + tan3 R, +tan3 R ̧ + tan3 R ̧

Tan R, tan R ̧ tan R2 = tan R sec3 S.

6. Shew that in an equilateral triangle tan R = 2 tan r.

7. 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.

8. If three small circles be inscribed in a spherical triangle having each of its angles 120°, 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°, and that the centres of the three small circles coincide with the angular points of the polar triangle.

VIII. AREA OF A SPHERICAL TRIANGLE.
SPHERICAL EXCESS.

96. To find the area of a Lune.

A Lune is that portion of the surface of a sphere which is comprised between two great semicircles.

CDEF

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 lunes having equal angles are equal. Then by a process similar to that used in the Sixth Book of Euclid it may be shewn that 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,

Suppose

the radius of the sphere, then the surface is 4πr3

(Integral Calculus, Chap. VII.); thus

97. To find the area of a Spherical Triangle.

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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 semi-circumference. 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 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

triangle ABC + BGDC lune ABDCA = 2Ar3,

triangle ABC + AHEC = lune BCEAB = 2Br2,

triangle ABC + triangle CDE = lune CAFBC= 2Cr2; hence, by addition,

twice triangle ABC + surface of hemisphere = 2 (A + B+ C) r2; therefore triangle ABC = (A + B + C − π) r2.

The expression A+B+C-T is called the spherical excess of

the result obtained may be thus enunciated, the area of a spherical triangle is the same fraction of the surface of the hemisphere as the spherical excess is of four right angles.

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 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. 89. 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 absolutely equal to the corresponding isosceles triangles of the

other set.

99. To find the area of a spherical polygon.

Let n be the number of sides of the polygon, Σ 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 ʼn triangles. Hence, by Art. 97,

area of polygon = (sum of the angles of the triangles – nπ) r2, and the sum of the angles of the triangles is equal to Σ together with the four right angles which are formed round the common vertex ; therefore

area of polygon = {≥ − (n − 2) π} roa.

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.

100. We shall now give some expressions for certain trigonometrical functions of the spherical excess of a triangle. We denote the spherical excess by E, so that E=A+B+ С − π.

101. Cagnoli's Theorem. To shew that

sin § # =√{sin s sin (s — a) sin (s – 6) sin (s — c)}

2 cosa cos bcos c

Sin E = sin (4 + B + C − π) = sin {} (A + B) − } (π — C')}

= sin † (A + B) sin

sin C cos C

cos c