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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 0, 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 A BC + BGDC = lune ABDCA = 2Ar*,
triangle ABC + AHEC = lune BCEAB= 2Br,

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

twice triangle ABC + surface of hemisphere = 2 (A + B+C) 76 ; therefore triangle ABC = (A + B + C-Tre.

The expression A+B+C -7 is called the spherical excess of the triangle; and since

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the result obtained may be thus enunciated : 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.

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. 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 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, 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,

area of polygon = (sum of the angles of the triangles - NT) *, and the sum of the angles of the triangles is equal to together with the four right angles which are formed round the common vertex; therefore

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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+C -7.

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101. Cagnoli's Theorem. To shew that
sin } E = {sin 8 sin (8 – a) sin (s—b) sin (s—c)}

s s 6(
E

2 cos } a cos } b cosc
Sin E = sin }(A + B+C – 7) = sin {}(A + B) – 4 (7-C)}

= sin }(A + B) sin } C - cos } (A + B) cos C
sin C cos C
cos c

{cos } (a - b) - cos } (a+b)}, (Art. 54),
sin C sin } a sin } ]

cosc sin ; a sin 1 b

8 } sin a sin 6°{sin s sin (8- a) sin (8 – 6) sin (8-c)} cos c

69
[{sin s sin (s – a) sin (s b) sin (8—c)}.

2 cos } a cos } b cosc

2

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102. Lhuilier's Theorem. To shew that

tan E={tan s tan } (8 – a) tan } (8-6) tan }(8–c)}.
Tan 1E =

sin (A+B+C –)
cos (A + B + C-7)
sin } (A + B) - sin (T-C)

(Plane Trig. Art. 84),
cos } (A + B) +cos }(T-C)'
sin } (A + B) – cos C
cos (A + B) + sin C
cos} (a - b) - cos } c cos C

(Art. 54). cos } (a + b) + cost o sin C' Hence, by Art. 45, we obtain

sin | (c+a-b) sin 1 (c+b-a) sin s sin (8-)) tanE

cos } (a + b + c) cos 1 (a + b-c) V sin(8-a) sin (8-6)) =/{tan s tan } (s - a) tan } (8-6) tan } (s -c)}.

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103. We may obtain many other formulæ involving trigonometrical functions of the spherical excess. Thus, for example,

cos 1 E=cos {} (A + B)-1 (1-C)}

= cos 1 (1+B) sin C + sin }(4 +B) cos C
= {cos } (a+b) sinoJC +cos }(a – 6) cos'{C}sec £c, (Art. 54),
= {co
cos 1 a cos 16 (cos' 1 C+sin' 1C)
+ sin ; a sin 1 b (cos' {C-sin' 1 C) sec o

6 " }– C}£ {cos , a cos 16+ sin ; a sin 1 b cos C} sec .........(1).

=

с

Again, it was shewn in Art. 101, that
sin | E=sin C sin } a sin 1 b sec 1c;

sin } a sin fb sin C therefore tan 1 E

cos a cos 16+ sin a sin fb cos C

.....(2)

Again, we have from above

cos } E={cos } a cos $b+ sin ţa sin ļ b cos c; sec

с

E

=

secc

a

(1 + cos a)(1 + cos 6) + sin a sin b cos C 4 cos } a cos } b cos c

6 1 + cos a + cos b + cos C cos* }a + cos* 16+cos 10-1 4 cos a cos 1 b cos c 2 cos a cos 1 b cosc

.(3)

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In (3) put 1 - 2 sin'? E for cos. E ; thus

1 + 2 cos įa cos } b cos c sin E=

- cos* ja - cos* 1 b- cos}c 4 cos a cos } b cos c

By ordinary development we can shew that the numerator of the above fraction is equal to

4 sin s sin } (8 – a) sin (8 6) sin 3 (8-c);

therefore

sin sin } (8 – a) sin }(s b) sin } (8–c) sin (E =

cos a cos } b cosc

...(4) .

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+

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Similarly
cos' {E=
cos is cos (8-a) cos } (8 6) cos }(s-c)

....(5).

5 cos }a cos }b cos }c . Hence by division we obtain Lhuilier's Theorem.

Again, sin (C-1E) sin } E

= sin C cot } E - cos C

cos } a cos } b + sin } a sin } b cos C =sin C

- cos C, by (2) sin } a sin } b sin C =cot } a cot} b; therefore, by Art. 101, sin (C-E)=

{{sin s sin (s – a) sin (8 — b) sin (s—c)}

2 sin } a sin } b cosc Again, cos (C-1E) = cos C cos } E+sin C sin } E (1+cosa)(1+cos b) cos C+sin asin bcos'

+ sin’C sin a sin } bsecc 4 cos } a cos } b cosc (1 + cos a)(1 + cos 6) cos C + sin a sin b

4 cos } a cos } b cosc

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=

a

a

= cos ș a cos £b cos C + sin ţa sin }sec } c

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a

COS C

II

sin a sin b cos C + 4 sin'}a sino 1 6

4 sin } a sin} b costo
cos a cos b + (1 - cos a)(1 – cos b)

4 sin } a sin }b cos c
1 + cos C - Cosa - cos' 1c-cosola- 2 - cos® 1 6+1
4 sin a sin 16 cosc 2 sin } a sin 1b cosc

- cos 6

=

--(6).

COS

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