SPHERICAL TRIGONOMETRY CHAPTER X FUNDAMENTAL FORMULAS 117. The spherical triangle.* Spherical trigonometry treats of the relations between the various parts of a spherical triangle and of the methods of solving the spherical triangle. The sides of a spherical triangle are always arcs of great circles. Having given a spherical triangle ABC, situated upon a sphere S, a triedral angle O-ABC may be formed by passing planes through O, the center of the sphere, and through the sides of the triangle. O It is known from geometry that the arc AB and the angle AOB contain the same number of degrees, and that the angle CAB and the diedral angle C-A0-B contain the same number of degrees. The sides and angles of a spherical triangle may have. any values between 0° and 360°. A triangle having one or more of its parts greater than 180° is called a general spherical triangle. A triangle having each of its parts less than 180° is called a spherical triangle. We shall consider only those triangles whose parts are each less than 180°. * For a course on the right spherical triangle read Arts. 117 and 126 and from Art. 128 to end of Chapter XI. 118. Law of sines. To find the relation between two sides of a spherical triangle and the angles opposite. Given a spherical triangle and its accompanying triedral angle; through the vertex P pass planes perpendicular to OR and OQ, intersecting in PF a line perpendicular to the plane ORQ. (1) = This demonstration applies to similar figures drawn for all possible cases,* hence the theorem is always true. 119. Law of cosines. To find the relation between the three sides and an angle. b<90°, c< 90°; a < 180°, a < 180°. Then AB =r tan c, OB=r sec c, AC=r tan b, OC=r sec b. From the triangle OBC, by Art. 90, 2 2 BC = (r sec b)2 + (r.sec c) - 2 (r secb) (r sec c) cos a. Likewise from the triangle ABC, (1) BC2 = (r tan b)2 + (r tan c)3 − 2 (r tan b) (r tan c) cos a. (2) Subtracting (2) from (1), we have 0=2(sec2 b-tan2 b) + 2(sec2 c-tan2 c) - 2 r2 sec b sec c cos a +22 tan b tan c cos α, which reduces to or Also and 01 secb sec c cos a + tan b tan c cos α, cos a = cos b cos c + sin b sin c cos a. * The following seven cases can arise : (5) (4) 1 side < 90°, 2 angles< 90° (5) 1 side 90°, 1 angle <90° (6) 1 side < 90°, 0 angle <90° 0 angle < 90°. It is to be understood that all parts not mentioned are greater than 90°. 120. To extend the law of cosines. In the derivation of the formula cos a = cos b cos c + sin b sin c cos α, b and c were less than 90°, while a and a were less than 180°. To show that the formula is true in general, it is necessary to consider two additional cases: 1st. Both b and c greater than 90°. 2d. Either b or c greater than 90°. Since ẞ and y do not enter the formula, they may have any value consistent with the above conditions. B 180 α First. Given the triangle ABC in which b> 90° and c> 90°. Extend the sides b and c of the triangle ABC, forming the lune whose angle is a. Then in the triangle A'BC, the sides A'B and A'C are each less than 90°, hence by Art. 119 cos a = cos (180° — b) cos (180° — c) + sin (180° — b) sin (180° — c) cos α, or cos a = cos b cos c + sin b sin c cosa. Hence the law of cosines holds when both b and c are greater than 90°. Second. Given the triangle ABC, in which b< 90° and c > 90°. Extend the sides a and c of the triangle ABC, forming the lune whose angle is B. Then in the triangle AB'C, the sides AB' and AC are each less than 90°, and the angle B'AC is equal to 180° - α. Then, by Art. 119, cos (180° — a) = cos b cos (180° — c) B с β 180°- c A B 180-a a a + sin b sin (180°c) cos (180° — α), Hence the law of cosines holds when either b or c is greater than 90°. The law of cosines is therefore true in general. 121. To find the relation between one side and the three angles. Let a, b, and c be the sides of any spherical triangle, and a', b', c' the sides of its polar triangle. Applying the law of cosines to the polar triangle, we have cos a' cos b' cos c' + sin b' sin c' cos a'. α cos (180° — α) = cos (180° — ẞ) cos (180° — y) b B a' a γλ + sin (180° – ẞ) sin (180° — y) cos (180° — a), COS α=— cos B cos y + sin ẞ sin γ cos a. or (1) 122. The sine-cosine law. To find the relation between three sides and two angles. Eliminating cos a by substitution, cos b = cos b cos2 c + sin b sin c cos c cos a + sin c sin a cos B. Transposing and factoring, cos b (1 — cos2 c) = sin b sin c cos e cos a + sin a sin c cos ß. Replacing (1 cos2 c) by sin2 c, and dividing by sinc, we have cos b sin c = sin b cos c cos a + sin a cos ẞ. Rearranging terms, sin a cos ẞ= cos b sin c — sin b cos c cos α. (3) |