(567) Secondly. To find the adjacent angle B; co. B is the middle part, and co. h and a are the adjacent parts. Hence, by (474), (569) {or sin. (co. B.) = cos. B tang. a cotan, h, Thirdly. To find the other leg b; co. h is the middle part, and a and b are the opposite parts. Hence, by (475), (568) cos. h cos, a cos. b; and, by (7), =tang. a tang. (co, h), 23. Scholium. The question is impossible by (505), (570) when the given value of the hypothenuse differs more from 90° than that of the leg. (570/ 24. Solution. When h and a are both equal to 90°, it may be shown, as in (563), that the values of B and b are indeterminate. EXAMPLE. Given, in the spherical right triangle (fig. 2), a=141° 11', and h=127° 12'; to solve the triangle. Ans. A 128° 7', B= 52° 22′, 25. Problem. To solve a spherical right triangle, when one of its legs and the opposite angle are known. Solution. Let ABC (fig. 2.) be the triangle; a the given leg, and A the given angle. First. To find the hypothenuse h; a is the middle part, and co. h and co. A are the opposite parts. Hence, by (475), sin. a = sin. h sin. A; and, by (7), Secondly. To find the other angle B; co. A is the middle part, and a and co. B are the opposite parts. Hence, by (475), cos. A cos. a sin. B; sin. a cosec. A. sin. b = sec. a cos. A. Thirdly. To find the other leg b; b is the middle part, and a and co. A are the adjacent parts. Hence, by (474), tang, a cotan. A. Fig. 3 E (571) (572) a h 4.6 C 26. Scholium. There are two triangles ABC and A'BC (fig. 3.), formed by producing the sides AB and AC, to the point of meeting A', both of which satisfy the conditions of the problem. For the side BC or a, and the angle A, or by art. 2 its equal A', belong to both the triangles. (576) Now ABA and ACA' are semicircumferences, since the line AA' joining their points of intersection (573) (574) (575) (577) is the line of intersection of their planes, and therefore passes through the centre of the sphere and is a diameter. Hence h', the hypothenuse of A'BC, is the (578) supplement of h; b' is the supplement of b; and A'BC is the supplement of ABC. One set of values, then, of the unknown quantities, given by the tables, as in (562), correspond to the triangle ABC, and the other set to A'BC, (580) (579) 27. Corollary. When the given values of a and A sin. h 1 B = 90°, b = 90°; (581) or, by (158), h=90°, as in (522). (582) 28. Corollary. When a and A are equal to 90°, the values of b and B are indeterminate, as in (563). 29. Scholium. The problem is, by (517), impossible, when the given values of the leg and its opposite angle are such that one surpasses 90° while the other (583) does not, or that one is equal to 90° while the other differs from 90°; and, by (519), it is impossible when the given value of the angle differs more from 90° than that of the leg. Given, in the spherical right triangle 35° 44′ and A = 37° 28'; to solve the 30. Problem. To solve a spherical right triangle, when one of its legs and the adjacent angle are known. Solution. Let ABC (fig. 2.) be the triangle; a the given leg, and B the given angle. First. To find the hypothenuse h; co. B is the middle part, and co. h and a are adjacent parts. Hence, by (474), cos. B tang. a cotan. h; and, by (7), Secondly. To find the other angle A; co. A is the middle part, and co. B and a are opposite parts. Hence, by (475), cos. A cos. a sin. B. tang. b Thirdly. To find the other leg b; a is the middle part, and co. B and b are adjacent parts. Hence, by (474), sin. a = tang. b cotan. B; sin. a cotan. B cotan. a cos. B. sin. a tang. B. EXAMPLE. Given, in the spherical right triangle, (fig. 2.), a = 118° 54′ and B = 12° 19′; to solve the triangle. Ans. h = 118° 20', A = 95° 55', b = 10° 49'. (584) (595) (586) (587) (488) (589) (590) (591) (592) 31. Problem. To solve a spherical right triangle, when its two legs are known. Solution. Let ABC (fig. 2.) be the triangle, a and b the given legs. First. To find the hypotheneuse h; co. h is the middle part, a and b are opposite parts. Hence, by (475), and, by (7), cos. h = cos. a cos. b, Secondly. To find one of the angles, as A; bis the middle part, and co. A and a are adjacent parts. Hence, by (474), sin. b tang. a cotan. A; cotan. A In the same way cotan. B cotan. b sin, a. EXAMPLE. Given, in the spherical right triangle (fig. 2.), a=1° and b 100°; to solve the triangle. Ans. h=100°, = A 32. Problem. To solve a spherical right triangle, when the two angles are given.. Solution. Let ABC (fig. 2.) be the triangle, A and B the given angles. First. To find the hypothenuse h; co. h is the middle part, and co. A and co. B are adjacent parts. Hence, by (474), |