2. Given, in the spherical triangle ABC, a = 45° 54′, B = 125° 37′, and C 98° 44′; to solve the triangle. = 40. Problem. To solve a spherical triangle when two sides and an angle opposite one of them are given. Solution. Let ABC (figs. 4. and 5.) be the triangle; a and c the given sides, and C the given angle. From B let fall on AC the perpendicular BP. Ans. A 138° 34', c = 126° 26'. First. To find PC. We know, in the right triangle PBC, the side a and the angle C. Hence, by (474), tang. PC cos. C tang. a. Secondly. To find AP. If, in the triangle PBC, co. a is the middle part, CP and PB are the opposite parts; and, if, in the triangle ABP, co. c is the middle part, AP and PB are the opposite parts. Hence, by (605), cos. a cos. c:: cos. PC: cos. AP. b and in the other (fig. 5.) Thirdly. To find b. There are, in general, two triangles which resolve the problem, in one of which (fig. 4.) PC+AP, But, if AP is greater than PC, there is but one triangle, as in (fig. 4.), and b is obtained by (640); or, (642) if the sum of AP and PC is greater than 180°, there is but one triangle, as in (fig. 5.), and b is obtained by (641). (643) (644) Fourthly. A and B are found by (598). sin. c sin. C:: sin. a: sin. A 41. Scholium. In determining PC and AP by (645) (638) and (639), the signs of the several terms must be carefully attended to by means of (496). The two values of A, given by (643), correspond respectively to the two triangles which satisfy the problem. And the one, which belongs to each trian(646) gle, is to be selected, so that the angle BAP, which is the same as A in (fig. 4.) and the supplement of A in (647) (fig. 5.), may be obtuse if C is obtuse, and acute if C is acute. For BP is the side opposite BAP in the (648) right triangle ABP, and the side opposite C in the triangle BCP; and therefore, by (517), BP, BAP, and C are all at the same time less than 90°, or all greater than 90°. Of the two values of B, given by (644), the one (649) which belongs to each triangle is to be determined by means of (623–626). с 42. Scholium. The problem is, by (772), impossible, when the given value of c differs more from 90° than that of a; if, at the same time, the value of one (650) of the two quantities, c and C, is greater than 90° while that of the other is less than 90°. And in this case we should find that AP was larger than PC, and at the same time that the sum of AP and PC was more than 180°. Solution. 1. Given, in the spherical triangle ABC, α = 35°, c = 142°, C = 176°; to solve the triangle. By (638), 176°. α = 35°. = EXAMPLES. 3° 44'. - 37° 56′ = 107° 8'. sin. (ar. co.) 10.21066 sin. 8.84358 sin. 9.75859 sin. 8.81283 (652) to solve the triangle. sin. (ar. co.) 10.21066 8.84358 9.98029 = Ans. b 73° 16', B = Ans. or 9.03453 = A B = 2. Given, in the spherical triangle ABC, 12°; = = b = A = B = 107° 8', 3° 44', 6° 13'. 43. Problem. To solve a spherical triangle when two angles and a side opposite one of them are given. Solution. Let ABC (figs. 4. C and 5.) be the triangle; A and C Fig.4 33° 32', 153° 19', 17° 51'. First. PC may be found, as in (638), tang. PC B A P B From B let fall on AC the per(651) pendicular BP. This perpendicular will, by (647), fall within the triangle, if A and C are either both obtuse or both acute; but it will fall without if one is obtuse and the other acute. Fig.5 a C Secondly. To find AP. If, in the triangle PBC, PC is the middle part, co. C and PB are the adjacent parts; and, if, in the triangle ABP, AP is the middle part, co. BAP and BP are the adjacent parts. Hence, by (604), cotan. C cotan. BAP :: sin. PC : sin. AP. Thirdly. To find b. We have b= PC+AP, (fig. 4.) (fig. 5.) b = PC · AP. Fourthly. c and B are found by (538). sin. A sin. a: sin. C: sin. c; (653) (654) (655) (656) (657) 44. Scholium. In determining PC by (652), the signs of the several terms must be attended to by means of (496). Either value of AP, given by (653), may be used, and there will be two different triangles solving the problem, except when AP + PC (fig. 4.) is greater than 180°, or PC (fig. 5.) is less than AP. It may (658) be that both values of AP satisfy the conditions of the problem, or that only one value satisfies them, or (659) that neither value does; in which last case the problem is impossible. Of the values of c, determined by (655), the true value must be ascertained from the right triangle ABP, by (495) and (517); or since, as in (648), PB (660) and C are both greater than 90° or both less than 90° at the same time; it follows, from (495), that when C and AP are both greater or both less than 90°, that c is less than 90°; but when one of them is (661) greater and the other less than 90°, c is greater than 90°. |