Fifthly. The side b is found by the proportion of § 43, : sin. A sin. a sin. B: sin. b. (316) 54. Scholium. We must carefully attend to the signs of the several functions, in the use of (312), (314), and (315), in order to determine, by Pl. Trig. § 62, whether CBP, BAP, and c are acute or obtuse. The side b, being found by means of its sine, cannot be thus determined; but if we take that one of the two given angles which differs the most from 90° as the angle B, the side b must, by § 45, proposition VI, be acute or obtuse according as B is acute or obtuse. 55. Corollary. If we denote by A'B'C' the triangle which is polar to ABC, A' being the pole of the side a, B' of b, and C' of c, we have, by Geometry, for the sides and angles of the polar triangle, We have given, then, in the problem of § 53, the sides b' and c' and the included angle A' of the triangle A'B'C'; so that we might find the unknown parts of the triangle ABC of that problem by solving its polar triangle A'B'C' by § 46. In fact, we are thus led to the same formulas as those of § 53. It is evident that the great circle of which B'P' is an arc passes through the point B, since it is perpendicular to c', of which B is the pole, and that it is perpendicular to c, since it passes through B', the pole of c. The relations of the segments A'P' and P'C' to the angles CBP and PBA are, then, easily determined. 56. Corollary. If formula (306) is applied to the polar triangle A'B'C', it gives or cos. a' cos. b' cos. c' + sin. b' sin. c' cos. A', or, by (98) and (99), C) cos. B) ( cos. C)+sin. B sin. C (-cos. a), (317) and this equation, which may be applied to either of the three angles of a spherical triangle, is another fundamental equation of Spherical Trigonometry. 57. Corollary. In like manner, we can obtain, from the six equations represented by (307 a), six equations of the form cot. c sin. a cot. C sin. B+ cos. B cos. a; but, as they can also be obtained from the equations (307 a) by simple transposition, they do not constitute a distinct group of equations. 58. Corollary. In the same way (308) gives for A'B'C', if we change c to a', a to b', b to c', and C to A', or cos. a' cos. (b' + c') + 2 sin. b' sin. c' (cos. A')2, cos. (180° A) cos. (360° - [B + C]) +2 sin. (180° B) sin. (180° C) (cos. [90° — § a])2, or, by (98), (99), and (123), and since cos. (90° - a) = sin. § a, cos. (B+ C) 2 sin. B sin. C (sin. cos. A a)2; (318)· which, like (317), may be applied to either of the three angles, and which may be used, like (309), in connexion with Table XXIII, to find the value of the unknown angle in the problem of § 53. In like manner, (309) gives, for A'B'C', cos. a' cos. (b'. -c') — 2 sin. b′ sin. c' (sin. † A')3, or, by (98), (99), and (111), or (36), cos. Acos. (C — B) + 2 sin. B sin. C (cos. a)2 ——cos. (B — C) +2 sin. B sin. C (cos. a); (319) which may also be applied to either of the three angles, and which may be used to find the value of the unknown angle in the problem of § 53. 59. EXAMPLES. 1. Given in a spherical triangle one side equal to 175° 27', and the two adjacent angles equal to 126° 12' and 109° 16'; to solve the triangle. Solution. I. Observing that B should denote the angle which differs most from 90°, let a = 175° 27', B=126° 12', C109° 16'. Since CBPB, the perpendicular falls within the triangle, as in fig. 32. Hence, by (313), PBA 126° 12' - 19° 19′ 24′′ 106° 52° 36". II. The third angle is thus calculated by (318) and Table XXIII. III. The third angle is thus calculated by means of (319). 2. Given in a spherical triangle one side 45° 54′, and the two adjacent angles 125° 37' and 98° 44'; to solve the triangle. The other two sides 138° 34′ 17′′, and=126° 26′ 11′′. 60. Problem. To solve a spherical triangle, when two sides and an angle opposite one of them are given. [B., p. 439.] Solution. Let ABC (fig. 32 or 33) be the triangle, a and c the given sides, and C the given angle. From B let fall on AC the perpendicular BP. First. To find CP. side a and the angle C. We know, in the right triangle PBC, the Hence, by Napier's Rules, as in § 46, tang. CP cos. C tang. a. (320) Secondly. To find PA., If, in the triangle PBC, co. a is the middle part, CP and BP are the opposite parts; and if, in the triangle ABP, co. c is the middle part, PA and PB are the opposite Hence, by Bowditch's Rules, parts. cos. a cos. ccos. CP: cos. PA. (321) Thirdly. To find b. It is evident that this problem, like the corresponding one of Plane Trigonometry § 75, may have two solutions for the same values of the given parts, in one of which the perpendicular falls within the triangle, as in fig. 32, and while in the other the perpendicular falls without the triangle, as in fig. 33, and But, if PA is greater than CP, the second solution is impossible, and, if CP + PA is greater than 180°, that is, if PA is greater than the supplement of CP, the first solution is impossible. Fourthly. A and B may be found by the proportions sin. c sin. Csin. a: sin. A sin. c sin. Csin. b: sin. B. (324) (325) 61. Scholium. In determining CP and PA by (320) and (321), the precepts of Pl. Trig. § 62 concerning the signs of the trigonometric functions must be carefully attended to. Since A is found by means of its sine, two supplementary values of A are given by (324). These two values correspond to the two solutions of the problem, since BAP, which is equal to A in fig. 32 and its supplement in fig. 33, must have the same value in both solutions. Also, as PB is the leg opposite C in the right triangle BPC, and the leg opposite BAP in the right triangle BPA, therefore, by § 14, C, PB, and BAP are, in both solutions, either all acute or all obtuse. Hence, in the first solution (fig. 32) C and A are both acute or both obtuse, and in the second solution (fig. 33) one of them is acute and the other is obtuse. The two values of b are to be substituted separately in (325); and, for each value of b, (325) gives two values of B, of which the correct one is to be selected by the rules of § 45. But, instead of using (325), we can find CBP by (312), PBA by (315), and B by the equation B CBP PBA. of c differs less from 90° than only one solution of the problem 62. Scholium. If the given value that of a; that is, if sin. c> sin. a; is possible, since, by § 45, prop. VI, a and A must be both acute or both obtuse. In this case, (321) gives, since cos. c<cos. a, if absolute values only are considered, cos. PA cos. CP. |
