If, then, a and C are alike; that is, both acute or both obtuse; CP - CP. is, by (320), acute; so that PA> CP, and PA <180° But, if a and C are unlike; that is, one acute and the other obtuse; CP is, by (320), obtuse; so that PACP, and PA 180° - CP. Therefore, in the former case, the first solution is the possible one, and, in the latter case, the second solution is the possible one.

If c differs more from 90° than a; that is, if sin. c<sin. a; (321) gives, in absolute value,

cos. PA cos. CP.

If, then, c and C are unlike, the first ratio of (321) is opposite in sign to the second member of (320) and, therefore, to cos. CP; so that (321) gives cos. PA negative; that is, PA is obtuse; and we have PA>CP, PA> 180° — CP, and neither solution of the problem is possible. This also appears from § 45, prop. VI. But if c and C are alike, PA is acute; and we have PA < CP, PA<180° – CP, and both solutions are possible.

If, however, sin. c <sin. a sin. C; that is, if sin. c <sin. BP; (324) gives sin. A>1; so that the problem is impossible. In this case, (321) will also give cos. PA>1.

63. Scholium. The various cases of this problem may be compared with those of Pl. Trig. § 75; a, c, and C, in this problem, corresponding respectively with b, a, and A, in that.

1. Given in a spherical triangle one side = 35°, a second side= 142°, and the angle opposite the second side = 176°; to solve the triangle.

Solution.

Let a = 35°, c = 142°,

C=176°.

Since c differs less from 90° than a, while a and C are unlike, the second solution alone is possible. Then, by (320) and (321),

b = 145° 3' 56". · 37° 56′ 30′′ 107° 7′ 26".

2. Given in a spherical triangle one side = 120°, a second side = 135°, and the angle opposite the second side = 155°; to solve the triangle.

Ans. The third side = 94° 38′ 18" or 17° 58′ 54′′.

The first angle = 142° 14′ 22′′ or = 37° 45′ 38′′.

The third angle = 135° 11′ 14′′ or = 12° 36′ 31′′.

3. Given in a spherical triangle one side = 54°, a second side= 22°, and the angle opposite the second side = 30°; to solve the triangle.

Ans. The question is impossible.

65. Problem. To solve a spherical triangle, when two angles and a side opposite one of them are given. [B., p. 440.]

Solution. Let ABC (fig. 32 or 33) be the triangle, A and C the given angles, and a the given side.

From B let fall on AC the perpendicular BP. This perpendicular must fall within the triangle, if A and C are either both obtuse or both acute; but it falls without, if one is obtuse and the other acute. This is evident from § 61.

First. CBP may be found, as in § 53, by (312).

Secondly. PBA may be found by (314); that is,

Fourthly. The sides c and b may be found by the proportions

Since PBA is

(329)

(330)

66. Scholium. found by means of its sine, two values of PBA, supplements of each other, are given by (326). These two values correspond to two solutions of the problem. For, if the triangle CBA' (fig. 72) is a solution of the problem, and if the angles PBA' and PBA" are supplements of each other, CBA" is likewise a solution. For, if A'B and A'C are produced till they meet at A, they form a lunary surface, in which the angle A" the angle A'; and, since A""BP = the supplement of PBA' — PBA", the right triangles APB and A"PB are symmetrical, and BA"P = BA'P the given angle A; so that CBA" contains the given values of A, C, and a. But both solutions are alike in respect to the position of the perpendicular within or without the triangle.

=

=

If, when the perpenaicalar falls within the triangle, either value of PBA is greater than the supplement of CBP, or if, when the perpendicular falls without, either value is greater than CBP, that value must be rejected, and the corresponding solution is impossible.

Two supplementary values of c are given by (329), which evidently correspond to the two solutions of the problem; and since, by § 14, C and PB are alike, and PBA and PA are alike, therefore, in each solution, by § 12, C, PBA, and c must be either all acute or else one acute and the other two obtuse.

Each value of B gives two values of b, by (330), of which the true value is to be selected by the rules of § 45. But, instead of using (330), we may find CP by (298), PA by (301), and b by (322) or (323).

67. Scholium. If A differs less from 90° than C; that is, if sin. A> sin. C; we have cos. BAP<cos. C, and (326) gives

sin. PBA <sin. CBP.

Hence the acute value of PBA is less, and the obtuse value of PBA is greater, than CBP and than the supplement of CBP; so that, whether BP falls within or without, the problem has but one solution ; namely, that in which PBA is acute. This also follows from § 45, prop. VI.

If A differs more from 90° than C; that is, if sin. A <sin. C; (326) gives

sin. PBA sin. CBP.

Hence, if CBP is acute; that is, by (312), if a and Care alike; both values of PBA are greater than CBP and less than its supplement but, if CBP is obtuse; that is, if a and Care unlike; both values of PBA are less than CBP and greater than its supplement. But B is found by (327), when A and C are alike, and by (328), when A and C are unlike. Therefore, if a and A are alike, both solutions are possible; and, if a and A are unlike, neither solution is possible.

If, however, sin. A <sin. C sin. a; that is, if sin. A <sin. BP ; the problem is impossible, since (329) gives sin. c>1. In this case, (326) also gives sin. PBA>1.

68. Scholium. The formulas of § 65 might also be obtained by applying those of § 60 to the polar triangle A'B'C', supposing a', c', and A' to be known; and the various cases of the two problems correspond to each other. The ambiguity of solution arises, in the former problem, from doubt whether the perpendicular BP falls within or without the triangle, and, in this, from doubt whether PBA is acute or obtuse. Nevertheless, the two solutions of the former problem correspond to the two solutions of this.

69. EXAMPLES.

=

1. Given in a spherical triangle one angle 95°, a second angle = 104°, and the side opposite the first angle = 138°; to solve the triangle.

Solution. Let A = 95°, C = 104°,

a = 138°.

Since A and C are both obtuse, the perpendicular falls within the triangle; and, since A differs less from 90° than C, PBA is acute, and there is only one solution. Then, by (312) and (326),

B = 18° 32′ 49′′ † 6° 34′ 49′′ = 25° 7′ 38′′.

2. Given in a spherical triangle one angle 135°, a second

=

=

angle 60°, and the side opposite the first angle 155°; to solve the triangle.

Ans. The third angle = 94° 38' 18" or =

70. Problem.

sides are given.

17° 58' 54".

142° 14' 22".

12° 36′ 31".

To solve a spherical triangle, when its three [B., p. 440.]

Solution. Equation (305) gives, by transposition and division,

cos. ccos, a cos. b

cos. C

sin. a sin. b

(331)

whence the value of the angle C may be calculated; and in the same way either of the other angles.

71. Corollary. An equation more easy for calculation by logarithms may be obtained from (308), which gives, by transposition and division,

2 (cos. C)2=

cos. ccos. (a + b)
sin. a sin. b

(332)

Now, letting s denote half the sum of the sides, or

if we make, in the general formula (42),

M=(a+b+c) = s,

N=(a+b-c) = s―c;