A leg and the opposite angle given.

35. EXAMPLE.

Given in the spherical right triangle, (fig. 30), a = 35°44′, and A = 37° 28'; to solve the triangle.

36. Problem. To solve a spherical right triangle, when one of its legs and the adjacent angle are given.

Solution. Let ABC (fig. 30) 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

cos. B tang. a cotan. h;

and, by (6),

cotan. h =

cos. B tang. a

cotan. a cos. B.

co. A is the middle Hence

Secondly. To find the other angle A;

part, and co. B and a are opposite parts.

cos. A cos, a sin. B.

Thirdly. To find the other leg b; a is the middle part, and co. B and b are adjacent parts.

Hence

sin. a tang. b cotan. B ;

and, by (6),

tang. b =

sin, a cotan. B

sin. a tang. B.

The legs given.

37. EXAMPLE.

Given in the spherical right triangle (fig. 30), a=118°54′, and B-12° 19'; to solve the triangle.

38. Problem. To solve a spherical right triangle, when its two legs are given.

Solution. Let ABC (fig. 30) be the triangle, a and b the given legs.

First. To find the hypothenuse h; co. h is the middle part, a and b are opposite parts. Hence

cos. hcos, a cos. b.

Secondly. To find one of the angles, as A; b is the middle part, and co. A and a are adjacent parts. Hence

The angles given.

39. EXAMPLE.

Given in the spherical right triangle (fig. 30), a 1°, and b 100°; to solve the triangle.

40. Problem. To solve a spherical right triangle, when the two angles are given.

Solution. Let ABC (fig. 30) 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.

cos. hcotan. A cotan. B.

Hence

Secondly. To find one of the legs, as a; co. A is the middle part, and co. B and a are the opposite parts. Hence cos. A = cos. a sin. B;

41. Scholium. The problem is, by § 20, impossible, when the sum of the given values of A and B is less than 90°, or

The angles given.

greater than 270°, or when their difference is greater than

90°.

42. EXAMPLE.

Given in the spherical right triangle (fig. 30), A=91° 11', and B=111° 11'; to solve the triangle.

Sines of sides proportional to sines of opposite angles.

CHAPTER III.

SPHERICAL OBLIQUE TRIANGLES.

43. Theorem. The sines of the sides in any spherical triangle, are proportional to the sines of the opposite angles. [B. p. 437.]

Proof. Let ABC (figs. 32 and 33) be the given triangle. Denote by a, b, c, the sides respectively opposite to the angles A, B, C. From either of the vertices let fall the perpendicular BP upon the opposite side AC. Then, in the right triangle ABP, making BP the middle part, co. c and co. BAP are the opposite parts. Hence, by Napier's Rules,

sin. BP

sin. c sin. BAP = sin, c sin. A. For BAP is either the same as A, or it is its supplement, and in either case has the same sine, by (98).

Again, in triangle BPC, making BP the middle part, co. a and co. C are the opposite parts. Hence, by Napier's Rules,

sin. BP sin. a sin. C;

and, from the two preceding equations,

sin, c sin. A = sin. a sin. C,

which may be written as a proportion, as follows:

sin. a sin. A sin. c : sin. C.

In the same way,

sin, a sin. A sin. b ; sin. B.