(589) 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), 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, by (474), EXAMPLE. Given, in the spherical right triangle (fig. 2.), a=1° and b= 100°; to solve the triangle. Ans. h: = 100°, 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), cos. h = cotan. A cotan. B. (593) 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, by (475), 33. Scholium. The problem is, by (540), impossible when the sum of the given values of A and B (597) is less than 90°, or greater than 270°, or when their difference is greater than 90°. EXAMPLE. Given, in the spherical right triangle (fig. 2.), CHAPTER III. Spherical Oblique Triangles. SECTION I. Theorems for the Solution of Spherical Oblique Triangles. 34. Theorem. The sines of the sides in any (598) spherical triangle are proportional to the sines of the opposite angles. (599) (600) (601) (602) Demonstration. Let ABC(figs.4. B and 5.) be the given triangle. De- Fig.4 P 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 (195). Again, in triangle BPC, making BP the middle part, co. a and co. C are the opposite parts. Hence, by (475), sin. BP = sín. a sin. C; and, from (599) and (600), sin. c sin. A = sin. a sin. C, which may be written as a proportion, as follows; In the same way sin. a sin. A :: sin. b: sin. B. 35. Theorem. Bowditch's Rules for Oblique Triangles. If, in a spherical triangle, two right triangles are formed by a perpendicular let fall from one. of its verticles upon the opposite side; and if, in the two right triangles, the middle parts are so taken that the perpendicular is an adjacent part in both of them; then (603) The sines of the middle parts in the two triangles are proportional to the tangents of the adjacent parts. (604) But, if the perpendicular is an opposite part in both the triangles, then The sines of the middle parts are proportional to (605) the cosines of the opposite parts. Demonstration. Let M denote the middle part in one of the right triangles, A an adjacent part, and O an opposite part. Also let m denote the middle part (606) in the other right triangle, a an adjacent part, and o an opposite part; and let p denote the perpendicular. First. If the perpendicular is an adjacent part in both triangles, we have, by (474), sin. M tang. A tang P, sin. m = tang. a tang. p. The quotient of (607), divided by (608), is (607) (608) sin. M tang. A tang. p tang. A (611) (612) Secondly. If the perpendicular is an opposite part in both the triangles, we have, by (475), sin. M = cos. O cos. p, sin. m = cos. o cos. p. The quotient of (611) divided by (612) is (615) SECTION. II. Solution of Spherical Oblique Triangles. 36. Problem. To solve a spherical triangle when two of its sides and the included angle are known. Solution. Let ABC (figs. 4. and 5.) be the triangle; a and b the given sides, and C the given angle. From B let fall on AC the perpendicular BP. First. To find PC, we know, in the right triangle BPC, the hypothenuse a and the angle C. Hence, by means of (474), tang. PC cos. C tang. a. Secondly. AP is the difference between AC and PC, that is, (616) (fig. 4.) AP-b— PC, or (fig. 5.) APPC — b. Thirdly. To find the side c. If, in the triangle BPC, co. a is the middle part, PC and PB are |