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cos. hcotan. A cotan. B.
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.),
Spherical Oblique Triangles.
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.
Demonstration. Let ABC(figs.4.
and 5.) be the given triangle. De- Fig.4
ly opposite to the angles A, B, C. A
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 sin. 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
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. Mtang. A tang p,
The quotient of (607), divided by (608), is
sin. m = tang. a tang. p.
Secondly. If the perpendicular is an opposite part
in both the triangles, we have, by (475),
sin. M = cos. O cos. p,
sin. mcos. o cos. p.
The quotient of (611) divided by (612) is
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),
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 PC and PB are
opposite parts; and if, in the triangle ABP, co. c is the middle part, BP and AP are the opposite parts. Hence, by (605),
cos. PC: cos. AP : : sin. (co. a) : sin. (co. c),
cos. PC: cos. AP:
Cos. a cos. c.
Fourthly. To find the angle A. If, in the triangle BPC, PC is the middle part, co. C and BP are adjacent parts; and if, in the triangle ABP, AP is the middle part, co. BAP and BP are adjacent parts. Hence, by (604),
sin. PC sin. PA: ; cotan. C: cotan, BAP, and BAP is the angle A (fig. 4.), when the perpen- (619) dicular falls within the triangle; or it is the supplement of 4 (fig. 5.), when the perpendicular falls without the triangle.
Fifthly. B is found by means of (598)
sin. c sin. C: sin. b: sin. B.
37. Scholium. In determining PC, c and BAP,
by (615), (617), and (618), the signs of the several (621) terms must be carefully attended to; by means of (496).
But to determine which value of B, determined by (620), is the true value, regard must be had to the (622) following rules which will be demonstrated hereafter.
I. The greater side of a spherical triangle is always (623) opposite to the greater angle (768).