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44. Theorem. Bovditch'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.
But, if the perpendicular is an opposite part in both the triangles, then
The sines of the middle parts are proportional to the cosines of the opposite parts. [B. p. 437.]
Proof. 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 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 Napier's Rules,
Secondly. If the perpendicular is an opposite part in both the triangles, we have, by Napier's Rules,
45. Problem. To solve a spherical triangle, when two of its sides and the included angle are given. [B. p. 438.]
Solution. Let ABC (figs. 32 and 33) 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 Napier's Rules, tang. PC = cos. C tang. a.
Secondly. AP is the difference between AC and PC,
(fig. 32) AP=b- PC, or (fig. 33) AP= PC- 6. (299)
Thirdly. To find the side c. If, in the triangle BPC, co, a is the middle part, 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 Bowditch's Rules,
cos. PC: cos. AP = sin. (co. a) : sin. (co. c),
Rules for acute or obtuse angles and sides.
Fourthly. To find the angle A. If, in the triangle BPC, PC is the middle part, co. Cand BP are adjacent parts; and if, in the triangle ABP, AP is the middle part, co. BAP and BP are adjacent parts. Hence, by Bowditch's Rules,
sin. PC: sin. PA= cotan. C: cotan. BAP; (301) and BAP is the angle A (fig. 32), when the perpendicular falls within the triangle; or it is the supplement of A (fig. 33), when the perpendicular falls without the triangle. Fifthly. B is found by means of § 43, sin. c: sin. C= sin b: sin. B.
(302) 46. Scholium. In determining PC, c, and BAP, by (298), (300), and (361,) the signs of the several terms must be carefully attended to; by means of Pl. Trig. § 62.
But to determine which value of B, determined by (302), is the true value, regard must be had to the following rules, which are proved in Geometry
I. The greater side of a spherical triangle is always opposite to the greater angle.
II. Each side is less than the sum of the other two.
III. The sum of the sides is less than 360°.
IV. Each angle is greater than the difference between 180°, and the sum of the other two angles.
There are, however, cases in which these conditions are all satisfied by each of the values of B. "In any such case this angle can be determined in the same way in which the angle A was determined, by letting fall a perpendicular from the Fundamental equation.
vertex A on the side BC. But this difficulty can always be avoided, by letting fall the perpendicular upon that of the two given sides which differs the most from 90°.
47. Corollary. By (299), (111), and (35), we have cos. AP = cos. (6 PC) = cos. (PC — b)
= cos. b cos. PC + sin. b sin. PC, (303) which, substituted in (300), gives
cos. PC:cos. b cos. PC+sin. b sin. PC=cos. a: cos.c.
Dividing the two terms of the first ratio by cos. PC, we have by (7),
1: cos. b + sin. 6 tang PC = cos. a : cos. C. (304)
The product of the means being equal to that of the extremes, we have
cos. c = cos. A cos. b + sin. b ços, a tang. PC. (305) But by (298),
cos. C sin. a tang. PC = cos. C tang. a =
cos, a tang. PC= cos. C sin. a;
which, substituted in (305), gives
cos. c = cos, a cos. b + sin. a sin. 6 cos. C,
which is one of the fundamental equations of Spherical
cos. C= -1 + 2 (cos. 1 C),
Column of Log. Rising of Table XXIII.
which, substituted in (307), gives, by (34),
cos. c= cos. (a+b) + 2 sin, a sin. b (cos. } C), (308) from which the value of the side c can readily be found by using the table of Natural Sines.
49. Corollary. We have, by (56),
cos. C=1 – 2 (sin. 1 C)2,
which, substituted in (307), gives, by (35),
cos. c = cos. (a - b) — 2 sin, a sin. 6 (sin. 1 C)2, which can be used like formula (308),
50. Corollary. The use of formula (309) is much facilitated by means of the column of Rising in Table XXIII of the Navigator. This column contains the values of log. 2 (sin. C)2 = 2 log. sin. 1 C + log. 2
= 2 log. sin. 1 C+0.30103. (310)
But the decimal point is supposed to be changed so as to correspond to the table of Natural Sines, that is, 5 is added to the logarithm ; and 20 is to be subtracted from the value of 2 log. sin. 1 C, which is given by table XXVII, as is evident from Pl. Trig. § 31. So that the column Rising of Table XXIII is constructed by the formula
log. Ris. C 2 log. sin. 1 C + 5.30102 - 20
= 2 log. sin. ; C – 14.69897, (311) which agrees with the explanation in the Preface to the Navigator.