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Secondly. To find AP. If, in the triangle PBC, PC is the middle part, co. C and PB are the adjacent parts; and, if, in the triangle ABP, AP is the middle part, co. BAP and BP are the adjacent parts. Hence, by (604),
cotan. C cotan. BAP :: sin. PC: sin. AP.
sin. a sin. A:: sin. b: sin.B.
44. Scholium. In determining PC by (652), the signs of the several terms must be attended to by means of (496).
Either value of AP, given by (653), may be used, and there will be two different triangles solving the problem, except when AP + PC (fig. 4.) is greater than 180°, or PC (fig. 5.) is less than AP. It may (658) be that both values of AP satisfy the conditions of the problem, or that only one value satisfies them, or (659) that neither value does; in which last case the problem is impossible.
Of the values of c, determined by (655), the true value must be ascertained from the right triangle ABP, by (495) and (517); or since, as in (648), PB (660) and Care both greater than 90° or both less than 90° at the same time; it follows, from (495), that when C and AP are both greater or both less than 90°, that c is less than 90°; but when one of them is (661) greater and the other less than 90°, c is greater than
From the two values of B (656) the true value must be selected by means of (623 – 627).
45. Scholium. The problem is impossible, by (772), when A differs more from 90° than does C, and when at the same time, one of the two quantities a and A is (663) less than 90°, while the other is greater than 90°. But this case is precisely the same as the impossible case of (659). .
Solution. Let ABC (figs. 4. and 5.) be the triangle; a, b, and c being the given sides.
From B let fall on AC the perpendicular BP.
Then, in the right triangle PBC, if co. C is the middle part, co. a and PC are the adjacent parts. Hence, by (474), cos. C
=cotan. a tang. PC.
If, in the triangle BPC, co. a is the middle part, BP and PC are the 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. a cos. c:: cos. PC: cos. AP.
(666) (fig. 4.) AP-b-PC, and (fig. 5.) AP = PC — b. Hence, by (116) and (202),
cos. AP =
PC) =cos. (PC b), = cos. b cos. PC + sin. b sin. PC;
which, substituted in (665), gives
(669) cos. a: cos. c:: cos. PC: cos. b cos. PC+sin. b sin. PC. Dividing the two terms of the last ratio of this proportion by cos. PC, and reducing by (10), we
cos, a cos. c:: 1: cos. b + sin. b tang. PC. Make the product of the means equal that of the extremes, and we have
cos, a cos. b + cos. a sin. b tang. PC cos. c; by transposition
cos. a cos. b.
cos. a sin. b tang. PC = cos. c
whence the value of the angle C may be calculated, and in the same way either of the other angles.
47. Corollary. The equation (674) may be brought into a form more easy for calculation by logarithms,
as follows. Add unity to both its members and it
1+ cos. C
which, substituted in the numerator of (675), gives
Now we have (122)
cos. (MN) cos. (M+N)
and letting s denote half the sum of the sides or
sin. a sin. b
2 sin. M sin. N; (678)
SM+N= a + b,
and (678) becomes
cos. c— cos. (a + b) 2 sin. s sin. (s — c);
which, substituted in (677), gives