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65. Problem. To solve a spherical triangle when its three angles are given.
Solution. Let ABC (figs. 4. and 5.) be the triangle, the angles A, B, and C being given.
From B let fall on AC the perpendicular BP.
Then, if, in the right triangle PBC, co. a is made the middle part, co. C and co. PBC are the adjacent parts. Hence, by (474),
cos. a cotan. C cotan. PBC.
If, in the right triangle PBC, co. C is the middle part, co. PBC and PB are the opposite parts; and, if, in the triangle ABP, co. BAP is the middle part, co. PBA and PB are the opposite parts. Hence, by (605),
cos. C cos. BAP :: sin: PBC: sin. PBA. (779)
(fig. 4.) BAP = A, and (fig. 5.) BAP = 180° — A ; (780)
(fig. 4.) sin. PBA = sin. (B — PBC)
sin. B cos. PBC
cos. B sin. PBC,
(fig. 5.) sin. PBA= sin. (PBC — B)
sin. B cos. PBC+cos. B sin. PBC;
whence (779) becomes (fig. 4.)
cos. C: cos. A :: sin. PBC
: sin. B cos. PBC-cos. B sin. PBC;
and, (fig. 5.)
cos. A sin. PBC
: sin. B cos PBC + cos. B sin. PBC, which becomes the same as (785) by changing the signs of the second and fourth terms.
Divide the two terms of the second ratio of (785) by sin. PBC and reduce, by (11),
(787) cos. C; cos. A :: 1 sin. B cotan. PBC cos. B. Make the product of the means equal that of the
extremes, and we have
(788) sin. B cos. C cotan. PBC
cos. B cos. C: =cos. A ;
sin. B cos. C cotan. PBC=cos. A+ cos. B cos. C. Divide by sin. B sin. C, and reduce by (11)
(790) cotan. C cotan. PBC
cos. A + cos. B cos. C
which, subtituted in (778), gives
Cos. a =
sin. B sin. C
cos. Acos. B cos. C
sin. B sin. C
whence the value of the side a may be calculated, and in the same way either of the other sides.
66. Corollary. The equation (791) may be brought into a form more easy for calculation by logarithms, as follows.
Subtract each member from unity and it becomes
1 COS. a
-cos. B cos. C+ sin. B sin. C ;
which, substituted in the numerator of (792), gives
Now we have by (121), changing the signs,
and letting S denote half the sum of the angles or
if we make in (795),
S M = 1⁄2 (A + B + C) = S,
N = 1 (− A + B + C) = S — A ;
67. Corollary. The sides b and c may be found
by the two following equations which are readily deduced from (803),
68. Corollary. The equation (791) can be brought into another form equally simple for calculation. Add each member to unity and it becomes
(809) cos. (M+N) + cos. (M — N) = 2 cos. M cos. N;
=2 cos. (SB) cos. (SC);
which, substituted in (808), gives