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)

and

But

(fig. 4.) BAP = A, and (fig. 5.) BAP = 180° — A ; (780) also,

(fig. 4.)

PBA B-PBC,

(fig. 5.) Hence, and by (190), (fig. 4.) (fig. 5.) cos. BAP

cos. BAP

also by (781), (91), and (202),

=

=

PBA PBC — B.

(fig. 4.) sin. PBA = sin. (B — PBC)
sin. B cos. PBC cos. B sin. PBC,
(fig. 5.) sin. PBA= sin. (PBC — B)
sin. (BPBC)

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;

cos. A,
cos. A;

(778)

(781)

(782)

(783)

(784)

(785)

(786)

cos. C: 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 ;

(790)

(791)

and, (fig. 5.)

by transposition,

(789)

sin. B cos. C cotan. PBC=cos. A+ cos. B cos. C. Divide by sin. B sin. C, and reduce by (11)

cotan. C cotan. PBC

cos. A cos. B cos. C
sin. B sin. C

"

(793)

(794)

(792) 1

which, subtituted in (778), gives

COS. a =

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.

COS. a

cos. Acos. B cos. C

"

Subtract each member from unity and it becomes sin. B sin. C― cos. A cos. B cos. C sin. B sin. C But, by (104), changing the signs cos. (B+C)= cos. B cos. C + sin. B sin. C' ; which, substituted in the numerator of (792), gives — cos. (B+C) — cos. A sin. B sin. c

1 cos. α =

Now we have by (121), changing the signs,

N)

-cos. (MN) cos. (M
2 cos. M cos. N;

and letting S denote half the sum of the angles or

S=(A + B + C);

(796)

if we make in (795),

we have

S M = 1⁄2 (A + B + C) = S,
{ N = } ( A + B + C)= S — A;

=

sin. a =

2 cos. S cos. (S
sin. B sin. C

cos. a = 2 (sin. † a)2;

- A)

2 cos. S cos. (S — A)
sin. B sin. C

– A),

A)

67. Corollary.

The sides b and c may be found

by the two following equations which are readily deduced from (803),

(795)

(797)

(798)

(799)

(800)

(801)

(802)

(803)

(804)

(805)

(810)

(811)

(812)

(813)

(814)

sin.c

(808) 1 + cos. a=

68. Corollary. The equation (791) can be brought into another form equally simple for calculation. Add each member to unity and it becomes

(806) 1 + cos. a=

But, by (116), (807) cos. (BC)

=

Now we have, by (121),

(809) cos. (M+N) + cos. (MN):

and if we make

we have

which, substituted in (806), gives

cos. S cos. (S
sin. A sin. B

=

cos. A cos. B cos. C + sin. B sin. C sin. B sin. C

and (809) becomes

cos. B cos. C+ sin. B sin. C';

SM+N=A,
M-NB-C;

=

cos. A + cos. (B —- C)
sin. B sin. C

1 + cos. a=

B)

M (A + B — C) = SC,
N
1⁄2 (A - B+ C) = S

Ᏼ ;

C)

which, substituted in (808), gives

2 cos. (S

But, by (140) we have

cos. A cos. (B — C)

=2 cos. (SB) cos. (SC);

2 cos. M cos. N;

B) cos. (SC)

sin. B sin. C

1 + cos. a= 2 (cos. 1⁄2 a)2;

tang. c =

2 cos. (S

69. Corollary. duce the following equations,

cos. b =

cos. (S — B) cos. (S — C)
sin. B sin. C

B) cos. (SC)

In the same way we might de

COS. (SA) cos. (S
sin. A sin. C

sin. B sin. C

cos. (S- A) cos. (S
sin. A sin. B

cos. (S

cos. (S

70. Corollary. The quotient of (803) divided by (816) is

tang. a =

In the same way

tang. b =

cos. S cos. (S

--

cos. S cos. (S- A)

cos. (SB) cos. (S — C) '

cos. S cos. (S

B)

A) cos. (SC)

C')

B)

Ans.

C)

4) cos. (SB)

=

A 89°, B = 5°, C 88°;

=

to solve the triangle.

EXAMPLE. Given, in the spherical triangle ABC, (figs 4. and 5.),

(815)

(816)

(817)

(818)

(919)

(820)

(821)