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(662) 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).

EXAMPLES.

1. Given, in the spherical triangle ABC,
A = 95°, C = 104°, and a

138° ;

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b = 12° 17′ + 4° 17′ = 16° 34′.

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2. Given, in the spherical triangle ABC,

A = 104°, C = 95°, and a = 138° ;

to solve the triangle.

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46. Problem. To solve a spherical triangle when its three sides are given.

Solution. Let ABC (figs. 4. and 5.) be the trian

gle; a, b, and e 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.

*

(664) (665)

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),

But

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),

(667)

(668)

cos. AP = cos. (b - 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 have

(670)

(671)

(672)

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 sin. b tang. PC = cos. c - cos. a cos. b.

Divide by sin. a sin. b and reduce the first member

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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

becomes

1 + cos. C

But, by (104),

cos. c - cos. a cos. b + sin. a sin. b
sin. a sin. b

cos. (a + b) = cos. a cos. b - sin. a sin. b, which, substituted in the numerator of (675), gives

COS. c 1 + cos. C

cos. (a + b)

sin. a sin. b

Now we have (122)

1

(675)

(676)

(677)

cos. (M - N) - cos. (M + N) = 2 sin. M sin. N; (678)

and letting s denote half the sum of the sides or

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cos. c - cos. (a + b) = 2 sin. s sin. (s - c);

which, substituted in (677), gives

(681)

(682)

(683)

1 + cos. C =

2 sin. s sin. (s - c)

sin. a sin. b

But, by (140),

1+cos. C = 2 (cos. + C)2,

(684)

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cos. C =

(686)

sin. a sin. b

48. Corollary. The angles A and B may be found by the two following equations which are easily deduced from (686),

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49. Corollary. The equation (674) can be brought into another form equally simple in calculation. Sub

tract each member from unity

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(690) cos. (a - b) = cos. a cos. b + sin. a sin. b,

which substituted in the numerator of (689) gives

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(692) COS. (MN) - cos. (M + N) = 2 sin. M sin. N;

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(695)

(a-b+c) = = s - b,

N=(a+b+c) = s—a;

{M+N=&

and (692) becomes

cos. (a - b) - cos. c = 2 sin. (s-a) sin. (s - b);

which, substituted in (691), gives

2 sin. (s - a) sin. (s - b)

1 - cos. C

(696)

sin. a sin. b

But, by (141),

(697)

1-cos. C

2 (sin. + C)2;

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