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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).

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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 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.

(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) cos, a cos. c :: 1: cos. b + sin. b tang. PC.

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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 by (11),

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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

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1+ cos. C

cos. (MN) - cos. (M+N)

which, substituted in the numerator of (675), gives

Now we have (122).

·

and letting s denote half the sum of the sides or

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= 2 sin. M sin. N; (678)

(679)

{ N =

(a + b —c) = s — c;

(680)

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(687)

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. Subtract each member from unity

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(690)

cos. (a - b)

cos. a cos. b + sin. a sin. b,

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which substituted in the numerator of (689) gives

Now, if in (678),

(692) COS. (MN)

SM = 1⁄2 (a - b + c)

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cos. (M + N)

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(693)

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(695)

and (692) becomes

cos. (a - b) — cos. c = 2 sin. (s—a) sin. (s — b) ; which, substituted in (691), gives

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