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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),

But

(778)

cos. C:cos. BAP :: sin: PBC : sin. PBA. (779) (fig. 4.) BAP = A, and (fig. 5.) BAP = 180° A; (780) also,

(fig. 4.) PBA=B PBC,

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

(fig. 5.) sin. PBA=sin. (PВС В)
sin. (B - PBC)

=

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;

(783)

(784)

(785)

and, (fig. 5.)

cos. C:-cos. A:: sin. PBC

(786)

: - 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; 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)

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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.

(792) 1

Subtract each member from unity and it becomes sin. B sin. C-cos. A cos. B cos. C

cos. a

sin. B sin. C

But, by (104), changing the signs

(793) - cos. (B+C)=-cos. B cos. C + sin. B sin. C;

which, substituted in the numerator of (792), gives

1

cos. (B+C) - cos. A

(794)

- cos. a

sin. B sin. c

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

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

2 cos. M cos. N;

and letting S denote half the sum of the angles or

S = + (A + B + C);

(795)

(796)

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67. Corollary. The sides b and e may be found by the two following equations which are readily deduced from (803),

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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),

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

sin. B sin. C

(807) cos. (B - C) = cos. B cos. C + sin. B sin. C;

which, substituted in (806), gives

(808) 1 + cos. a

cos. A + cos. (В С)

sin. B sin. C

Now we have, by (121),

(809) cos. (M+N) + cos. (M-N) = 2 cos. M cos. N;

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

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EXAMPLE. Given, in the spherical triangle ABC,

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c = 53° 8'.

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