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Ex. 1. Given c = 140°, a = 20°; find b, A, B.

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Ex. 2. Given c = 72° 30', a = 45° 15'; find b, A, B.

Ans. b = 64° 42′ 52′′, A = 48° 7' 44".5, B

= 71° 27' 15".

203. Case III.-Given a side a and the adjacent angle B; to find A, b, c.

By (10), (6), (4) of Art. 185, we have

COS A = cos a sin B, tan b = sin a tan B, cot c = cot a cos B.

Check formula,

cos Atan b cot c.

In this case there is evidently no ambiguity.

Ex. 1. Given a=31° 20′ 45′′, B=55° 30′ 30"; find A, b, c.

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Ex. 2. Given a=112° 0′ 0′′, B=152° 23′ 1′′.3; find A, b, c. Ans. A = 100°, b = 154° 7' 26".5, c = 70° 18' 10".2.

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204. Case IV. Given a side a and the opposite angle A; to find b, c, B.

By (7), (3), (10) of Art. 185, we have

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In this case there is an ambiguity, as the parts are determined by their sines, and two values for each are in general admissible. But for each value of b there will, in general, be only one value for c, since c and b are connected by the relation cos c = cos a cos b (Art. 185); and at the same time only one admissible value for B, since cos c = cot A cot B. Hence there will be, in general, only two triangles having the given parts, except when the side a is a quadrant and the angle A is also 90°, in which case the solution becomes indeterminate.

It is also easily seen from a figure that the ambiguity must occur in general (Art. 188).

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When a A, the formulæ, and also the figure, show that b, c, and B are each 90°.

Ex. 1. Given a = 46° 45', A = 59° 12'; find b, c, B.

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Ans. b = 154° 7'26".5, c=70° 18'10".2,

or 25° 52' 33".5, or 109° 41'49".8,

205. Case V. Given the two sides a and b; to find A,

B, c.

By (7), (6), (1) of Art. 185, we have

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cot A

= cot a sin b, cot B

=

cot b sin a, cos c = cos a cos b.

Check formula,

cos ccot A cot B.

In this case there is no ambiguity.

Ex. 1. Given a = ÷ 54° 16′, b = 33° 12'; find A, B, c.

Ans. A = 68° 29′53′′, B = 38° 52′ 26′′, c = 60° 44′46′′.

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Ex. 2. Given a 56° 34', b = 27° 18'; find A, B, c.

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Ans. A = 73° 9′ 13′′, B = 31° 44′ 9", c = 60° 41′9".

206. Case VI.

b, and c.

By (10), (9), (8)

Given the two angles A and B; to find a,

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32°10'; find a, b, c.

28° 24′ 54′′, c = 63° 21′ 24′′.5.

Ex. 2. Given A 91° 11', B = 111° 11'; find a, b, c.
Ans. a 91° 16' 8", b = 111° 11' 16", c = = 89° 32′ 29′′.

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207. Quadrantal and Isosceles Triangles. Since the polar triangle of a quadrantal triangle is a right triangle (Art. 184), we have only to solve the polar triangle by the formulæ of Art. 185, and take the supplements of the parts thus found for the required parts of the given triangle; or we can solve the quadrantal triangle immediately by the formulæ of Art. 189.*

A biquadrantal triangle is indeterminate unless either the base or the vertical angle be given.

An isosceles triangle is easily solved by dividing it into two equal right triangles by drawing an arc from the vertex to the middle of the base.

The solution of triangles in which a+b=π, or A + B = π, can be made to depend on the solution of right triangles. Thus (see the second figure of Art. 191) the triangle B'AC has the two equal sides, a' and b, given, or the two equal angles, A and B', given, according as a+b=π or A + B in the triangle ABC.

EXAMPLES.

Solve the following right triangles :

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3. Given c=55° 9′32′′, | a=22°15' 7";

find b=51°53', A=27°28'37'.5, B=73° 27'11".1.

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* Quadrantal triangles are generally avoided in practice, but when unavoidable, they are readily solved by either of these methods.

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12. Given A=67° 54' 47", ·| B=99° 57′35′′; find a=67°33'27", b=100° 45',

13. Solve the quadrantal triangle in which

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c=94°5'.

112°10'20", a = 46°31′36′′.

14. Solve the quadrantal triangle in which

a = 174° 12′ 49′′.1, b = 94° 8' 20", c= = 90°.

Ans. A = 175° 57′10′′, B = 135° 42′ 55′′, C=135°34′8′′.

SOLUTION OF OBLIQUE SPHERICAL TRIANGLES.

208. The Solution of Oblique Spherical Triangles presents Six Cases; as follows:

I. Given two sides and the included angle, a, b, C.

II. Given two angles and the included side, A, B, C.
III. Given two sides and an angle opposite one of them,

a, b, A.

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