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If B differs more from 90° than A, b must be of the same species as B, and there can be but one solution; but if B differs less from 90° than A, there may be two solutions.

Art. 179.)

(Prop. VII.

Or, if only one of the supplementary values of b makes (a+b) of the same species as (A+B), there can be but one solution; but if both values of b fulfil that condition, there will be two solutions. (Prop. VIII. Art. 179.)

EXAMPLES.

1. Given, in an oblique-angled spherical triangle, the angle A equal to 135°, the angle B equal to 60°, and the side a equal to 155°; to find the other parts.

Ans. C, 98° 3' 4" or 16° 57' 1"; b, 31° 10′ 17′′ or 148° 49'

43"; c, 143° 42′ 57′′ or 10° 2′ 6′′.

2. Given, in an oblique-angled spherical triangle, two angles equal to 97° 26′ 30′′ and 65° 33′ 10′′, and the side opposite to the first equal to 100° 49′ 30′′; to find the other parts.

CASE III.

183. Given two sides and the included angle.

Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the sides a and b, and the included angle C; to solve the triangle.

To find A and B. By means of Napier's analogies (186) and (187), we have

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cos (a - b) cot C,
(a + b)

cos

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sin † (a — b) cot † C ;
sin (a+b)

log tan (A+B)= log cos (a-b) — log cos(a+b)

1⁄2 1 log tan (A-B) = log sin (a—b) — log sin

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(a+b)

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values gives A, and the second subtracted from the first gives B.

To find c. We use equation (240), as in the first two cases. The value of c might also be obtained by (147) or (149); but as it is thus determined from its sine, it would be necessary to remove the ambiguity by means of the principles contained in Art.

179.

As A, B, and c may all be found by means of tangents, there can be but one value for each. It will be observed that ₫ (A+B) must always be of the same species as (a+b). (Prop. VIII. Art. 179.)

EXAMPLES.

1. Given, in an oblique-angled spherical triangle ABC, the side a equal to 70°, the side b equal to 38° 30', and the included angle equal to 31° 34′ 26′′; to solve the triangle.

Solution.

(a + b) = 54° 15', (a — b) = 15° 45', and ≥ α= 15° 47′ 13′′; then,

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Ans. Angle A, 130° 3′ 11′′; angle B, 30° 28′ 11′′; side c, 40°.

2. Given, in an oblique-angled spherical triangle, an angle equal to 48° 36', and the two adjacent sides equal to 112° 22′ 58′′ and 89° 16′ 53′′; to find the other parts.

CASE IV.

184. Given two angles and the included side.

Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A and B, and the ineluded side c; to solve the triangle.

To find a and b. By means of Napier's analogies (188) and (189), we have

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log tan (a+b)=log cos (A-B) — log cos (A+B)

log tan (ab) = log sin (A-B) — log sin

which determine

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To find C. We use equation (239), as in the first two cases; but (147) or (149) may be employed, as in the last case.

This case is analogous to Case III., and gives rise to no ambiguity.

EXAMPLES.

1. Given, in an oblique-angled spherical triangle ABC, the angles A and B equal to 119° 15′ and 70° 39', and the side c equal to 52° 39′ 4′′; to solve the triangle.

Ans. Sides a and b, 112° 22′ 58′′ and 89° 16' 53"; angle C, 48° 36'.

2. In an oblique-angled spherical triangle, given two angles equal to 130° 3′ 11′′ and 31° 34′ 26′′, and the included side equal to 38° 30'; to find the other parts.

CASE V.

185. Given the three sides.

Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the sides a, b, and c; to solve the triangle.

To find A, B, and C, we have, by (164), (165), and (166),

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log sin4=

log sin (s-b)+log sin (s—c)-log sin blog sin c

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(246)

2

log sin B

log sin (s—c)+log sin (s—a)—log sin c-log sin a

=

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(247)

log sin C

log sin (s-a)+log sin (s—b)—log sin a-log sin b

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(248)

2

A, B, and C can also be determined by formulæ (167), (168), and (169) for the cosine of half an angle, and by formulæ (170), (171), and (172) for the tangent of half an angle.

Since the half-angles must be less than 90°, there is no ambiguity in determining the angles by any of these formulæ.

EXAMPLES.

1. Given, in an oblique-angled spherical triangle, the side a equal to 70°, the side b equal to 38°, and the side c equal to 40°;

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A, log sin 9.959686B, log sin 9.405047 C, log sin 9.425238

A = 65° 41′ 33′′.7 1⁄2 B = 14° 43′ 18′′ ₫ C=15° 26′ 21′′.7
Ans. A, 131° 23′ 7′′; B, 29° 26' 36"; C, 30° 52′ 43′′.

2. Given, in an oblique-angled spherical triangle, the sides equal to 112° 22′ 59′′, 89° 16′ 53′′, and 52° 39′ 4′′; to solve the triangle.

CASE VI.

186. Given the three angles.

Let there be given, in the oblique-angled spherical triangle ABC (Fig. Art. 181), the angles A, B, and C ; to solve the triangle.

To find a, b, and c, we have, by (175),

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log cos S+log cos (S-A)-log sin B-log sin C

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, (249)

log cos S+log cos (S—B)—log sin C-log sin A

log sin b=

2

, (250)

log cos S+log cos (S-C)-log sin A-log sin B

log sin c=

T.

(251)

2

a, b, and c can also be determined by formulæ (176) for the cosine of half an angle, and by formulæ (177) for the tangent of half an angle.

Here a, b, and c are determined without ambiguity.

EXAMPLES.

1. Given, in an oblique-angled spherical triangle, the angle A equal to 120° 43' 37", the angle B equal to 109° 55′ 42′′, and the angle equal to 116° 38' 33"; to find the sides.

Ans. a, 115° 13′ 26′′; b, 98° 21′ 39′′; c, 109° 50′ 20′′.

2. Given the angles in an oblique-angled spherical triangle equal to 131° 23′ 7′′, 29° 26′ 36′′, and 30° 52′ 43′′; to solve the triangle.

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