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Formulæ (184) and (185) may be thus expressed:

cos (a+b): cos(ab):: cot

sin(a+b) sin (a - b) :: cot That is,

C: tan (A+B), (186).

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C: tan (4- B). (187)

The cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference as the cotangent of half the included angle is to the tangent of half the sum of the other two angles.

The sine of half the sum of two sides of a spherical triangle is to the sine of half their difference as the cotangent of half the included angle is to the tangent of half the difference of the other two angles.

By applying (186) and (187) to the polar triangle, Art. 150, we obtain

cos (A+B): cos ≥ (A — B) : : tan sin 1 (A+B); sin ¿ (A — B) : : tan

That is,

c: tan † (a+b), (188)

c: tan (a — b). (189)

The cosine of half the sum of two angles of a spherical triangle is to the cosine of half their difference as the tangent of half the included side is to the tangent of half the sum of the other two sides.

The sine of half the sum of two angles of a spherical triangle is to the sine of half their difference as the tangent of half the included side is to the tangent of half the difference of the other two sides.

The above four proportions are called, from their inventor, Napier's Analogies.

RELATIONS BETWEEN THE SIDES AND ANGLES OF RIGHT-ANGLED SPHERICAL TRIANGLES.

157. The sine of either oblique angle is equal to the sine of the opposite side, divided by the sine of the hypothenuse.

Let A B C be any spherical trian

gle, right-angled at C.

By means of (146) we have

B

h

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C

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158. The cosine of either oblique angle is equal to the tangent

of the adjacent side, divided by the tangent of the hypothenuse.

By means of (161) we have

cot h sin b = cot C sin A+ cos

but, if α= 90°, then cot C=0, and

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159. The tangent of either oblique angle is equal to the tangent of the opposite side, divided by the sine of the adjacent side.

By means of (156) we have

cot p sin b = cot A sin C+ cos b cos C';

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160. The sine of either oblique angle is equal to the cosine of the other, divided by the cosine of its opposite side.

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which, by making C=90°, becomes

cos Bcos b sin A,

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161. The cosine of the hypothenuse is equal to the product of the

cosines of the other two sides.

By means of (152) we have

cos h = cos p cos b + sin p sin b cos 0,

which, by making C

=

90°, becomes

cos hcos p cos b.

(198)

162. The cosine of the hypothenuse is equal to the product of the cotangents of the two oblique angles.

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163. The preceding formulæ may readily be remembered from their similarity to the corresponding ones for plane triangles; and, for convenience of reference, they are brought together in the following

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SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES.

164. The solution of spherical triangles is the process by which, when the values of a sufficient number of their six elements are given, we calculate the values of the remaining elements.

In order to solve a right-angled spherical triangle, two of its elements, other than the right angle, must be given.

165. The formulæ requisite for the solution of right-angled spherical triangles are readily furnished by means of the relations demonstrated in the foregoing articles. Thus,

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enable us to determine every case of right-angled spherical triangles. For every one of these ten equations is a distinct combination, involving three of the five quantities, p, b, h, A, B; and five quantities, taken three at a time, can be combined only. in ten different ways.

NAPIER'S CIRCULAR PARTS.

166. If, in any right-angled spherical triangle, the right angle be left out of consideration, the two sides adjacent to the right angle, and the complements of the hypothenuse and of the two other angles, are called the five circular parts of the triangle.

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