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Denoting the polar triangle by A'B'C', we have by Art. 136, (f):

C' 90°, A' 112° 22', B' = 131° 10'; to find a', b', and c'.

=

=

By Art. 144, the formulæ for the solution are

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Then in the given quadrantal triangle, we have

A = 180° - a' = 59° 38.2',

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XII. SPHERICAL OBLIQUE TRIANGLES.

GENERAL PROPERTIES OF SPHERICAL TRIANGLES.

155. In any spherical triangle, the sines of the sides are proportional to the sines of their opposite angles.

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Let ABC be any spherical triangle, and draw the arc CD perpendicular to AB.

There will be two cases according as CD falls upon AB (Fig. 1), or upon AB produced (Fig. 2).

In the right triangle ACD, in either figure, we have by Art. 144,

Also, in Fig. 1,

sin CD

sin A=

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156. In any spherical triangle, the cosine of either side is equal to the product of the cosines of the other two sides, plus the continued product of their sines and the cosine of their included angle.

In the right triangle BCD, in Fig. 1 of the preceding article, we have by Art. 144,

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cos a = cos c cos AD cos CD + sin c sin AD cos CD.

But in the right triangle ACD, by Art. 144,

cos AD cos CD= cos b.

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157. Let ABC and A'B'C' be a pair of polar triangles.

Α'

B

a

Applying the theorem of Art. 156 to the side a' of the triangle A'B'C', we obtain

cos a' = cos b' cos c'+ sin b' sin c' cos A'.

Putting for a', b', c', and A' their values as given in Art. 136, (f), we have

cos (180° — A) = cos (180° — B) cos (180° — C)

+ sin (180° B) sin (180°-C) cos (180°-a).

Whence by Art. 47,

cos A = ( — cos B) (−cos C)+ sin B sin C ( — cos a).

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That is,

cos A

In like manner,

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and

COS B cos C=

cos C cos A+ sin C sin A cos b,
cos A cos B+ sin A sin B cos c.

(91)

(92)

The above proof illustrates a very important application of the theory of polar triangles in Spherical Trigonometry.

If any relation has been found between the elements of a triangle, an analogous relation may be at once derived from it, in which each side or angle is replaced by the opposite angle or side, with suitable modifications in the algebraic signs.

158. To express the sines, cosines, and tangents of the halfangles of a spherical triangle in terms of the sides of the triangle.

or,

or,

From (87), Art. 156, we obtain

sin b sin c cos A = cos a cos b cos c,

COS A: =

cos acos b cos c
sin b sin c

(B)

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sin2+4=

2 sin } [a+ (b − c)] sin 1⁄2 [a−(b−c)],

sin b sin c

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sin(a+b-c) sin (a − b + c)
sin b sin c

Denoting a+b+c by 2s, so that s is the half-sum of the sides of the triangle, we have

and

a+b-c=(a+b+c)-2c=2s2c2(sc), a−b+c=(a+b+c)-2b=2s-2b=2 (s-b).

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