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or, sin side req. > sin hypoth.

to find the rest;

sin a sin B;

sin opp. angle

B

tanc tan a COS B;

or, tan side req.tan hypoth. x cos included angle, cot c cos a tan B;

or, cot angle req. cos hypoth. x tan given angle. In this case there can be nothing ambiguous; for, in applying the first form, it is known that the angle and the opposite side are always of the same affection; and in the two latter the rules for the changes of sines in the different quadrants (chap. iv. 9), will determine to which the result belongs.

Case II.

27. Given the hypothenuse a, and one of the sides b; to find the rest.

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28. Given the two sides including the right angle, namely, b and c; to find the rest.

cosa cos b cos c;

or, cos hypoth. rectangle cosines of the sides..

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Case IV.

29. Given a side b, and its opposite angle в; to find the rest.

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30. Given a side c, and its adjacent angle в; to find the rest.

tan b tan B sin c;

or, tan side req. = tan opp. angle × sin given side.

tan a =

tan c
COS B

; or, tan hypoth. =

tan given side
cos given angle

or, cot a cos B cot c;

that is, cot hypoth. = cos given angle x tan given side. cos ccos c sin B;

or, cos angle req. = cos opp. side x sin given angle.

Case VI.

31. Given the two oblique angles B and c; to find the rest.

cos ccot B cot c;

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or, cos hypoth. rectangle cot's given angles.

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Note 1. Here the rule of the signs (chap. iv. 9) serves

all along to determine the kind or affection of the unknown parts.

Note 2. In working by the logarithms, the student must observe, that when the resulting log. is the log. of a quotient, 10 must be added to the index: when it is the log. of a product, 10 must be subtracted from the index. This is done in conformity with the rule (chap. iv. 17), to make the terms homogeneous by multiplying or dividing by the powers of radius.

SECTION III.

Resolution of Oblique angled Spherical Triangles. 32. This may be effected by means of four general cases; each comprehending two or more problems.

Case I.

Given three of these four things, viz. two sides b, c, and their opposite angles B, C; to find the fourth.

This case comprehends two problems, in one of which the unknown quantity is an angle, in the other a side. They are both solved by means of equa. (1) of this chapter from which we have

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33. Of these four things, viz. the three sides a, b, c, and an angle, any three being given, to find the fourth. This case comprises three problems.

1. When the three sides are given, to find an angle. Here from equa. (2) we have

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But these are not fitted for logarithmic computation. Recurring, therefore, to (chap. iv. equa. t'), we have 1+ cos A = 2 cos2 A, and 1 - cos a = 2 sin2 þa. Hence,

2 cos2 A =

sin b sin c

+

sin b sin c

cos a cos b cos c

sin b sin e

=

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The latter of these divided by the former, gives,

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Whence, from the 4th of equa. (u) chap. iv.

tan1 A =

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

Hence we have, for the tangents of the half angles, these three symmetrical equations:

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The expressions for the sines of the half angles might be deduced with equal facility. As they are symmetrical we shall put down but one, viz.

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Expressions for the cosines and cotangents of the half angles, may be readily found from the above, by

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Cor. When two of the sides, as b and c, become equal,

sin a

the expression for sin A, becomes sin a = sin b

Cor. 2. If a = b = c = 90°, then sin A = = √2 = sin 45°; and A = B = C = 90°.

Leaving other corollaries to be deduced by the student, let us proceed to the next problem in this case.

2. To find the side c opposite to the given angle ; that is, given two sides and the included angle, to find the third side.

Find from the data a dependent angle 4, such that tan 4 cos c tan b (12.)

....

Substitute for cos c in the third of equa. (2) its value

in this, it will become

1

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Note. The equa. (12) obviously reduces to

.COS C tan p

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analogous to equa. (8). So that b is the hypothenuse and one leg of a right angled triangle. The above transformation, therefore, is equivalent to the division. of the proposed triangle into two, by an arc from the vertical angle a falling perpendicularly upon the opposite side a.

3. To find the side a, not opposite to the given angle; b, c, and c, being given.

Here find, as before, by equa. (12): then from equa. (13) we have

cos (a — 4) =

COS C COS

cos b

. (14.)

Hence a is known by adding 9.

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