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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.
Resolution of Oblique angled Spherical Triangles.
32. This may be effected by means of four general cases; each comprehending two or more problems.
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 sincsin B sincino.
83. 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 cosa-cosocosc.]
COS A - sin b sinc cos b – cosa cosc COS B = —sin a sinc cos c – cos a cosb COs C - -
sin a sin à J
But these are not fitted for logarithmic computation. Recurring, therefore, to (chap. iv. equa. U'), we have 1 + cos A = 2 cos” #A, and I — cos A = 2 sin” A. Hence,
Hence we have, for the tangents of the half angles, these three symmetrical equations:
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.
Expressions for the cosines and cotangents of the half angles, may be readily found from the above, by
sin cos the forms cos = ..., co tan Siro
Cor. When two of the sides, as b and c, become equal, the expression for sin A, becomes sin AA = #.
Cor. 2. If a = b = c = 90°, then sin 3A = ** = } v2 = 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 c : that is, given two sides and the included angle, to find the # 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 cos c = cos a cos b + sin a cost tan 4
H-cose – To or cot 4 cos c = cot b; with analogous to equa. (8). So that b is the hypothenuse and 4 one leg of a right angled triangle. The above transformation, therefore, is equivalent to the division of the Fo 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 o, as before, by equa. (12): then from equa. (13) we have cos c cos p
. . . . (14.) Hence a is known by adding p.
34. Of these four things, viz. two sides a and c, and two angles B and c, one opposite, the other adjacent; three being given, to find the fourth.
This case presents four problems.
1. Given a, c, B; to find c.
Determine an arc 4' by this condition,
cote = cot & cosm, or # = cott'.... (15)
Substitute this value of cot c for it in the third of
equa. (4); it will become
= (cot o' sin a – cosa) cot B; *::::=2.... (16) -sun P Note. The equa. (15) is akin to equa. (8); showing that the operation here performed is equivalent to letting fall a perpendicular arc from the angle A to the base a ; the subsidiary arc & being the segment adjacent to the angle B. - 2. Given B, c, c; to find a. Here of must be found by equa. (15), and then from equa. (16) we have
. . . . (17.) Whence a becomes known. 3. Given B, c, a ; to find c.
Substitute this value of cot c for it in the third of equa. (4), and it will reduce to cot a cos (B — o') —: $"
Cot c = .... (19.)
4. Given a, c, c; to find B. Determine p" from equa, (18); then
35. Of these four things, viz. the three angles and a side, suppose c, any three being given, to find the fourth.
This comprises three problems.
1. Given the three angles; to find a side.
Suppose the first of equa. (11) to be applied to the solution of the supplemental triangle, by changing a, b, c, and c, into a', bo, c', and c'. Then, to bring it back to the triangle proposed, let there be substituted for a', b', c', and c', the corresponding values 180° – A, 180° — B, 180° – c. 180°– c. Those equations will then be transformed into the following, applicable to the present problem.
- - cos # (A + B – c.) cos # (A – R + c)Y
cot a = V=######3
cos #(B + c – A) coso (A + B – c)
The following are the expressions for the sines of the half angles, viz. coso (A + b + c) coso (b.4 c – A)]
sin #a E - — sin B sinc - cos # (A + b + c) cos # (A + c – B) sin #% - —HI- (22.)
- , scos #(A + b + c) cos # (A + b – c) sin #6 = Veleto;;−2.
—sin Asin B