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Case I. 26. Given the hypothenuse a C and an angle B; to find the rest; sin b = sin a sin B; B A. or, sin side req. = sin opp. angle C × sin hypoth. tan c = 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. cos a = cos & cos c; or, cos hypoth. = rectangle cosines of the sides. - tan b * - tan c. tan B = H. ... tan c T tanb’

tan opp. side

or, tan angle req. = tan adj, side

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30. Given a side c, and its adjacent angle B; to find

the rest. tan b = tan B sin c; or, tan side req. = tan opp. angle x sin given side. tan c tan given side tan a = ...; or, tan hypoth. = tan given side. cos B cos given angle or, cot a = cos B cot c ; that is, cot hypoth. = cos given angle x tan given side. cos c = cos 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 c = Cot B cot C ; 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.


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 _ cosa – cost cos c Y COS A = sin b sinc cos b – cos a cos c sin a sin c cos c —-cos a cos b sin a sin &

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But these are not fitted for logarithmic computation.

Recurring, therefore, to (chap. iv. equa. U'), we have

1 + cos A = 2 cos” #A, and 1 — cos A = 2 sin” #A.

sin b sinc cosa – cost cosc

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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.

* - I A — sin à (a + b - c) sin #(a + c - 3). sin #A = sin b sin c Expressions for the cosines and cotangents of the half angles, may be readily found from the above, by si cos

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Cor. When two of the sides, as b and c, become equal, sin #a sin à

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the expression for sin #A, becomes sin #A =

= } 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 cosb tan 4

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# ‘p • COS C - H. or cot 4 cos c = cot b, which is 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 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 p, as before, by equa. (12): then from equa. (13) we have

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Hence a is known by adding p.

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