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CHAPTER III.

RIGHT TRIANGLES.

32. Problem. To solve a right triangle, when the hypothenuse and one of the angles are known. [B. p. 38.]

Solution. Given (fig. 4) the hypothenuse h and the angle A, to solve the triangle.

First. To find the other acute angle B, subtract the given angle from 90°.

Secondly. To find the opposite side a, we have by (1),

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log. a = log. h + log. sin. A.

Thirdly. To find the side b, we have by (4)

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(21)

(22)

log.blog.h + log. cos. A.

33. Problem. To solve a right triangle, when a leg and the opposite angle are known. [B. p. 39.]

Solution. Given (fig. 4) the leg a, and the opposite angle A, to solve the triangle.

First. The angle B is the complement of A.

Secondly. To find the hypothenuse h, we have by (21)

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Thirdly. To find the other leg b, we have by (4)

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34. Problem. To solve a right triangle, when a leg and the adjacent angle are known. [B. p. 39.]

Solution. Given (fig. 4) the leg a and the angle B, to solve the triangle.

First. The angle A is the complement of B.

Secondly. The other parts may be found by (23) and (24), or from the following equations, which are readily deduced from equations (4) and (6),

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35. Problem. To solve a right triangle, when the hypothenuse and a leg are known. [B. p. 49.]

Solution. Given (fig. 4) the hypothenuse h and the leg a, to solve the triangle.

First. The angles A and B are obtained from equation (4),

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log. sin A= log. cos. B = log. a +(ar. co.) log. h.

Secondly. The leg b is deduced from the Pythagorean property of the right triangle, which gives

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log. b = 1⁄2 log. (h2 — a2) = } [log. (h+a) + log. (h — a)].

36. Problem. To solve a right triangle, when the two legs are known. [B. p. 40.]

Solution. Given (fig. 4) the legs a and b, to solve the triangle. First. The angles are obtained from (4)

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log. tang. A

(30)

log. cotan. Blog. a + (ar. co.) log. b.

Secondly. To find the hypothenuse, we have by (28)

h = √(a2 + b2).

(31)

Thirdly. An easier way of finding the hypothenuse is to make

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1. Given the hypothenuse of a right triangle equal to 49.58, and one of the acute angles equal to 54° 44'; to solve the triangle.

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Solution. The other angle 90° 54° 44' 35° 16'. Then making h=49.58, and A= 54° 44'; we have, by (21) and (22),

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2. Given the hypothenuse of a right triangle equal to 54.571, and one of the legs equal to 23.479; to solve the triangle.

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3. Given the two legs of a right triangle equal to 44.375, and

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* To avoid negative characteristics, the logarithms are retained as in the tables, according to the usual practice with the logarithms of decimals, as in B., p. 29.

4. Given the hypothenuse of a right triangle equal to 37.364, and one of the acute angles equal to 12° 30′; to solve the triangle.

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5. Given one of the legs of a right triangle equal to 14.548, and the opposite angle equal to 54° 24';, to solve the triangle.

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6. Given one of the legs of a right triangle equal to 11.111, and the adjacent angle equal to 11° 11'; to solve the triangle.

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7. Given the hypothenuse of a right triangle equal to 100, and one of the legs equal to 1; to solve the triangle.

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8. Given the two legs of a right triangle equal to 8.148, and 10.864; to solve the triangle.

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