30. By the formulas here given, a complete table of sines, cosines, &c., may be calculated. Such tables have been actually calculated, but generally by methods more convenient in practice than that explained in this chapter. Table XXIV of the Navigator is a table of sines and cosines calculated to five places of decimals; and Table XXVII gives the five-place logarithms of all the trigonometric functions.

The trigonometric functions themselves are called natural, as in Table XXIV, to distinguish them from their logarithms, which are more often used, and which are sometimes called the artificial sines, &c.

Table XXIV is constructed on the system of § 21, the radius being

so that this table is reduced to the present system by dividing each number by this radius; that is, by putting the decimal point five places back, or prefixing it to each number as it is given in the table.

so that this table is reduced to the present system by subtracting from each number the logarithm of this radius, which is 10; that is, by subtracting 10 from each characteristic.

These values of the radius are taken in order to avoid printing the decimal point in the first case, and to avoid negative characteristics in the second case.

The method of using these two tables is fully explained in pp. 33-35 (given at the end of the Useful Tables) and pp. 391, 392 of the Navigator. It is supposed to be understood in the remainder of this book.

If we disregard the "Hour" columns in Table XXVII, with which at present we have nothing to do, and the insertion of angles greater than 90°, which will be explained in a future chapter, this table corresponds precisely to the series of right triangles described in § 4, the two opposite angles being always complements of each other, and the six principal columns giving the values of the six trigonometric ratios, each of which, as in § 8, bears complementary relations to the two angles.

CHAPTER III.

RIGHT TRIANGLES.

31. The general formulas which are obtained for the solution of triangles should in each case, as far as possible, express the unknown parts in terms only of those which are given at the outset; but it is occasionally better, for practical reasons, in the working of a numerical example, to compute certain of the unknown parts first, and then use these in finding the others.

Two classes of problems in the solution of right triangles may be distinguished; the first class including those in which an acute angle and a side are given; the second, those in which two sides are given.

In problems of the first class, the general method of finding either unknown side is to see what trigonometric function of the known angle is represented by the ratio of the side sought to the given side, find its value in the table, and multiply it by the given side. The unknown angle is the complement of the known angle.

In problems of the second class, the general method of finding either acute angle is to see what function of this angle is represented by the ratio of the given sides, find the value of this ratio, and look out the corresponding value of the angle in the table. The unknown side may be found by the Pythagorean Proposition.

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.

from 90°.

To find the other acute angle B, subtract the given angle

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

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 (4),

= =

log. h log. a+ log. cosec. A log. a + (ar. co.) log. sin. A.

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

34. Problem.

"b cotan. A=

-

a

ba cotan. A;

log. blog. a+log. cotan A.

(24)

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. We have for h and b, from (4) and (6),

35. Problem.

log. h = log. a + log. sec. B,

log. blog. a + log. tang. B.

To solve a right triangle, when the hypothe

nuse and a leg are known. [B., p. 40.]

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),

or, by logarithms,

α

sin. A cos. B

;

(27)

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

b =

log. b =

√ (h2 — a2) = √ [(h+a) (h —− a)];

log. (h2 — a2) = 1 [log. (h+a) +log. (h—a)].

(28)

(29)

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),

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

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

Thirdly. A practically better way of finding the hypothenuse is to make use of (23) or (25),

ha cosec. A≈ a sec. B;

log. h = log. a+ log. cosec. A: log. a+ log. sec. B.

(32)

37. EXAMPLES.

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.

Solution. The other angle = 90°· - 54° 44′ 35° 16'. Then making h49.58, and A 54° 44'; we have, by (21) and (22),

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.

3. Given the two legs of a right triangle equal to 44.375, and 22.165; to solve the triangle,

*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.