CHAPTER III.

FORMULE FOR THE SOLUTION OF TRIANGLES.

We shall here repeat the enunciations of the two propositions established in Chapter I.

PROP. I.

In any right-angled plane triangle,

I. The ratio which the side opposite to one of the acute angles has to the hypothenuse, is the sine of that angle.

2o. The ratio which the side adjacent to one of the acute angles has to the hypothenuse, is the cosine of that angle.

3o. The ratio which the side opposite to one of the acute angles has to the side adjacent to that angle, is the tangent of that angle.

Thus, in any right-angled triangle ABC,

In any plane triangle, the sides are to each other as the sines of the angles opposite to them.

We shall, henceforth, in treating of triangles, make use of the following notation. We shall denote the angles of the triangle by the large letters at the angular points, and the sides of the triangle opposite to these angles, by the corresponding small letters.

Thus, in the triangle ABC, we shall denote the angles BAC, CBA, BCA, by the letters A, B, C, respectively, and the sides BC, AC, AB, by the letters a, b, c, respectively.

According to this, we shall have, by the proposition,

"

a

..(3)

B

PROP. III.

In any plane triangle, the sum of any two sides, is to their difference, as the tangent of half the sum of the angles opposite to them, is to the tangent of half their difference.

Let ABC be any plane triangle, then, by proposition, II

To express the cosine of an angle of a plane triangle in terms of the sides of

the triangle.

Let ABC be a triangle; A, B, C, the three angles;

a, b, c, the corresponding sides.

1. Let the proposed angle (A) be acute.

From C draw CD perpendicular to AB, the base of the triangle.

It will be seen that this result is identical with that which we deduced in the last case, so that, whether A be acute or obtuse, we shall have,

To express the sine of an angle of a plane triangle in terms of the sides of the triangle.

Let A be the proposed angle; then by last prop.,

1

sin. A =2bc Va+b+cxb+c—a)(a+c—b)(a+b—c)...(3)

The above expression, for the sine of an angle of a triangle in terms of the sides, is sometimes exhibited under a form somewhat different.

8

Let s denote the semiperimeter, that is to say, half the sum of the sides of the triangle; then

8 =

s—a =

a+b+c, and,

2
ca
2
a+c-b

b+c

8- b =

=

...

...

... 2 (sc) = a+b-c

a),...... for a+b+c, b + c — a,...... in the ex

Proceeding in the same manner for the other angles, we shall

ab

√s (s — a) (s—b) (s — c)

Dividing the formulæ marked () by those marked (a), we have