CASE II. When the included angle A is obtuse.

In Fig. 3, BD = AD+c.

Squaring, and adding CD to both members,

But,

Also,

BD2 + CD2 = AD2 + CD2 + c2 + 2 c × AD.

BD+CD a2, and AD2 + CD2 = b2.

=

117. To express the cosines of the angles of a triangle in terms of the sides of the triangle.

118. To express the sines, cosines, and tangents of the half-angles of a triangle in terms of the sides of the triangle.

Denoting a+b+c by 2s, so that s is the half-sum of the sides of the triangle, we have

and

ab+c=(a+b+c)-2b=2s-2b-2 (s—b), a+b-c=(a+b+c)-2c=2s-2c=2(s-c).

Again, adding both members of (55) to unity, we have

b+ca=(b+c+a)-2a=2(s-a).

cos2 A =

and

Hence,

or,

In like manner,

and

48 (s- a)

4 bc

bc

са

cos C=

ab

(63)

Note. Since each angle of a triangle is less than 180°, its half is

less than 90°; hence the positive sign must be taken before the radical in each of the formulæ of Art. 118.

AREA OF AN OBLIQUE TRIANGLE.

119. CASE I. Given two sides and their included angle.

There will be two cases, according as the included angle A is acute (Fig. 1), or obtuse (Fig. 2).

In each case let CD be drawn perpendicular to AB. Then, denoting the area of the triangle by K, we have by Geometry,

=2bc sin cos A (Art. 74).

Dividing by 2, and substituting the values of sin A and cos A from Art. 118, we have