Draw AD perpendicular to BC, or to BC produced.

It follows, therefore, from the symmetrical nature of this

sin C

Q.E.D.

Taking the figures of the last article, we have

(1.) When C is an acute angle

By Euc. II., 13, AB = BC + AC-2 BC. CD.

Hence we have, AB2 BC2 + AC2 2 BC. AC cos C,

=

... cos C

(2.) When C is an obtuse angle, as in the second figure— By Euc. II., 12, AB2 = BC2 + AC2 + 2 BC. CD; AC cos ACDAC cos (180°

=

and CD Hence, AB2

BC2 + AC2 + 2 BC (

or, c2 = a2 + b2 2 ab cos C.

C) :

=

AC cos C

AC cos C);

as before.

32. To express the sine of any angle of a triangle in terms

of the sides.

We have

=

(a + b + c) (b + c − a) (a + c − b) (a + b −c).

(2 bc)2

Hence, taking the square root, and taking the positive sign, because (Art. 11) the sin A is always positive when A is the angle of a triangle, we have—

Sin A

1=

1

2 bc

√(a + b + c) (b + c − a) (a + c − b) (a + b −c).

Let a + b + c = 2 s, then b + c -a = 2

The results in (1), (2), (3) express the area of a triangle in terms of two sides, and the included angle.

We will now express the area in terms of the three sides. We have

34. To express the sine, cosine, and tangent of half an angle of a triangle in terms of the sides.

35. In any triangle ABC, a = b cos C + c cos B. Using the figures of Art. 30, we have

(1.) When C is acute

BD

=

AB cos B, and DC = AC cos C. .. BD + DC AC cos C+ AB cos B, or a = b cos C + c cos B,

=

(2.) When C is obtuse—

and DC

=

BD AB cos B,

AC cos ACDAC cos (180° — C) – – AC cos C. ... BD DC AC cos C+ AB cos B,

or a =

=

b cos C + c cos B, as before.

The form of this result gives us also

b = a cos C + c cos A, c = a cos B + b cos A. COR. 1. Hence sin (B + C)

=

b

=

sin B cos C + cos B sin C.

For we have, 1

a

a

COR 2. Hence also

sin (B - C)

= sin B cos C- - cos B sin C. For, since this result has been proved for any angles of a triangle, it will be also true for a triangle which contains an angle supplementary to B; that is, it will be true, if we put 180°-B for B. We then have—

Sin (180° - B+ C) = sin (180° - B) cos C
+cos (180° B) sin C.

==

=

=

But sin (180° - B + C) = sin (180° - B-C) sin (180° - B) sin B, cos (180° – B) Hence, sin (BC) sin B cos C

=

sin (B-C),

cos B.

cos B cos C.

NOTE. The results of Cor. 1 and Cor. 2 have been proved only for angles less than two right angles. We shall see in Vol. II. they are generally true.