Plane trigonometry, by S.L. Loney.

252

TRIGONOMETRY.

Also a2 + b2 - 2ab cos B = c2 + d2 - 2cd cos D,
so that
a2 + b2 - c2 - d2 = 2ab cos B - 2cd cos D......(2).
Squaring (1) and (2) and adding, we have
16A2 + (a2 + b2 - c2 - d2)2 = 4a2b2 + 4c2d2
- 8abcd (cos B cos D - sin B sin D)
= 4a2b2 + 4c2d2 - 8abcd cos (B + D)
= 4a2b2 + 4c2d2 - 8abcd cos 2a
= 4a2b2 + 4c2d2- 8abcd (2 cos2 a - 1)
= 4 (ab + cd)2 - G6abcd cos2 a,
so that
16A2 = 4 (ab + cd)2 - (a2 + b2 - c2 - d2)2 16-abcd cos2 a............(3 ).
But, as in Art. 219, we have
4 (ab + cd)2 - (a2 + b2 - -d2)2
= 2 (s - a). 2 (s - b). 2 (s - c). 2 (s - d)
= 16 (s - a) (s - b) (s - c) (s - d).
Hence (3) becomes
2 = (s - a) (s - b) (s - c) (s - d) - abcd cos2 a,
giving the required area.
Cor. 1. If d be zero the quadrilateral becomes a
triangle, and the formula above becomes that of Art. 198.
Cor. 2. If the sides of the quadrilateral be given in
length, we know a, b, c, d and therefore s. The area A is
hence greatest when abcd cos2 a is least, that is when cos2 a
is zero, and then a= 90~. In this case the sum of two
opposite angles of the quadrilateral is 180~ and the figure
inscribable in a circle. (Euc. III. 22.)