In the triangle ABC, the lengths of three sides opposite to angles A, B, and C are denoted by a, b and c respectively. If ab cos C + bc cos A + ca cos B = c^2, what is the area of ABC?
The cosine rule says that
\(\displaystyle \cos C= \frac{a^{2}+b^{2}-c^{2}}{2ab}.\)
Make that substitution and, in the same way, substitute for cos B and cos A.
Tidy up and you find that
\(\displaystyle a^{2}+b^{2}=c^{2},\)
so ABC is a right-angled triangle with hypotenuse c.
