Question

Prove that $cosA=\frac{b^2+c^2-a^2}{2bc}$

Answer 1

$a^2={(c-x)}^2+h^2{a}^2=c^2+h^2+x^2-2cx$

But,$h^2+x^2=b^2$

$\therefore a^2=c^2+b^2-2cx$
$\frac{x}{b}=\cos A\Rightarrow x=b\cos A\therefore a^2=c^2+b^2-2cb\cos A\therefore \cos A=\frac{c^2+b^2-ac}{2bc}$