Question

If in a $△ABC,$
$\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ca}$
Then, the value of the $\angle A$ is .... degree.

Answer 1

Given $\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ca}$
We know that, cos A = $\frac{b^2+c^2+a^2}{2bc}$
cos B = $\frac{c^2+a^2-b^2}{2ac}$
and cos C = $\frac{a^2+b^2-c^2}{2ab}$
On putting these values in Eq. (i), we get
$\frac{2(b^2+c^2-a^2)}{2abc}+\frac{c^2+a^2-b^2}{2abc}$
$+\frac{2(a^2+b^2-c^2)}{2abc}=\frac{a}{bc}+\frac{b}{ca}$
$\Rightarrow \frac{2(b^2+c^2-a^2)+c^2+a^2-b^2+2(a^2+b^2-c^2)}{2abc}=\frac{a^2+b^2}{abc}$
$\Rightarrow 3b^2+c^2+a^2=2a^2+2b^2$
$\Rightarrow b^2+c^2=a^2$
Hence, the angle A is $90^{\circ }$.