Question

If in a $△ABC$
$\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ac}$, then $b^2+c^2$ is equal to

• A. $a^2$
• B. $ac$
• C. $bc$
• D. None of these

Answer 1

Answer: (A)

$a^2$

We have, $\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ac}$
On multiplying both sides by $abc$
$\Rightarrow 2bc\cos A+ac\cos B+2ab\cos C=a^2+b^2$
$\Rightarrow (b^2+c^2-a^2)+\frac{(c^2+a^2-b^2)}{2}+(a^2+b^2-c^2)=a^2+b^2$
$\Rightarrow (c^2+a^2-b^2)=2a^2-2b^2$
$\Rightarrow b^2+c^2=a^2$.