Question

In a triangle ABC, if $\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ca}$, then the value of the angle A is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

$\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{b^2+c^2-a^2}{abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{abc}=\frac{a^2+3b^2+c^2}{2abc}$

But given $\frac{a^2+3b^2+c^2}{2abc}=\frac{a^2+b^2}{abc}$

$a^2+3b^2+c^2=2a^2+2b^2⟹a^2=b^2+c^2⟹A={90}^{\circ }$