Question

In a $\Delta ABC,\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ca}$ then angle A is :

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

Answer 1

The correct option is C ${90}^{\circ }$

$\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}$ (using cosine rule)

$=\frac{2(b^2+c^2-a^2)}{2abc}+\frac{(a^2+c^2-b^2)}{2abc}+\frac{2(b^2+a^2-c^2)}{2abc}$

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

$\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$

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

$\Rightarrow 3b^2+c^2+a^2=2a^2+2b^2$

$b^2+c^2=a^2$

$\Rightarrow \cos A=0$ $(\cos A=\frac{b^2+c^2-a^2}{2bc})$

$\Rightarrow \angle A={90}^o$.