If in a triangle ABC, $\frac{(2\cos A)}{a}+\frac{(\cos B)}{b}+\frac{(2\cos C)}{c}=\frac{a}{bc}+\frac{b}{ca}$, then the value of angle A in degrees is :

90° 135° 45° None of these

90°

Combine first and third and put the value of cos B. $∴\frac{2}{ac}.(b)+\frac{1}{b}\frac{c^2+a^2-b^2}{2ca}=\frac{a^2+b^2}{abc}$ or $4b^2+c^2+a^2-b^2=2a^2+2b^2$ $∴b^2+c^2=a^2$ $∴∠A=90°$