In a triangle ABC,frac{2cos A}{a} + frac{cos | vedclass

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(C) Given the equation: $\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$Using the cosine rule $\cos A = \frac{

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(C) Given the equation: $\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:$\frac{2(b^2 + c^2 - a^2)}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{2(a^2 + b^2 - c^2)}{2abc} = \frac{a^2 + b^2}{abc}$Multiplying by $2abc$ on both sides:$2(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2) = 2(a^2 + b^2)$$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$$b^2 - a^2 + c^2 = 0$$a^2 = b^2 + c^2$Since $a^2 = b^2 + c^2$,by the converse of the Pythagorean theorem,$\angle A = 90^o$.

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 $A$ is .....$^o$

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