Question

If in a triangle ABC,\[\frac{2 \cos A}{a} + \frac{\cos B}b + \frac{2\cos C}c = \frac{a}{bc}+\frac{b}{ca}\]Find \(\angle A\).

Answer 1

Easy problem. Try it.

Answer 2

\[b^2+c^2-a^2+\frac{1}{2}(a^2+c^2-b^2) + a^2+b^2-c^2 = a^2 + b^2\]\[b^2 - a^2 + \frac{1}2(a^2+c^2-b^2) = 0\]\[2b^2 - 2a^2 + a^2 + c^2 - b^2=0\]\[b^2 + c^2 = a^2\]