In a $\Delta ABC$ if $\frac{2\cos A}{a} +\frac{\cos B}{b} +\frac{2\cos C}{c}=\frac{a}{bc} +\frac{b}{ca}$, find $A$

Question

$$\frac{2bc\cos A + ac\cos B +2ab \cos C}{abc}= \frac{a^2+b^2}{abc}$$

$$b^2+c^2-a^2+a^2+b^2-c^2+ac\cos B =a^2+b^2$$

$$ac\ cos B = a^2-b^2$$

How do I find angle $A$ from here?

Answer 1

By the Cosine Rule, $2ac \cos B = a^2 + c^2 - b^2$, so from your equation above ($ac \cos B = a^2-b^2$), we obtain \begin{equation*}a^2 + c^2 - b^2 = 2(a^2-b^2)\end{equation*} and thus \begin{equation*}c^2 = a^2-b^2\end{equation*} This can be rewritten as $b^2 + c^2 - a^2 = 0$, so by the Cosine Rule again, $\cos A = 0$. As $A$ is an angle in a triangle, we must have $A = 90^{\circ}$.

Answer 2

Use Law of sines

then Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

Observe that as $0<A,B,C<\pi, \sin A,\sin B,\sin C>0$

and finally use $\sin(A+B)=\cdots=\sin C$

to find $$\sin B\cos A=0\implies?$$