# Solution for the value of angle $A$ of a triangle

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

This is a quiz bee problem sent to me by my friend in FB. He asked me if I can do a solution for it. Well I tried several ways but I am out of idea now. The answer is 90 degree but what he asked, and I am also asking it now, is the solution for it.

Thank you.

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Using $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ etc.,

we get, $$\frac2a\frac{b^2+c^2-a^2}{2bc}+\frac1b\frac{a^2+c^2-b^2}{2ac}+\frac2c\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2}{abc}$$

or, $$2(b^2+c^2-a^2)+(a^2+c^2-b^2)+2(a^2+b^2-c^2)=2(a^2+b^2)$$

or $b^2+c^2=a^2$ as $abc\ne0$ $a,b,c$ being the sides of triangle.

So, $\cos A=0\implies A=(2n+1)\frac\pi2$ where $n$ is any integer.

As $0<A<\pi,A=\frac\pi2$

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I don't understand how the cosine law arrive to the quantity $b^2 + c^2 = a^2$. – Romel Verterra Nov 21 '12 at 12:36
@RomelVerterra, please find the edited answer – lab bhattacharjee Nov 21 '12 at 12:44
Would you please verify the original question.I think the coefficient of $\frac{\cos B}b$ was 1, but I could not find that version . – lab bhattacharjee Nov 21 '12 at 12:46
Thank you. And yes the coefficient of $\frac{cos B}{b}$ is $1$, not $2$. I corrected the problem. – Romel Verterra Nov 21 '12 at 12:50