Question

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

Answer 1

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