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 .... degree.

Answer 1

Given $ \frac{ 2 \, cos \, A }{ a} + \frac{ cos \, B}{ b} + \frac{ 2 \, cos \, C }{ c} = \frac{a}{ bc } + \frac{b}{ca}$
We know that, cos A = $ \frac{ b^2 + c^2 + a^2 }{ 2bc} $
cos B = $ \frac{ c^2 + a^2 - b^2 }{ 2ac} $
and cos C = $ \frac{a^2 + b^2 - c^2 }{ 2ab}$
On putting these values in Eq. (i), we get
$ \frac{ 2 (b^2 + c^2 - a^2 ) }{ 2abc } + \frac{ c^2 + a^2 - b^2 }{2abc } $
$$ $ + \frac{2 (a^2 + b^2 - c^2 )}{ 2abc} = \frac{a}{bc} + \frac{b}{ ca} $
$\Rightarrow \frac{ 2 (b^2 + c^2 - a^2 ) + c^2 + a^2 - b^2 + 2 (a^2 + b^2 - c^2 )}{ 2abc } = \frac{a^2 + b^2 }{ abc}$
$\Rightarrow \, 3b^2 + c^2 + a^2 = 2a^2 + 2b^2 $
$\Rightarrow b^2 + c^2 = a^2$
Hence, the angle A is $ 90^\circ$.