Question

In a $\triangle ABC$, if $\frac{\cos A }{ a }=\frac{\cos B }{ b }=\frac{\cos C }{ c }$ and the side $a =2$, then find area of the triangle.

Answer 1

$\frac{\cos A }{ a }=\frac{\cos B }{ b }=\frac{\cos C }{ c }$
$\Rightarrow \frac{\cos A }{ k \sin A }=\frac{\cos B }{ k \sin B }=\frac{\cos C }{ k \sin C }$
$\Rightarrow \cot A =\cot B =\cot C$
$\Rightarrow A = B = C $
$\Rightarrow$ equilateral triangle
$\therefore$ Area $=\frac{\sqrt{3}}{4}( a )^{2}=\frac{\sqrt{3}}{4}(2)^{2}=\sqrt{3}$