Question

In a $△ABC$, 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

Answer: 1.73

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