Question

If $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$and the side $a=2$ then find the value of ${\Delta }^2$, where $\Delta$ is the area of the triangle.

Answer 1

Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$
$\frac{\cos A}{2R\sin A}=\frac{\cos B}{2R\sin B}=\frac{\cos C}{2R\sin C}$

$\tan A=\tan B=\tan C$

Hence,$\Delta$ is an equilateral triangle

Now, area of an equilateral $\Delta =\frac{\sqrt{3}}{4}{a}^2=\sqrt{3}$ $(\because a=2)$

Hence ${\Delta }^2=(\sqrt{3}{)}^2=3$