Question

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

• A. $1$ sq unit
• B. $2$ sq unit
• C. $\frac{\sqrt{3}}{2}$ sq unit
• D. $\sqrt{3}$ sq unit

Answer 1

The correct option is D $\sqrt{3}$ sq unit

We have, $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$
$\Rightarrow \frac{\cos A}{k\sin A}=\frac{\cos B}{k\sin B}=\frac{\cos C}{k\sin C}$
$\Rightarrow \cot A=\cot B=\cot C$
$\Rightarrow A=B=C={60}^{\circ }$
Therefore, $△ABC$ is an equilateral triangle.
We know area of triangle $=\frac{\sqrt{3}}{4}{a}^2$
$=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}$ sq. units