$\text{In }\mathrm{△}\mathrm{ABC}\text{, if}\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}}$ and side a = 2, then area of the $\mathrm{△}\mathrm{ABC}$ (in sq. units) is

$\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}}$

$\begin{array}{r}\Rightarrow \frac{\cos A}{k\sin A}=\frac{\cos B}{k\sin B}=\frac{\cos C}{k\sin C}\\ \Rightarrow \cot A=\cot B=\cot C\\ \Rightarrow A=B=C={60}^{\circ }\end{array}$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}{\mathrm{a}}^2$

$=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}\text{ sq. units }$