Question:
In a $\Delta A B C$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$, then area of $\triangle A B C$ is equal to ___________________
Solution:
In a $\Delta A B C$
Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \ldots(1)$ and $a=2$
Using sine formula
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \quad \ldots(2)$
using above two, (1) and (2)
we get,
$\frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}$
$\Rightarrow \cot A=\cot B=\cot C$
$\Rightarrow$ angle $A=B=C=60^{\circ}$
$\Rightarrow \triangle A B C$ is an equilateral triangle
$\therefore$ Area of triangle $\mathrm{ABC}=\frac{\sqrt{3}}{4} a^{2}$
$=\frac{\sqrt{3}}{4} \times 4$
$=\sqrt{3}$ unit $^{2}$
or $\sqrt{3}$ Sq. unit