Question

If $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ and the side $a=2$, than find the area of the triangle.

Answer 1

We have,

$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\to \left(i\right)$

In any triangle, we have

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\left(\sin rule\right)+ic)$

Multiply equation (i) and (ii), we get

$\begin{array}{c}\frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}\\ \cot A=\cot B=\cot C\end{array}$

This is possible only one

$\begin{array}{c}A=B=C={60}^{\circ }.\\ \therefore triangleisequilateral.\\ andhence,a=b=c=2\\ \therefore area=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}\times {\left(2\right)}^2=\sqrt{3}.\end{array}$

Hence, this is the answer.