Question

In a ∆ABC, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and a = 2, then area of ∆ABC is equal to __________.

Answer 1

In a ∆ABC

Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}...\left(1\right)\mathrm{and}a=2$
Using sine formula
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}...\left(2\right)$

using above two, (1) and (2)
we get,
$\frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}$

⇒ cotA = cotB = cotC
⇒ angle A = B = C = 60^°
⇒ ∆ABC is an equilateral triangle

$\begin{array}{rcl}\therefore \mathrm{Area}\mathrm{of}\mathrm{triangle}\mathrm{ABC}& =& \frac{\sqrt{3}}{4}{a}^2\\ & =& \frac{\sqrt{3}}{4}\times 4\\ & =& \sqrt{3}{\mathrm{unit}}^2\end{array}$
$o\mathrm{r}\sqrt{3}\mathrm{Sq}.\mathrm{unit}$