Question

In a triangle ABC, if $\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$ and $a=2$, then its area is

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

Answer: (B)

Given, in $△ABC$,
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}...(i)$
From sine rule,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}...(ii)$
On dividing Eq. (ii) by Eq. (i), we get
$\Rightarrow \tan A=\tan B=\tan C$
$\Rightarrow A=B=C$
So, $△ABC$ is an equilateral triangle
$\left(\begin{array}{cc}\because \text{ in }\Delta ABC,& A+B+C=180^{\circ }\\ & \Rightarrow A+A+A=180^{\circ }\\ & \Rightarrow A=60^{\circ }=B=C\end{array}\right)$
Area of equilateral $△ABC$
$=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}(2{)}^2(\because a=2,$ given $)$
$=\frac{\sqrt{3}}{4}\times 4=\sqrt{3}$