Question

If $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and the side $a=2$ then area of the triangle is

• A. $1$
• B. $2$
• C. $\frac{\sqrt{3}}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

Since $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$
Using cosine rule, we get
$\Rightarrow \frac{b^2+c^2-a^2}{2bc.a}=\frac{c^2+a^2-b^2}{2ca.b}=\frac{a^2+b^2-c^2}{2ab.c}$
$\Rightarrow b^2+c^2-a^2=c^2+a^2-b^2=a^2+b^2-c^2$
From Relations $I$ and $II$ we have
$b^2+c^2-a^2=c^2+a^2-b^2$
$\Rightarrow 2b^2=2a^2$
$\Rightarrow b=a=2$ (given $a=2$ )
From relations $II$ and $III$ we have
$b^2+c^2-a^2=a^2+b^2-c^2$
$\Rightarrow -2b^2=-2c^2$
$\Rightarrow b=c=2$
Hence, $a=b=c=2$
Hence$△ABC$ is equilateral.
$\therefore$ Area$=\frac{\sqrt{3}}{4}\times s^2$
$=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}$sq.units