Question

If in a triangle ABC, $\angle A={30}^o$ and the area of triangle is $\frac{\sqrt{3a^2}}{4}$, then prove that either $B=4C$ or $C=4B$.

Answer 1

Area of triangle=$\frac{1}{2}bc\sin A\Rightarrow \frac{1}{2}bc\sin 30°$

$\therefore \frac{\sqrt{3}}{4}{a}^2=\frac{1}{2}bc\times \frac{1}{2}$ $\left(\sin 30°=\frac{1}{2}\right)$

So, $\sqrt{3}{a}^2=bc$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\frac{1}{2a}=\frac{\sin B}{b},\sin B=\frac{b}{2a},\sin C=\frac{c}{2a}$

$\sin B\times \sin C=\frac{bc}{4a^2}\Rightarrow \frac{bc\times \sqrt{3}}{4\times bc}$

$\sin B\sin C=\frac{\sqrt{3}}{4}$

So , either $B=4C$ or $C=4B.$