Question

In a $△ABC$, if $a=2b$ and $\angle A=3\angle b$, then angle $A$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

By sine formula, we have

$\frac{a}{\sin A}=\frac{b}{\sin B}$ $\Rightarrow \frac{a}{b}=\frac{\sin A}{\sin B}\Rightarrow 2=\frac{\sin 2B}{\sin B},$

$[\because a=2b\Rightarrow \frac{a}{b}=2$ and $\angle A=3\angle B]$

$\Rightarrow 2\sin B=3\sin B-4{\sin }^{3}B$ $\Rightarrow 4{\sin }^{3}B-\sin B=0$

$\Rightarrow \sin B(4{\sin }^{2}B-1)=0$ $\Rightarrow \sin B=0$ or ${\sin }^{2}B=\frac{1}{4}$

$\Rightarrow \sin B=\frac{1}{2}$ $[\because \sin B$ cannot be zero or negative$]$

$\Rightarrow \angle B={30}^0$

$\therefore \angle A=3\angle B=3\times {30}^0={90}^0$