Question

In a triangle ABC, if $B={30}^0$ and $c=\sqrt{3}b$, then A can be equal to :-

• A. ${45}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${120}^0$

Answer 1

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

In a triangle ABC

$\angle B={30}^o$

And

$c=\sqrt{3}b$

$\frac{c}{b}=\frac{\sqrt{3}}{1}$

$\Rightarrow \frac{b}{c}=\frac{1}{\sqrt{3}}$

Using sine rule and we get,

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

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

$\frac{\sin B}{\sin C}=\frac{1}{\sqrt{3}}$

If $B={30}^o$

Then,

$\frac{\sin {30}^o}{\mathrm{sinC}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \sin C=\sqrt{3}\sin {30}^o$

$\Rightarrow \sin C=\sqrt{3}\times \frac{1}{2}$

$\Rightarrow \sin C=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin C=\sin {60}^o$

$\Rightarrow \angle C={60}^o$

Then,

We know that, In any triangle $ABC$

$\angle A+\angle B+\angle C={180}^o$

$\angle A+{30}^o+{60}^o={180}^o$

$\angle A={90}^o$

Hence, this is the answer.