Question

In a $△ABC$, if $\left(\sqrt{3}-1\right)a=2b,A=3B,\mathrm{then}\angle C$ is equal to

• A. $60°$
• B. $120°$
• C. $30°$
• D. $45°$

Answer 1

The correct option is B

$120°$

Explanation for the correct answer:

Step 1: Finding $A$ and $B$:

We have, $\left(\sqrt{3}-1\right)a=2b$

$\therefore \frac{a}{b}=\frac{2}{\left(\sqrt{3}-1\right)}$

By sine rule, we know that,

$$\begin{gathered} \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B} \\ \Rightarrow \frac{a}{b}=\frac{\sin A}{\sin B} \\ \Rightarrow \frac{\sin A}{\sin B}=\frac{2}{(\sqrt{3}-1)}\left[\because \frac{a}{b}=\frac{2}{(\sqrt{3}-1)}\right] \end{gathered}$$

Divide the numerator ad denominator of $RHS$ with $2\sqrt{2}$, we have,

$$\begin{gathered} \Rightarrow \frac{\sin A}{\sin B}=\frac{{\displaystyle \frac{2}{2\sqrt{2}}}}{{\displaystyle \frac{(\sqrt{3}-1)}{2\sqrt{2}}}} \\ \Rightarrow \frac{\sin A}{\sin B}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{(\sqrt{3}-1)}{2\sqrt{2}}}} \\ \Rightarrow \frac{\sin A}{\sin B}=\frac{\sin 45°}{\sin 15°}\left(\because \sin 45°=\frac{1}{\sqrt{2}}\mathrm{and}\sin 15°=\frac{(\sqrt{3}-1)}{2\sqrt{2}}\right) \\ \therefore \mathrm{On}\mathrm{comparing}\mathrm{we}\mathrm{have}A=45°\mathrm{and}B=15° \end{gathered}$$

Step 2: Finding $\angle C$:

We know that sum of internal angles of a triangle is $180°$.

$$\begin{gathered} \therefore \angle A+\angle B+\angle C=180° \\ \Rightarrow \angle C=180°-45°-15° \\ \Rightarrow \angle C=120° \end{gathered}$$

Hence, the correct answer is option (B)