Question

Mark the correct alternative in each of the following:

In a ∆ABC, if a = 2, $\angle B=60°$ and $\angle C=75°$, then b =

(a) $\sqrt{3}$ (b) $\sqrt{6}$ (c) $\sqrt{9}$ (d) $1+\sqrt{2}$

Answer 1

It is given that a = 2, $\angle B=60°$ and $\angle C=75°$.

In ∆ABC,

$$\begin{gathered} \angle A+\angle B+\angle C=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow \angle A+60°+75°=180° \\ \Rightarrow \angle A=180°-135°=45° \end{gathered}$$

Using sine rule, we get

$$\begin{gathered} \frac{2}{\sin 45°}=\frac{b}{\sin 60°}\left(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right) \\ \Rightarrow b=\frac{2\times {\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{\sqrt{2}}}}=\sqrt{6} \end{gathered}$$

Hence, the correct answer is option (b).