Question:
Mark the correct alternative in each of the following:
In a $\triangle \mathrm{ABC}$, if $a=2, \angle B=60^{\circ}$ and $\angle C=75^{\circ}$, then $b=$
(a) $\sqrt{3}$
(b) $\sqrt{6}$
(c) $\sqrt{9}$
(d) $1+\sqrt{2}$
Solution:
It is given that $a=2, \angle B=60^{\circ}$ and $\angle C=75^{\circ}$.
In ∆ABC,
$\angle A+\angle B+\angle C=180^{\circ} \quad$ (Angle sum property)
$\Rightarrow \angle A+60^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow \angle A=180^{\circ}-135^{\circ}=45^{\circ}$
Using sine rule, we get
$\frac{2}{\sin 45^{\circ}}=\frac{b}{\sin 60^{\circ}} \quad\left(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right)$
$\Rightarrow b=\frac{2 \times \frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$
Hence, the correct answer is option (b).