Trigonometric Functions — MATHS STD 11 Science — Question

2. $\sqrt{6}$

Solution:

It is given that $\text{a}=2,\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ}.$

In $\triangle\text{ABC},$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ (Angle sum property)

$\Rightarrow\angle\text{A}+60^{\circ}+75^{\circ}=180^{\circ}$

$\Rightarrow\angle\text{A}=180^{\circ}-135^{\circ}=45^{\circ}$

Using sine rule, we get

$\frac{2}{\sin45^{\circ}}=\frac{\text{b}}{\sin60^{\circ}}$ $\Big(\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}\Big)$

$\Rightarrow\text{b}=\frac{2\times\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$

Hence, the correct answer is option (b).