Correct Answer: $90^{\circ}$

Solution :
$\angle$BIC = $\frac{\angle A}{2}$ + X
In $\triangle$ABC,
⇒ $\angle$A + $\angle$B + $\angle$C = $180^{\circ}$
⇒ $\angle$B + $\angle$C = $180^{\circ}$ – $\angle$A
⇒ $\frac{1}{2}$($\angle$B + $\angle$C) = $90^{\circ}$ – $\frac{\angle A}{2}$
In $\triangle$BIC,
⇒ $\frac{\angle B}{2}$ + $\frac{\angle C}{2}$ + $\angle$BIC = $180^{\circ}$
⇒ $90^{\circ}$ – $\frac{\angle A}{2}$ + $\angle$BIC = $180^{\circ}$
⇒ $\angle$BIC = $180^{\circ}$ – $90^{\circ}$ + $\frac{\angle A}{2}$ = $90^{\circ}$ + $\frac{\angle A}{2} = \frac{\angle A}{2}$ + X
⇒ X = $90^{\circ}$
Hence, the correct answer is $90^{\circ}$.