Given that, BI and CI are angle bisectors of \( \Large \angle B \) and \( \Large \angle C\), respectively.



Now, in \( \Large \triangle BIC, \)

\( \Large x ^{\circ} +\frac{B}{2}+\frac{C}{2}=180 ^{\circ} \) [let \( \Large \angle BIC = x ^{\circ}\)]

=> \( \Large x ^{\circ} = 180 ^{\circ} - \frac{1}{2} \left(B+C\right) \)

=> \( \Large x ^{\circ} = 180 ^{\circ} - \frac{1}{2} \left(180 ^{\circ} -A\right) \)

[\( \Large \because In \triangle ABC, A+B+C=180 ^{\circ} \)]

\( \Large x ^{\circ} = 180 ^{\circ} - 90 ^{\circ} +\frac{A}{2} \)

\( \Large \therefore \angle BIC = x ^{\circ} = 90 ^{\circ} +\frac{A}{2} \)