In $\triangle ABC, \angle A = 68^\circ$. If is the incentre of the triangle, then measure of $\angle BIC$ is:

124°

According to the concept,

\(\angle\)BIC = 90 + \(\frac{1}{2}\) x \(\angle\)BAC

= \(\angle\)BIC = 90 + \(\frac{1}{2}\) x 68

= \(\angle\)BIC = 90 + 34

= \(\angle\)BIC = \({124}^\circ\)

Therefore, \(\angle\)BIC is \({124}^\circ\).