I is the incentre of △ABC of ∠A = 46°, then ∠BIC =?

113°

In \(\Delta \)BIC

\(\angle\)BIC + \(\angle\)IBC + \(\angle\)ICB = 180

\(\angle\)BIC + \(\angle\)B/2 + \(\angle\)C/2 = 180

\(\angle\)BIC + 90 - \(\angle\)A/2 = 180

\(\angle\)BIC = 180 - 90 + \(\angle\)A/2

\(\angle\)BIC = 90 + \(\frac{46}{2}\) = 90 + 23 = \({113}^\circ\).

Therefore, \(\angle\)BIC = \({113}^\circ\).