I is the incentre of △ABC of ∠A = 46°, then ∠BIC =?
Answer & explanation
Correct answer: option 2
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 = \({113}^\circ\).