In ΔABC, ∠A = 52° and O is the orthocentre of the triangle (BO and CO meet AC and AB at E and F respectively when produced). If the bisectors of ∠OBC and ∠OCB meet at P, then the measure of ∠BPC is:

Solution (By Examveda Team)

$$\eqalign{ & \angle BOC = {180^ \circ } - {52^ \circ } = {128^ \circ } \cr & \angle BPC = {90^ \circ } + \frac{{\angle BOC}}{2} \cr & = {90^ \circ } + \frac{{128}}{2} \cr & = {90^ \circ } + {64^ \circ } \cr & = {154^ \circ } \cr} $$