ln ΔABC, ∠A = 66°. AB and AC are produced to points D and E, respectively. If the bisectors of ∠CBD and ∠BCE meet at the point O, then ∠BOC is equal to:

Solution (By Examveda Team)

$$\eqalign{ & \angle A = {66^ \circ } \cr & \angle BOC = ? \cr & \angle BOC = {90^ \circ } - \frac{{\angle A}}{2} \cr & = {90^ \circ } - \frac{{{{66}^ \circ }}}{2} \cr & = {90^ \circ } - {33^ \circ } \cr & = {57^ \circ } \cr} $$