In $\triangle ABC, \angle A = 66^\circ$. AB and AC are produced to points D and E, respectively. If the bisectors of angle CBD and angle BCE meet at the point O, then $\angle BOC$ is equal to:

66° 93° 57° 114°

57°

\(\angle\)BOC = 90 - \(\angle\)A/2 = \(\angle\)BOC = 90 - \(\frac{66}{2}\) = \(\angle\)BOC = 90 - 33 = \(\angle\)BOC = \({57}^\circ\) Therefore, \(\angle\)BOC is \({57}^\circ\).