In $\triangle ABC $, 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, and $\angle BOC= 57^\circ$, then $ \angle A$ is equals to:

66°

\(\angle\)BOC = 90 - (\(\angle\)BAC/2)

= \(\angle\)BAC/2 = 90 - 57

= \(\angle\)BAC/2 = \({33}^\circ\)

= \(\angle\)BAC = \({66}^\circ\)

Therefore, \(\angle\)BAC is \({66}^\circ\).