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:
Answer & explanation
Correct answer: option 2
\(\angle\)BOC = 90 - (\(\angle\)BAC/2)
= \(\angle\)BAC/2 = 90 - 57
= \(\angle\)BAC/2 = \({33}^\circ\)
= \(\angle\)BAC = \({66}^\circ\)
Therefore, \(\angle\)BAC is \({66}^\circ\).