In $\triangle \mathrm{ABC}, \mathrm{AB}$ and $\mathrm{AC}$ are produced to points $\mathrm{D}$ and $\mathrm{E}$ respectively. If the bisectors of angle $\mathrm{CBD}$ and angle $\mathrm{BCE}$ meet at point $O$, such that $\angle B O C=63^{\circ}$, then $\angle A=$ ?

54°

\(\angle\)BOC = \({90}^\circ\) - (\(\angle\)BAC/2)

= \(\angle\)BAC/2 = 90 - 63

= \(\angle\)BAC/2 = \({27}^\circ\)

= \(\angle\)BAC = \({54}^\circ\)

Therefore, \(\angle\)BAC is \({54}^\circ\).