The internal bisectors of $\angle ABC$ and $\angle ACB$ of $\triangle ABC$ meet each other at O. If $\angle BOC$ = $110^{0}$, then $\angle BAC$ is equal to

The internal bisectors of $\angle ABC$ and $\angle ACB$ of $\triangle ABC$ meet each other at O. If $\angle BOC$ = $110^{0}$, then $\angle BAC$ is equal to
1). $40^{0}$
2). $55^{0}$
3). $90^{0}$
4). $110^{0}$

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