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Suppose $$O$$ is the circumcentre of $$\triangle ABC$$ such that $$O$$ lies inside the triangle. If angle $$BOA$$ is $$110^\circ$$ and angle $$COA$$ is $$90^\circ$$ then angle $$BAC$$ is
Since $$O$$ lies inside the triangle, $$\angle BOC = 360^\circ - 110^\circ - 90^\circ = 160^\circ$$. The angle at the centre is twice the angle at the circumference on the same arc, so $$\angle BAC = \frac{160^\circ}{2} = 80^\circ$$.
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