Two chords $AB$ and $AC$ of a circle subtend angles equal to $90^{\circ}$ and $150^{\circ}$,respectively,at the centre. Find $\angle BAC$,if $AB$ and $AC$ lie on the opposite sides of the centre.

$(120^{\circ})$ Given that chords $AB$ and $AC$ subtend angles $\angle AOB = 150^{\circ}$ and $\angle AOC = 90^{\circ}$ at the centre $O$.
Since $AB$ and $AC$ lie on opposite sides of the centre,the angle subtended by the chord $BC$ at the centre is $\angle BOC = \angle AOB + \angle AOC = 150^{\circ} + 90^{\circ} = 240^{\circ}$.
The angle subtended by the chord $BC$ at the remaining part of the circle is $\angle BAC$.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,the reflex angle $\angle BOC = 2 \angle BAC$.
$240^{\circ} = 2 \angle BAC$.
$\angle BAC = \frac{240^{\circ}}{2} = 120^{\circ}$.