In the figure,$\angle ACB = 40^{\circ}$. Find $\angle OAB$.

$(50^{\circ})$ Since 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,we have:
$\angle AOB = 2 \angle ACB = 2 \times 40^{\circ} = 80^{\circ}$
In $\Delta OAB$,since $OA = OB$ (radii of the same circle),the angles opposite to equal sides are equal. Therefore,$\angle OAB = \angle OBA = p^{\circ}$.
Using the angle sum property in $\Delta OAB$:
$p^{\circ} + p^{\circ} + \angle AOB = 180^{\circ}$
$2p^{\circ} + 80^{\circ} = 180^{\circ}$
$2p^{\circ} = 180^{\circ} - 80^{\circ}$
$2p^{\circ} = 100^{\circ}$
$p^{\circ} = 50^{\circ}$
Hence,$\angle OAB = 50^{\circ}$.