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In the given figure, $PA$ is a tangent to the circle drawn from the external point $P$ and $PBC$ is the secant to the circle with $BC$ as diameter. If $\angle AOC = 130^{\circ}$, then find the measure of $\angle APB$, where $O$ is the centre of the circle.
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$\angle AOB = 180^{\circ} - 130^{\circ} = 50^{\circ}$
$\angle OAP = 90^{\circ}$
$\therefore \angle APB = 180 - (50^{\circ} + 90^{\circ}) = 40^{\circ}$
$\angle OAP = 90^{\circ}$
$\therefore \angle APB = 180 - (50^{\circ} + 90^{\circ}) = 40^{\circ}$