In the given figure, $PA$ and $PB$ are the tangents to the circle with center $O$.
If $\angle APB = 36^{\circ}$, then $\angle AOB = ?$
Since $PA$ and $PB$ are tangents to the circle at points $A$ and $B$ respectively, we know that $OA \perp PA$ and $OB \perp PB$.
Thus, $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
Now consider the quadrilateral $PAOB$. The sum of the angles in a quadrilateral is $360^{\circ}$.
Therefore, $\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$ $90^{\circ} + 36^{\circ} + 90^{\circ} + \angle AOB = 360^{\circ}$ $216^{\circ} + \angle AOB = 360^{\circ}$ $\angle AOB = 360^{\circ} - 216^{\circ} = 144^{\circ}$ Thus, $\angle AOB = 144^{\circ}$.