From an external point $A$, tangents $AP$ and $AQ$ are drawn to a circle with centre $O$. If $\angle APQ=40^{\circ }$, then find $\angle POQ$.

The tangents drawn from an external point to a circle are equal in length, so $AP=AQ$.
In $△APQ$, since $AP=AQ$, it is an isosceles triangle, which means:

$$\angle AQP=\angle APQ=40^{\circ }$$

The sum of angles in $△APQ$ is $180^{\circ }$:

$$\angle PAQ+\angle APQ+\angle AQP=180^{\circ }$$

$$\angle PAQ+40^{\circ }+40^{\circ }=180^{\circ }$$

$$\angle PAQ=180^{\circ }-80^{\circ }=100^{\circ }$$

The radius from the center to the point of tangency is perpendicular to the tangent line ($\angle OPA=\angle OQA=90^{\circ }$). In the quadrilateral $OPAQ$:

$$\angle POQ+\angle OPA+\angle OQA+\angle PAQ=360^{\circ }$$

$$\angle POQ+90^{\circ }+90^{\circ }+100^{\circ }=360^{\circ }$$

$$\angle POQ+280^{\circ }=360^{\circ }$$

$$\angle POQ=360^{\circ }-280^{\circ }=80^{\circ }$$