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

Method 1: Using Triangle Properties
1. The lengths of tangents drawn from an external point to a circle are equal. Therefore, $AP=AQ$.
2. In $△APQ$, since $AP=AQ$, the angles opposite to these sides are also equal:

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

3. The sum of angles in a triangle is $180^{\circ }$:

$$\angle PAQ=180^{\circ }-(\angle APQ+\angle AQP)=180^{\circ }-(40^{\circ }+40^{\circ })=100^{\circ }$$

4. The angle between two tangents from an external point and the angle subtended by the line segment joining the points of contact at the center are supplementary:

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

$$100^{\circ }+\angle POQ=180^{\circ }⟹\angle POQ=80^{\circ }$$

Method 2: Using Radius-Tangent Perpendicularity

1. The radius is perpendicular to the tangent at the point of contact, so $\angle APO=90^{\circ }$.
2. Given $\angle APQ=40^{\circ }$, we find $\angle OPQ$:

$$\angle OPQ=\angle APO-\angle APQ=90^{\circ }-40^{\circ }=50^{\circ }$$

3. In $△POQ$, $OP=OQ$ (radii of the same circle), which implies $\angle OQP=\angle OPQ=50^{\circ }$.
4. Therefore, the central angle $\angle POQ$ is:

$$\angle POQ=180^{\circ }-(\angle OPQ+\angle OQP)=180^{\circ }-(50^{\circ }+50^{\circ })=80^{\circ }$$