If TP and TQ are the two tangents to a circle with centre O so that $\angle POQ={110}^{\circ }$, then $\angle PTQ$ is equal to

Solution: We know tangent at any point $P$ is always perpendicular to line joining the center $O$ of that circle.
So, in the given figure,
$\angle OPT=\angle OQT={90}^{\circ }$
So, in rectangle $OPTQ$
$\angle POQ+\angle PTQ+\angle OPT+\angle OQT={360}^{\circ }$
$\angle PTQ+90+90+110=360$
$\angle PTQ=360-290={70}^{\circ }$