Question

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

• A. ${60}^o$
• B. ${70}^o$
• C. ${80}^o$
• D. ${90}^o$

Answer 1

Answer: (B)

${70}^o$

Given, $\angle POQ={110}^{\circ }$
We know,
$\angle OPT=\angle OQT={90}^{\circ }$ (Angle between the tangent and the radial line at the point of intersection of the tangent at the circle)
Now, in quadrilateral $POQT$
Sum of angles$=360$
$\angle OPT+\angle OQT+\angle PTQ+\angle POQ={360}^{\circ }$
$90+90+\angle PTQ+110=360$
$\angle PTQ=360-290$
$\angle PTQ={70}^{\circ }$