Question

In figure, 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. ${90}^o$
• B. ${70}^o$
• C. ${60}^o$
• D. ${80}^o$

Answer 1

Answer: (B)

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 $={360}^0$

$\angle OPT+\angle OQT+\angle PTQ+\angle POQ={360}^{\circ }$

${90}^0+{90}^0+\angle PTQ+{110}^0={360}^0$

$\angle PTQ={360}^0-{290}^0$

$\angle PTQ={70}^{\circ }$