Question

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

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

Answer 1

Answer: (B)

${70}^o$

Join $PQ$ in given figure then we have two triangles $\Delta$OPQ and $\Delta$ PTQ,

In $\Delta$OPQ ,$OP=OQ$ radius of triangle, and in $\Delta$ PTQ, $PT=TQ$ (Property of circle, two tangents drawn from a point to a circle are equal), so Both triangles are isosceles triangle

Now in $\Delta$OPQ, by Angle sum property of a triangle

$⟹\angle OPQ+\angle OQP+\angle POQ={180}^o$

$\angle OPQ=\angle OQP={35}^{\circ }$

Since $PT$ is tangent $\therefore \angle OPT={90}^{\circ }$

$\therefore \angle QPT={90}^{\circ }-{35}^{\circ }={55}^{\circ }=\angle PQT$
Now $\Delta$ PTQ,

$\angle PTQ+\angle TPQ+\angle TQP={180}^o$

$⟹\angle PTQ={180}^{\circ }-{110}^{\circ }={70}^{\circ }$