Question

In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If $\angle$ TPQ = ${70}^o$, find $\angle$ TRQ.

Answer 1

We have a circle having its centre at 0.

Now, OT and OQ are the radii of the circle.

PT and PQ are the tangents to the circle from an external point P.

We know that tangent to a circle is always $\perp$ to its radius at the point of contact.

Hence, $\angle OTP=\angle OQP={90}^o$.

Now, $\angle TPQ={70}^o$

In quadrilateral PQOT,

$\angle QOT+\angle OTP+\angle TPQ+\angle OQP={360}^o$ [Angle sum property]

$\angle QOT+{90}^o+{70}^o+{90}^o={360}^o$

$\angle QOT={110}^o$

We know that angle subtended by an arc at the centre is double the angle subtended by the same
arc at any pointy on the circle.

Consider the arc QT that subtends $\angle QOT$ at the centre and $\angle QRT$ at any point R on the circle.
Now, $\angle QOT=2\angle QRT$

${110}^o=2\times \angle QRT$

$\angle QRT={55}^o$