Question

In figure, PQ is a chord of a circle with centre O and PT is a tangent. If $\angle QPT={60}^{\circ },$, find $\angle PRQ$.

Answer 1

Given, O is the centre of the given circle

$\therefore$ OQ and OP are the radius of the circle.

$\therefore$ PT is a tangent

$\therefore OP\perp PT$
(Radius of a circle is perpendicular to the tangent through the point of contact)

So, $\angle OPT={90}^{\circ }$

$\angle OPQ={90}^{\circ }-\angle QPT$

$\angle OPQ={90}^{\circ }-{60}^{\circ }$

[Given,$\angle QPT={60}^{\circ }$]

$\angle OPQ={30}^{\circ }$

$\therefore \angle OQP={30}^{\circ }[\because \Delta OPQ$ is isosceles triangle]

Now,in $\Delta OPQ$

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

$\angle POQ+{30}^{\circ }+{30}^{\circ }={180}^{\circ }$

$\angle POQ={120}^{\circ }$

reflex $\angle POQ={360}^{\circ }-{120}^{\circ }={240}^{\circ }$

$\therefore \angle PRQ=\frac{1}{2}\text{reflex}\angle POQ$

[$\therefore$ The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.]

$\angle PRQ=\frac{1}{2}\times {240}^{\circ }$

Hence, $\angle PRQ={120}^{\circ }$