Question

$\text{In the given figure, PT is a tangent drawn}\text{from an external point T to the circle with}\text{centre O and PQ is the diameter of the}\text{circle. Point Q and T are joined by a line}\text{which is intersecting the circle at R. If}\angle POR={78}^{\circ },\text{then find the value of}\angle PTR.$

Answer 1

$\text{In,}△OQR,$
$\because OQ=OR\text{(Radius)}$
(Angles opposite to equal sides are equal.)
$\therefore \angle OQR=\angle ORQ$

By exterior angle property,
$\angle POR=\angle OQR+\angle ORQ\Rightarrow {78}^{\circ }=2\angle OQR\Rightarrow \angle OQR={39}^{\circ }\angle PQR=\angle OQR={39}^{\circ }$

We know that, tangent at any point is perpendicular to the radius at the point of contact.
$\therefore \angle TPQ={90}^{\circ }$

$\text{Now, consider}△TPQ,$
$\angle TPQ+\angle PQR+\angle QTP={180}^{\circ }$
$\angle QTP={180}^{\circ }-{39}^{\circ }-{90}^{\circ }$
$\therefore \angle QTP=\angle PTR={51}^{\circ }$