Question

In Fig. 3, PQ and PR are the tangents to the circle with center O such that $\angle QPR={50}^{\circ }.$
Then $\angle OQR$ is equal to:

• A. ${25}^{\circ }$
• B. ${30}^{\circ }$
• C. ${40}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

Answer: (A)

${25}^{\circ }$

In quadrilateral PQOR, $\angle P+\angle PQO+\angle PRO+\angle QOR={360}^{\circ }$

But $\angle PQO=\angle PRO={90}^{\circ }$ (Tangent is perpendicular to radius at point of contact)

Thus, ${50}^{\circ }+{90}^{\circ }+{90}^{\circ }+\angle QOR={360}^{\circ }$

So, $\angle QOR={130}^{\circ }$

In triangle OQR, $\angle OQR+\angle ORQ+\angle QOR={180}^{\circ }$

Hence, $2.\angle OQR+\angle QOR={180}^{\circ }$

(because, $\angle OQR=\angle ORQ$; (since $OQ=OR$, radii ))

Thus, $2.\angle OQR+{130}^{\circ }=\angle {180}^{\circ }$

So, $\angle OQR={25}^{\circ }$

Therefore option A is the answer.