Question

In the fig., if O is a center of the circle, PQ is a chord and tangent at P makes the angle of ${50}^{\circ }$ with PQ, then $\angle POQ=$

• A. ${100}^{\circ }$
• B. ${90}^{\circ }$
• C. ${80}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is A ${100}^{\circ }$

Since OP is perpendicular to PR

$\angle OPR$=${90}^{\circ }$

$\angle OPQ+\angle QPR={90}^{\circ }$

$\angle OPQ+50°={90}^{\circ }$

$\angle OPQ=\angle OQP={40}^{\circ }$ (Since OP = OQ = radii; so Isosceles traingle OPQ)

Again,

$\angle POQ+\angle OPQ+\angle OQP={180}^{\circ }$ ( Angle sum property of triangle)

$\angle POQ+{40}^{\circ }+{40}^{\circ }={180}^{\circ }$

$\angle POQ={180}^{\circ }–{40}^{\circ }–{40}^{\circ }$

= $100°$

option A is the answer.