Question

In fig 9.6, if O is the centre of a circle, PQ is chord and the tangent PR at P makes an angle of ${50}^0$ with PQ, then $\angle POQ$ is equal to

• A. ${100}^0$
• B. ${80}^0$,
• C. ${90}^0$
• D. ${75}^0$

Answer 1

The correct option is A ${100}^0$

as tangent is perpendicular to the circle

$\angle RPO=\angle RPQ+\angle QPO=90$

$\angle QPO=90-50=40$

$\angle QPO=\angle PQO$[since $OP=OQ$]

therefore in triangle POQ

$\angle POQ+\angle OQP\angle QPO=180$

$\angle POQ=180-40-40=100$