Question

Question 7
In figure, if 0 is the centre of a circle, PQ is a chord and the tangent PR makes an angle of ${50}^{\circ }$ with PQ, then $\angle$ POQ is equal to
(A) 100°
(B) 80°
(C) 90°
(D) 75°

Answer 1

Given: $\angle QPR={50}^{\circ }$
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore \angle OPR={90}^{\circ }\Rightarrow \angle OPQ+\angle QPR={90}^{\circ }\left[Fromfigure\right]\Rightarrow \angle OPQ={90}^{\circ }–{50}^{\circ }={40}^{\circ }[\because \angle QPR={50}^{\circ }]Now,OP=OQ=Radiusofcircle\therefore \angle OQP=\angle OPQ={40}^{\circ }[Since,anglesoppositetoequalsidesareequal]In\Delta OPQ,\angle O+\angle P+\angle Q={180}^{\circ }[Since,anglesoppositetoequalsidesareequal]\Rightarrow \angle O={180}^{\circ }-({40}^{\circ }+{40}^{\circ })[\because ,\angle P={40}^{\circ }=\angle Q]={180}^{\circ }-{80}^{\circ }={100}^{\circ }$