Question

O is the centre of the circle as shown in the figure. If $\angle ORP={35}^{\circ }$ and the distance between P and Q through 'O' is 4 cm, then what is the measure of $\angle ROQ$ ?

• A. ${55}^{\circ }$
• B. ${35}^{\circ }$
• C. ${105}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is D ${70}^{\circ }$



Since PQ is the diameter of circle, we have
$PQ=4\text{cm}=2\times OQ$, where OQ is the radius of the circle.
Join RQ.
We know that the angle subtended by an arc of a circle at its centre is double the angle subtended by it on any remaining part of the circle.
Since $\angle POQ={180}^{\circ },$ we have
$\angle PRQ=\frac{{180}^{\circ }}{2}={90}^{\circ }.$
$\therefore \angle ORQ={90}^{\circ }-{35}^{\circ }={55}^{\circ }$
$\text{But , OR = OQ = radii of the circle.}$
Since base angles of an isosceles triangle are equal, we have
$\angle ORQ=\angle OQR={55}^{\circ }.$
Using angle sum property, we have
$\therefore y={180}^{\circ }-({55}^{\circ }+{55}^{\circ })={70}^{\circ }.$