Question

In the given figure, O is the centre of a circle and PQ is a diameter. If $\angle ROS={40}^{\circ },$ then what is the value of $\angle RTS$?

Answer 1

Since PQ is the diameter, we have
$\angle PRQ={90}^{\circ }$ [angle in a semi circle]
But, $\angle PRQ+\angle TRQ={180}^{\circ }.$ [linear pair of angles]
$\therefore {90}^{\circ }+\angle TRQ={180}^{\circ }⟹\angle TRQ={90}^{\circ }$

Since the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any remaining point on the circle, we have
$\angle ROS=2\angle RQS$
i.e., ${40}^{\circ }=2\angle RQS$
$⟹\angle RQS=\frac{{40}^{\circ }}{2}={20}^{\circ }$

Using angle sum property in $△RQT,$ we have
$\angle RQT+\angle QTR+\angle TRQ={180}^{\circ }.$
i.e., ${20}^{\circ }+\angle RTS+{90}^{\circ }={180}^{\circ }$
$⟹\angle RTS={180}^{\circ }-{110}^{\circ }={70}^{\circ }$