Question

According to figure, PQ and XY are parallel lines. If $\angle {QRT=30}^{\circ },$ then find the value of $\angle TSY.$

Answer 1

Given :
$PQ\parallel XY$
and $\angle QRT={30}^{\circ }$
Diameter of circle of centre $O$ is $RS$
$\therefore OR\perp PR$ $OS\perp XY$
$\therefore \angle QRO={90}^{\circ }$
$\Rightarrow \angle QRO={90}^{\circ }$
$\angle RTS=\angle QRS-\angle QRT$
$={90}^{\circ }-{30}^{\circ }={60}^{\circ }$
$\therefore \angle TRS={60}^{\circ }$
Now $\angle TSY=\angle TRS$
$\Rightarrow \angle TSY={60}^{\circ }$