Question

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

Answer 1

In the figure, O is the centre of the circle, PQ is the diameter and $\angle ROS={40}^{\circ }$

Now we have to find $\angle RTS$

Are RS subtends $\angle ROS$ at the centre and $\angle RQS$ at the remaining part of the circle

$\therefore \angle RQS=\frac{1}{2}\angle ROS$

$=\frac{1}{2}\times {40}^{\circ }={20}^{\circ }$

$\because \angle PRQ={90}^{\circ }\left(Angleinasemicircle\right)$

$\therefore \angle QRT={180}^{\circ }-{90}^{\circ }={90}^{\circ }(\because PRTisastraightline)$

Now in $\Delta RQT$,

$\angle RQT+\angle QRT+\angle RTQ={180}^{\circ }\left(Anglesofatriangle\right)$

$\Rightarrow {20}^{\circ }+{90}^{\circ }+\angle RTQ={180}^{\circ }$

$\Rightarrow \angle RTQ={180}^{\circ }-{20}^{\circ }-{90}^{\circ }={70}^{\circ }$

or $\angle RTS={70}^{\circ }$