Question

In the given figure, $PQ$ is the diameter of the circle whose centre is $O$. Given $\angle ROS={42}^o$, calculate $\angle RTS$.

Answer 1

Join $PS.$

$PQ$ is a diameter.

$\Rightarrow$ $PSQ={90}^o$ [ Angle inscribe in a semicircle ]

$\Rightarrow$ $\angle PSQ+\angle PST={180}^o$ [ Linear pair ]

$\Rightarrow$ ${90}^o+\angle PST={180}^o$

$\Rightarrow$ $\angle PST={90}^o$

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

$\Rightarrow$ $\angle SPT=\frac{1}{2}\times {42}^o$

$\therefore$ $\angle SPT={21}^o$

In $△PST,$

$\Rightarrow$ $\angle SPT+\angle PTS+\angle PST={180}^o$

$\Rightarrow$ ${21}^o+\angle PTS+{90}^o={180}^o$

$\Rightarrow$ $\angle PTS={180}^o-{90}^o-{21}^o$

$\Rightarrow$ $\angle PTS={69}^o$

$\therefore$ $\angle PTS=\angle RTS={69}^o$