Question

In the circle shown above, chords $\overline{TR}$ and $\overline{QS}$ intersect at point P, which is the center of the circle, and the measure of $\angle PST$ is ${30}^{\circ }$. By the measurement of central angle calculate the measure of angle $RPS$

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$
• E. Cannot be determined from the given information

Answer 1

The correct option is C ${60}^{\circ }$

Consider triangle $PTS$.

Now, $PT=PS=$ radius of the circle.

Hence $△PTS$ is an isosceles triangle.

$\Rightarrow \angle PTS=\angle PST={30}^0$.

$\Rightarrow \angle TPS={180}^0-({30}^0+{30}^0)$...(angle sum property).
Hence, $\angle TPS={120}^0$

Now $\angle SPR$ and $\angle TPS$ form a linear pair.

Hence $\angle SPR={180}^0-\angle TPS={180}^0-{120}^0={60}^0$.