Question

$PQRS$ is a cyclic quadrilateral and $\overline{PQ}$ is the diameter of the circle. If $m\angle QPR={35}^{\circ },$ then the measure of $\angle PSR$ is equal to ?

• A. ${125}^{\circ }$
• B. ${135}^{\circ }$
• C. ${145}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

The correct option is A ${125}^{\circ }$

Given,

$\angle QPR={35}^{\circ }$

In $\Delta PRQ$

$\angle PRQ=90$ ( angle in a semicircle is right angle)

Also

$\angle QPR+\angle PRQ+\angle PQR=180$

$\angle PQR=180-35-90={55}^{\circ }$

As $PQRS$ is cyclic quadrilateral

$\therefore \angle PSR+\angle PQR=180⟹\angle PSR=180-55={125}^{\circ }$