Question

In the given figure, if $\overline{PQ}$ is a diameter, then the measure of $\angle SQP$ is equal to

Answer 1

$\measuredangle PSQ,{90}^0$

$\measuredangle SPQ,180-150={30}^0$

(Opp. $\measuredangle S$ of cyclic quadrilateral)

$\therefore$ $\measuredangle SQP=180-90-30={60}^0$