so it's going to be a long one, sorry i couldn't find a shorter one. But don't worry, the concept is easy.now in triangle PSO, PS = SO (given). ........1OP = SO (RADIUS) .......2from 1 and 2 PS=OP since all sides are equal, it's a equilateral triangle, which means all the angles the also equal (60°) .......3PM=MO this means, SM is median and is perpendicular to OP.now, angle SMO =90° angle OMR =90° ....... aIN triangle OSM angle OMR = 90° angle MOS = 60° (from 3)from these 2 angles, we can find angle OSM(OSR)= 30° since, OS = OR. (radius)angle OSR = angle ORS (angles opposite to equal sides)angle ORS = 30° ...........b in triangle OMR, from 'a' and 'b' angle POR = 60° angle PQR = 30° (angle formed by a segment on any POINT of circle is half the angle substended by it at CENTRE, (this therom is from class 10))now in triangle MQR WE, have angle QMR =90° angle. MQR = 30° then angle MRQ(QRS) =60° then finally, angle QRS - angle PQR = 60° - 30° = 30° and so your answer is 30° so you may MARK it as brainlist, if you are satisfied with the answer.thank you