Answer:

AngleQSR= 27° (as angleQRS is 90° )

AnglePSR= 54°

AnglePQS= 180°-(63+27+27)

= 180°-127

= 53°

Verified answer

Take SQ as the diameter.

Angle SPQ = Half of Angle SOQ

(angle SOQ is 180° due to a line. So angle SPQ is half of angle SOQ = 90° )

So, Angle SPQ = 90°

In ∆ SPQ, Angle PQS = 180° - (27°+90°) = 63°

Now, (Angle PSQ + Angle RSQ) + (Angle PQS + Angle RQS) = 180°

( because PQRS is a cyclic quadrilateral and we know that sum of opp. sides in a cyclic quad. is = 180°)

So, Angle PSR = 180° - Angle PQR

= 180°-126° = 54°

Hence, Angle PQS = 63° and angle PSR = 54°.

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