In a circle with center O, PQRS is a cyclic quadrilateral and PR is the diameter. Chords PQ and SR are Produced to meet at M. If ∠RPM = 34° and ∠M = 30° then ∠RQS is equal to:Choices:- 36° 26° 24° 34°

To solve this problem, we need to use the properties of cyclic quadrilaterals and circles.

Step 1: Since PR is the diameter of the circle, ∠PSR is a right angle (90°) because the angle in a semi-circle is a right angle.

Step 2: In triangle PRM, ∠RPM + ∠PRM + ∠PMR = 180° (sum of angles in a triangle). We know ∠RPM = 34° and ∠PRM = 90°, so we can find ∠PMR.

∠PMR = 180° - 90° - 34° = 56°

Step 3: ∠PMR = ∠PSR (angles subtended by the same chord in the same segment are equal). So, ∠PSR = 56°.

Step 4: In quadrilateral PQRS, ∠PSR + ∠Q = 180° (opposite angles of a cyclic quadrilateral sum to 180°). We know ∠PSR = 56°, so we can find ∠Q.

∠Q = 180° - 56° = 124°

Step 5: In triangle QMR, ∠Q + ∠QMR + ∠RQM = 180° (sum of angles in a triangle). We know ∠Q = 124° and ∠QMR = 30°, so we can find ∠RQM.

∠RQM = 180° - 124° - 30° = 26°

Step 6: ∠RQM = ∠RQS (angles subtended by the same chord in the same segment are equal). So, ∠RQS = 26°.

So, the answer is 26°.