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You're on a roll — Here's the solution for the geometry problem. Given: • PQRS is a cyclic quadrilateral. • TPX is a tangent to the circle at P. • O is the centre of the circle. • RQX is a straight line. • RPQ = 50^. • PRS = 25^. a) Angle PRQ Step 1: Identify angles subtended by the same arc. Angles subtended by the same arc at the circumference are equal. PRQ and PSQ are subtended by arc PQ. RPQ and RSQ are subtended by arc RQ. SPR and SQR are subtended by arc SR. QPS and QRS are subtended by arc QS. Step 2: Use the given information. We are given RPQ = 50^. PRQ and PSQ are subtended by arc PQ. SPR and SQR are subtended by arc SR. We are given PRS = 25^. Consider PQR. The sum of angles in a triangle is 180^. We need to find PRQ. We know RPQ = 50^. We also know that QPS = QRS (angles subtended by arc QS). And PQR + PSR = 180^ (opposite angles of a cyclic quadrilateral). Let's re-examine the diagram and given values. The diagram shows QPR = 50^ and PRS = 25^. The markings on the chords PQ and RS indicate that PQ = RS. If PQ = RS, then the angles subtended by these chords at the circumference are equal. So, PRQ = SPR. Also, PSQ = QPR = 50^ (angles subtended by arc PQ). And RPS = RQS (angles subtended by arc RS). Since PQ = RS, the arcs PQ and RS are equal. Therefore, the angles subtended by these arcs at the circumference are equal. PRQ = SPR. We are given RPQ = 50^ and PRS = 25^. In PRS, we have PRS = 25^. In PQR, we have RPQ = 50^. Let's use the property that angles subtended by the same arc are equal. PRQ is subtended by arc PQ. PSQ is subtended by arc PQ. So, PRQ = PSQ. We are given RPQ = 50^. We are given PRS = 25^. From the diagram, the chords PQ and RS are marked as equal. If chords PQ and RS are equal, then the angles they subtend at the circumference are equal. So, PRQ = SPR. Also, PSQ = QPR = 50^. And RPS = RQS. Since PQ = RS, then PRQ = SPR. In PQR, we have RPQ = 50^. In PRS, we have PRS = 25^. Let's use the fact that PSQ = PRQ (angles subtended by the same arc PQ). Also, RPS = RQS (angles subtended by the same arc RS). Since PQ = RS, then PRQ = RPS. So, PRQ = RPS. We know QPS = QPR + RPS = 50^ + RPS. We also know QRS = QRP + PRS = QRP + 25^. Since PQRS is a cyclic quadrilateral, QPS = QRS. So, 50^ + RPS = QRP + 25^. 25^ + RPS = QRP. Since PQ = RS, then PRQ = SPR. Let PRQ = x. Then SPR = x. In PQR, PQR + QRP + RPQ = 180^. PQR + x + 50^ = 180^. PQR = 130^ - x. In PRS, PSR + SPR + PRS = 180^. PSR + x + 25^ = 180^. PSR = 155^ - x. Since PQRS is a cyclic quadrilateral, PQR + PSR = 180^. (130^ - x) + (155^ - x) = 180^. 285^ - 2x = 180^. 2x = 285^ - 180^. 2x = 105^. x = 52.5^. So, PRQ = 52.5^. Reason: Angles subtended by equal chords (PQ and RS) are equal, and opposite angles of a cyclic quadrilateral sum to 180^. PRQ = 52.5^ b) Angle PSR Step 1: Use the result from part (a). From part (a), we found x = PRQ = 52.5^. We also established that PSR = 155^ - x. Substitute x = 52.5^: PSR = 155^ - 52.5^ = 102.5^. Reason: Opposite angles of a cyclic quadrilateral sum to 180^. Alternatively, PSR = PQR + QRS. We know PQR = 130^ - x = 130^ - 52.5^ = 77.5^. PSR = 180^ - PQR = 180^ - 77.5^ = 102.5^. PSR = 102.5^ c) Angle PXQ Step 1: Identify the relevant triangle. Consider PQX. We need to find PXQ. We know RPQ = 50^. We know RQX is a straight line. PQR = 77.5^ (from part b calculation). So, PQX = 180^ - PQR = 180^ - 77.5^ = 102.5^ (angles on a straight line). Step 2: Use the sum of angles in PQX. In PQX, PXQ + QPX + PQX = 180^. We need QPX. TPX is a tangent at P. QPX is the angle between the tangent PX and the chord PQ. By the Alternate Segment Theorem, QPX = PRQ. From part (a), PRQ = 52.5^. So, QPX = 52.5^. Now substitute the known angles into the sum of angles for PQX: PXQ + 52.5^ + 102.5^ = 180^. PXQ + 155^ = 180^. PXQ = 180^ - 155^ = 25^. Reason: Alternate Segment Theorem and sum of angles in a triangle. PXQ = 25^ d) Angle TPS Step 1: Apply the Alternate Segment Theorem. TPS is the angle between the tangent TP and the chord PS. By the Alternate Segment Theorem, TPS = PRS. We are given PRS = 25^. Reason: Alternate Segment Theorem. TPS = 25^ e) Angle POS Step 1: Relate the angle at the centre to the angle at the circumference. POS is the angle subtended by arc PS at the centre O. PRS is the angle subtended by arc PS at the circumference. The angle at the centre is twice the angle at the circumference subtended by the same arc. So, POS = 2 × PRS. We are given PRS = 25^. POS = 2 × 25^ = 50^. Reason: Angle at the centre is twice the angle at the circumference. POS = 50^ What's next?