Answer: Let ∠APQ = x and ∠AQP = y. Then ∠QAR = 180 - y and ∠PSA = 180 - x. Also, note that ∠QOS = 2y and ∠SOR = 90 - y. Then, using the exterior angle theorem in triangle APS, we have ∠SAR = x + y, and ∠ROQ = 180 - (x + y). Finally, we have: ∠SOR + ∠ROQ + ∠QOS = (90 - y) + (180 - (x + y)) + 2y = 270 - x But we know that ∠SOR + ∠ROQ + ∠QOS = 360 (angle sum property). Therefore, we have: 270 - x = 360 Solving for x, we get x = 90. Therefore, ∠APQ = 90 and ∠AQP = 90, which implies that PAQ is a straight line, and hence SOR PAQ = 180°.

<p>To prove that ∠SOR + ∠PAQ = 180°, we need to show that the angles are supplementary. Here's a step-by-step proof:</p><p><br></p><ol><li>Given that PQ is a diameter of the circle, it implies that ∠POQ is a right angle (90°), as the diameter subtends a right angle at any point on the circumference.</li><li>Since PS and QR are chords of the circle, they intersect externally at point A.</li><li>By the Exterior Angle Theorem, ∠SOR is equal to the sum of the opposite interior angles: ∠SOR = ∠OSQ + ∠QRA.</li><li>Similarly, ∠PAQ is equal to the sum of the opposite interior angles: ∠PAQ = ∠PAS + ∠SQR.</li><li>Notice that ∠OSQ is a central angle subtended by the chord QR, and ∠QRA is an inscribed angle subtended by the same chord. By the Inscribed Angle Theorem, these angles are supplementary: ∠OSQ + ∠QRA = 180°.</li><li>Similarly, ∠PAS and ∠SQR are supplementary angles, as they are inscribed angles subtended by the chord PS: ∠PAS + ∠SQR = 180°.</li><li>Combining the above results, we have:</li><li>∠SOR = ∠OSQ + ∠QRA = 180°,</li><li>∠PAQ = ∠PAS + ∠SQR = 180°.</li><li>Thus, ∠SOR + ∠PAQ = 180°, which proves that SOR PAQ is a straight angle.</li></ol><p><br></p>