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In an acute-angled triangle PQR, let O be the orthocentre and C the circumcentre. The line PO is extended to meet side QR at S. Given angle QCR = 130 deg and angle PQS = 60 deg, find angle RPS.

1. 35 deg
2. 30 deg
3. 50 deg
4. 60 deg

Correct answer: 35 deg

Solution

Central angle QCR = 2 * angle QPR => angle QPR = 130/2 = 65 deg. Since O (orthocentre) lies on the altitude from P, line PO meets QR perpendicularly, so angle PSQ = 90 deg. In triangle PSQ: angle QPS = 180 - 90 - 60 = 30 deg. Then angle RPS = angle QPR - angle QPS = 65 - 30 = 35 deg.

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