Question
O and C are the respectively orthocentre and
circumcentre of ∆PQR. The line PO is extended which intersect line QR at S. such that, ∠QCR = 140°, ∠PQS = 58°. Find ∠RPS = ?Solution
In ∆ PQS ∠QPS + ∠PQS + ∠PSQ = 180° ∠QPS + 58° + 90° = 180° ∠QPS = 32° 
∠QCR = 2∠QPR (∠QCR)/2 = ∠QPR 140/2 = ∠QPS + ∠RPS 70° = 32° + ∠RPS ∠RPS = 38°
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