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In Fig below, PQ is tangent at point R of the circle with center O. If ∠TRQ = 30°, find ∠PRS. |
Answer» Given, ∠TRQ = 30°. At point R, OR ⊥ RQ. So, ∠ORQ = 90° ⟹ ∠TRQ + ∠ORT = 90° ⟹ ∠ORT = 90°- 30° = 60° It’s seen that, ST is diameter, So, ∠SRT = 90° [ ∵ Angle in semicircle = 90°] Then, ∠ORT + ∠SRO = 90° ∠SRO + ∠PRS = 90° ∴ ∠PRS = 90°- 30° = 60° |
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