1.

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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