In the given figure, QR is the diameter of the circle. P and S are points on the circle such that ∠RPS = 50°. Find ∠SRQ.
Explanation:
Given QR is the diameter. Let QS be a chord.
Diameter subtends an angle 90° anywhere on the circle
∴ ∠RPQ = 90°.
Given, ∠RPS = 50°.
⇒ ∠SPQ = 90° - 50° = 40°.
Also, a chord subtends same angle in the same segment of the circle.
∴ ∠SRQ = ∠SPQ = 40°
Hence, option (b).
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