Mathematics — CBSE Class 10 board question

Option (B) 60°

Since MN is a diameter, ∠MLN = 90° (angle in a semicircle). In △MNL, ∠NLM = 90° − 30° = 60°. Since RS is tangent at L, OL ⊥ RS, so ∠RLM = 90° − ∠NLM = 90° − 60° = 60°.

Source: Chapter 10, Tangent to a Circle

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• Angle in a semicircle: ∠MLN = 90° (angle subtended by diameter at any point on circle).
• In △NML: ∠NLM = 180° − 90° − 30° = 60°.
• Tangent ⊥ radius (Theorem 10.1): The radius OL ⊥ RS, so ∠OLR = 90°. Since OL lies along the diameter direction (N–M), ∠RLM = 90° − ∠NLM = 30°...

Correction (careful reasoning): OL is the radius to L, perpendicular to RS. The angle ∠RLN + ∠NLM + ... Note that ∠RLM = 90° − ∠NLM only if N is on the R-side. From the figure, ∠RLM = 90° − ∠(between OL and LM). Since ∠NML = 30°, arc reasoning gives ∠RLM = ∠NML = 60° by the tangent-chord angle theorem (alternate segment theorem): the angle between tangent and chord equals the inscribed angle in the alternate segment.

Key rule to remember: Tangent-chord angle = inscribed angle in alternate segment. ∠RLM = ∠LNM = 60° (since ∠LNM = 90° − 30° = 60°). Answer: B) 60°.