Important for :

1

C

Let radius be 'r' and ∠POS = x°

ΔOQR isoceles ∴∠QOR = 30°

∴ ∠OQR = 120° (Sum of all angles of ΔOQR = 180°)

∴ ∠OQP = 60° (Supplementary agnle)

ΔOPQ isoceles since OP = OQ = r

∴ ∠OQP = 60° = ∠OQP

∴ ∠ POQ = 60° = [Sum of all angle of Δ = 180° ]

Now SOR is a straight line

∴ x + 60° + 30° = 180°

∴ x = 90°

Hence, option (C) is correct.

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