Discussion

Explanation:

In ∆AOB, AB = 5 = OA = PB

∴ ∆AOB is an equilateral triangle.

⇒ ∠AOB = 60°

In quadrilateral AOBP,

∠A = ∠B = 90°

∴ ∠AOB + ∠A + ∠B + ∠P = 360°

⇒ 60 + 90 + 90 + ∠P = 360

⇒ ∠P = 120°

Since PA || OC and PB || OD

⇒ ∠COD = ∠APB = 120°

Hence, option (b).

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