Answer is (A) 100°
Given : AB = CD
Radius of circle = OB = OD
∴ ∠OBA = ∠ODC = 40°
∆OCD in OD = OC (Radius of circle)
∴ ∆OCD is an isosceles triangle.
∠OCD = ∠ODC = 40°
In ∆OCD,
∠ODC + ∠OCD + ∠COD = 180°
40° + 40° + ∠CQD = 180°
∠COD = 180° – 40° – 40°
∠COD = 180° – 80°
∠COD = 100°