Question

In given fig., O is center of the circle. Chord AB = CD and ∠OBA = 40°, then ∠COD will be :

(A) 100°

(B) 80°

(C) 180°

(D) 90°

Answer 1

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°