In given fig., O is center of the circle. Chord AB = CD and ∠OBA = 4

1.	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» 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°

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»

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°

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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°

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