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In fig, if O is center of circle, then ∠AOB will be :(A) 70°(B) 110°(C) 120°(D) 140° |
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Answer» Answer is (D) 140° Join CO In ∆AOC, AO = CO (Radius of same circle) ⇒ ∠OAC = ∠OCA ∴ ∠OCA = 30° In ∆OCA ∠AOC + ∠OCA + ∠OAC = 180° [∵ Sum of all the angles of triangle is 180°] ⇒ ∠AOC + 30° + 30° = 180° ⇒ ∠AOC + 60° = 180° ∴ ∠AOC = 180° – 60° = 120° Similarly, OB = OC ∠OBC = ∠OCB = 40° In ∆OBC ∠BOC + ∠OBC + ∠OCB = 180° ⇒ ∠BOC + 40° + 40° = 180° ∴ ∠BOC = 180° – 80° = 100° ∴ ∠AOB = 360° – (∠AOC + ∠BOC) = 360° – (120° + 100°) = 360° – 220° ∠AOB = 140° |
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