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In figure, if ∠DAB = 60° , ∠ABD = 50°, then ∠ACB is equal to

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In figure, if ∠DAB = 60° , ∠ABD = 50°, then ∠ACB is equal to

(A) 60º (B) 50º (C) 70º (D) 80º

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(C) 70º

Given, ∠DAB = 60°, ∠ABD = 50°

Since, ∠ADB = ∠ACB …(i)
[angles in same segment of a circle are equal]

In ΔABD, ∠ABD + ∠ADB + ∠DAB = 180° [by angle sum property of a triangle]

50°+ ∠ADB + 60° = 180°

=> ∠ADB = 180° – 110° = 70°

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