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Parallel Lines with Transversal

In the given ...

Question

In the given figure, ∠DAB = 60° and ∠ABD = 50°, then ∠ACB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

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Solution

(c) 70°

In ΔABD, we have:
∠DAB + ∠ABD + ∠BDA = 180°
⇒ 60° + 50° + ∠BDA = 180°
⇒ ∠BDA = (180° - 110°) = 70°
⇒ ∠ACB = ∠BDA = 70° (Angles in the same segment)

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