Question:
In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°.
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(c) 70°
∠BDC = ∠BAC = 60° (Angles in the same segment of a circle)
In Δ BDC, we have:
∠DBC + ∠BDC + ∠BCD = 180° (Angle sum property of a triangle)
∴ 50° + 60° + ∠BCD = 180°
⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°
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