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Question
In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80o and angle ACE = 10o. Calculate : Angle BCD
Solution
DC || EB
∴ DCE = ∠BEC = 50° (Alternate angles)
∴ ∠ AOB = 80°
⇒ ∠ ACB = `1 /2` ∠AOB = 40°
(Angle at the center is double the angle at the circumference subtended by the same chord) We have,
∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° +10°+ 50° = 100°
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