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Question

In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If ∠CBD = 60°, calculate ∠CDE.

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Solution


Angles in the same segment of a circle are equal.
i.e., CAD = CBD = 60°
We know that an angle in a semicircle is a right angle.
i.e., ADC = 90°
In ΔADC, we have:
ACD + ADC + CAD = 180° (Angle sum property of a triangle)
ACD + 90° + 60° = 180°
ACD = 180° – (90° + 60°) = (180° – 150°) = 30°
CDE = ACD = 30° (Alternate angles as AC parallel to DE)
Hence, CDE = 30°

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