In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If $\angle CBD={60}^{\circ }$,calculate $\angle CDE.$

NSWER:
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 Δ ADC, 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 °