Question

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

Answer 1

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