Question

In the figure shown, chord $ED$ is parallel to diameter $AC$ of a circle. If $\angle CBE={60}^{\circ }$, then $\angle DEC$ must be

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${10}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

Answer: (B)

${30}^{\circ }$

Let EB be perpendicular to ED.
The measure of the angle CBE will remain ${60}^0$ irrespective of its position as angle in the same arc are equal.
So,if EB be perpendicular to ED and hence AC,we can say that AC will divide the isosceles triangle BCE into two equal parts and AC is the angle bisector of the angle BCE.
$\therefore \angle ACE=\angle DEC$ [alternate interior angles]
$\therefore \angle ACE={30}^0$ and $\angle DEC={30}^0$