Question

In the given figure, chord ED is parallel to the diameter AC of the circle. Given, $\angle CBE={65}^{\circ }$, find $\angle DEC$.

• A. ${35}^{\circ }$
  
• B. ${115}^{\circ }$
• C. ${25}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is C ${25}^{\circ }$

$\angle CAE=\angle CBE$ ($\because$ angles in the same segment of arc CDE)
$\Rightarrow \angle CAE={65}^{\circ }(\because \angle CBE={65}^{\circ })$
Since, AC is the diameter and the angle in a semi - circle is a right angle.
$\therefore \angle AEC={90}^{\circ }$
Now,in $\Delta ACE$,
$\angle ACE+\angle AEC+\angle CAE={180}^{\circ }$
$\Rightarrow \angle ACE={180}^{\circ }-(90+{65}^{\circ })={25}^{\circ }$
Since, $AC||DE\therefore \angle DEC$ and $\angle ACE$ are alternate angles.
$\Rightarrow \angle DEC=\angle ACE={25}^{\circ }$