Question

In the given figure, chord ED is parallel to the diameter AC of the circle. If ∠CBE= 65°, then ∠DEC =

• A. 15°
• B. 25°
• C. 55°
• D. 35°

Answer 1

The correct option is B 25°

Consider the arc CDE.
The angles in the same segment are:$\angle CBEand\angle CAE$
$\therefore \angle CAE=\angle CBE$
$\Rightarrow \angle CAE={65}^{\circ }[\angle CBE={65}^{\circ }]$

Since AC is the diameter of the circle and the angle in a semi - circle is a right angle
$\therefore \angle AEC={90}^{\circ }$.

Now, in $\Delta ACE$, we have
$\angle ACE+\angle AEC+\angle CAE={180}^{\circ }$
$\Rightarrow \angle ACE+{90}^{\circ }+{65}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ACE={25}^{\circ }$.



As, $\angle DEC$ and $\angle ACE$ are alternate angles, because $AC||DE$.
$\therefore \angle DEC=\angle ACE={25}^{\circ }$.