Question

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

Answer 1

Join $OE$. Arc $EC$ subtends $\angle EOC$ at the centre and $\angle EBC$ at the remaining Part of the circle.

$\angle EOC=2\angle EBC=2\times {65}^{\circ }={130}^{\circ }$

now in $△OEC,OE=OC$(radii of the same circle)

$\therefore \angle OEC=\angle OCE$

But in $△EOC$

$\angle OEC+\angle OCE+\angle EOC={180}^{\circ }$(angles of a triangle)

$\Rightarrow \angle OCE+\angle OCE+\angle EOC={180}^{\circ }$

$\Rightarrow 2\angle OCE+\angle EOC={180}^{\circ }$

$\Rightarrow 2\angle OCE={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

$\Rightarrow \angle OCE=\frac{{50}^{\circ }}{2}={25}^{\circ }$

$\therefore AC\parallel ED$(given)

$\therefore \angle DEC=\angle OCE$

$\Rightarrow \angle DEC={25}^{\circ }$