Question

For the given figure find the measure of angle ∠BEC.

degrees

Answer 1

In the given figure, consider the arc DA.


Arc DA subtends angle $\angle DCA$ at C and $\angle ABD$ at B.
As the measure of all angles subtended by an arc on the same segment of the circle are equal, $\angle DCA=\angle ABD={22}^o.$

$\angle DFA$ and $\angle CFB$ are vertically opposite angles.
$\therefore \angle DFA=\angle CFB={65}^o$

For triangle $△CFB,$
$\angle FCB+\angle CBF+\angle CFB={180}^o$
$\Rightarrow \angle FCB+\angle CBF={180}^o-{65}^o={115}^o$

For triangle $△ECB,$
$\angle ECB+\angle EBC=(\angle DCF+\angle FCB)+(\angle CBF+\angle ABF)$
$={22}^o+(\angle FCB+\angle CBF)+{22}^o$
$={22}^o+\left({115}^o\right)+{22}^o$
$={159}^o$

In $△BEC$
Also, $\angle ECB+\angle EBC+\angle BEC={180}^o$
$\Rightarrow \angle BEC={180}^o-{159}^o={21}^o$