Question

Determine the angle ∠FCE for the regular polygon ABCDEF.


degrees.

Answer 1

$ABCDE$ is a hexagon.

Sum of interior angles $=(n-2)\times {180}^o$
where $n$ is the total number of sides.

$\therefore$ Sum of interior angles $=(6-2)\times {180}^o$
$={720}^o$

For a regular polygon, the measure of each interior angle
$=\frac{{720}^o}{6}$
$={120}^o$

$\therefore \angle D={120}^o$

$CD=DE$ (regular polygon)
$\therefore$ For $△CDE,$ the opposite angles are equal in measure.
$\Rightarrow \angle DCE=\angle DEC$
$\therefore 2\angle DCE={180}^o-\angle D(\because \angle DCE+\angle DEC+\angle D={180}^o)$
$\Rightarrow \angle DCE=\frac{{180}^o-{120}^o}{2}={30}^o$

$CF$ is a mirror line of the polygon $ABCDEF$ that means it divides the hexagon into two equal parts.
$\therefore \angle DCF=\frac{\angle DCB}{2}=\frac{{120}^O}{2}={60}^o$

Also, $\angle DCF=\angle DCE+\angle ECF$
$\Rightarrow {60}^o={30}^o+\angle ECF$
$\Rightarrow \angle ECF={60}^o-{30}^o={30}^o$