Question

Find the angle ∠CAE of the given equiangular polygon ABCDE.

degree

Answer 1

ABCDE is an equiangular polygon.

We know, sum of interior angles $=(n-2)\times {180}^o$
where $n$ is the total number of sides, here it is 5.

$\therefore$ Sum of interior angles $=(5-2)\times {180}^o$
$=3\times {180}^o$
$={540}^o$

Let the interior angle be $x.$
$\therefore 5x={540}^o$
$\Rightarrow x=\frac{{540}^o}{5}={108}^o$

From figure, $\angle BCA={180}^o-{144}^o={36}^o$
For $△ABC,$
$\angle BAC+\angle B+\angle BCA={180}^o$
$\Rightarrow \angle BAC+{108}^o+{36}^o={180}^o$
$\Rightarrow \angle BAC={180}^o-{108}^o-{36}^o={36}^o$

$\angle BAE=\angle BAC+\angle CAE$
$\Rightarrow {108}^o={36}^o+\angle CAE$
$\Rightarrow \angle CAE={108}^o-{36}^o={72}^o$