Question

Find the angle $\angle FBA$ if $ABCDE$ is a regular polygon.

• A. ${30}^o$
• B. ${36}^o$
• C. ${72}^o$
• D. Insufficient data

Answer 1

The correct option is B ${36}^o$

$ABCDE$ is a pentagon.

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

Subsituting $n=5,$
sum of interior angles $=(5-2)\times {180}^o$
$={540}^o$

Also, a regular polygon has all the angles as equal in measure.
$\therefore$ Measure of each interior angle
$=\frac{{540}^o}{5}$
$={108}^o$

$\therefore \angle BAE={108}^o$

$\angle BAF$ is exterior to $\angle BAE.$
$\therefore \angle BAF={180}^o-\angle BAE$
$\Rightarrow \angle BAF={180}^o-{108}^o={72}^o$

As a regular polygon has all the sides as equal in measure, $BC=BA=5\text{units}$

For $△BFA,BF=BA=5\text{units}$
$\therefore$ The opposite angles will be also equal in measure.
$\Rightarrow \angle BAF=\angle BFA={72}^o\therefore \angle FBA={180}^o-2\times \angle BAF\Rightarrow \angle FBA={180}^o-2\times {72}^o={36}^o$