Question

In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of $\angle ABE$

• A. ${36}^{\circ }$
• B. ${45}^{\circ }$
• C. ${27}^{\circ }$
• D. ${54}^{\circ }$

Answer 1

The correct option is A ${36}^{\circ }$

In a regular pentagon ABCDE, BE is a diagonal.
$\angle BAE$ is the interior angle of the polygon.
Each interior angle of a regular polygon is = $\frac{{180}^o(n-2)}{n}$
where n = number of sides of polygon, $n=5$
$\angle BAE=\frac{{180}^o(5-2)}{5}={108}^o$
Now, $△ABE$ is an isosceles triangle in which AB $=$ AE.
So, $\angle ABE=\angle AEB$
$\angle ABE+\angle AEB+\angle BAE={180}^o$
$=>2\angle ABE={180}^o-{108}^o$
$=>\angle ABE=\frac{{72}^o}{2}={36}^o$