[tex]m\angle BAE=180^o-\dfrac{360^o}{5}=180^o-72^o=108^o[/tex]
We know, in a triangle, the three interior angles always add to 180° and in
an isosceles triangle are two equal angles.
Therefore in ΔEBA
[tex]m\angle AEB=\dfrac{180^o-108^o}{2}=\dfrac{72^o}{2}=36^o.[/tex]
