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How do you solve #3?

How do you solve #3?-example-1

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m\angle BAE=180^o-(360^o)/(5)=180^o-72^o=108^o

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


m\angle AEB=(180^o-108^o)/(2)=(72^o)/(2)=36^o.
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