Pentagon Problem

First of all, all the angles in any n-sided polygon add up to $(n - 2)\times180°$ . For any pentagon, that would be $3\times180° = 540°$ . For a regular pentagon, the five angles are equal, so each one is 540/5 = 108°.

Now, look at isosceles triangle ABC, with an angle of 108° at B. The other two angles are equal: call each $x$ . $108 + x + x = 180$ , which leads to $x = 36°$ .

So, ∠BAC = 36° and ∠ECD = 36°, which means that ∠ACE = 36°

$\text{Given that ABCDE is a regular pentagon. The points A, B, C, D, E are on circle with center F (See figure above).}$

$\text{Therefore,}$ $\angle ACE = \frac{\angle AFE}{2}$ $\text{where}$ $\angle AFE = 72 ^ \circ$ $\text{(because}$ $\angle AFE$ $\text{is the central angle of the regular pentagon ABCDE).}$

So $\angle ACE = \frac{72 ^ \circ}{2} = 36 ^ \circ$ .