Calculate the angle in the regular pentagon

Question

I want calculate the angle of $DAC$ in the regular pentagon. But I'm not sure with my attempt. Is it right my attempt?

Since $\angle EAB$ divide into $\angle EAD$, $\angle DAC$, $\angle CAB$ and $ABCDE$ is regular pentagon so $\angle EAD = \angle DAC = \angle CAB$. We know in the regular pentagon have interior angle $\angle EAB = 108^\circ$, so we have $\angle DAC=108^\circ/3= 36^\circ$. Is it right?

Answer 1

Your solution is correct. Perhaps an even easier way to see this is to note that central angle $\angle DOC$ measures $\frac{360^\circ}{5} = 72^\circ$ due to symmetry, and then $\angle DAC$ is the corresponding inscribed angle, so it measures $\frac{1}{2} (72^\circ) = 36^\circ$.

Answer 2

Recall too that constructing a regular pentagon requires first constructing an isosceles triangle in which each base angle is double the angle at the vertex (Euclid, Elements, IV, 10). This is $\triangle DAC$, so $\angle DAC=36^o$.