Notes
pentagon with two semi-circles solution

Solution to the Pentagon with two Semi-Circles Puzzle

Solution by Angle Between a Radius and Tangent and Angles in a Regular Pentagon

In the above diagram, the points labelled $B$ and $H$ are the centres of their semi-circles.

Since the angle between a radius and tangent is $90^\circ$, angles $B \hat{E} D$ and $F \hat{E} H$ are both $90^\circ$. Angle $D \hat{E} F$ is the interior angle in a regular pentagon so is $108^\circ$. Since the angles at a point add up to $360^\circ$, this leaves $72^\circ$ for angle $H \hat{E} B$.

Triangles $A B E$ and $A E G$ are both isosceles, so angles $B \hat{A} H$ are $H \hat{E} B$ are equal and therefore the angle between the bases of the semi-circles is $72^\circ$.