Notes
two semi-circles in an equilateral triangle solution

Solution to the Two Semi-Circles In an Equilateral Triangle Puzzle

Solution by Angle Between a Tangent and Radius, Angle at the Circumference is Half the Angle at the Centre, and Angles in a Triangle

In the diagram above, $O$ and $Q$ are the centres of their respective semi-circles, $P$ is the point of tangency and $F J$ is the tangent to both circles at $P$.

Since the angle between a tangent and radius is $90^\circ$, the angles in an equilateral triangle are $60^\circ$, and the angles in a triangle add up to $180^\circ$, angle $A \hat{Q} E$ is $30^\circ$.

Then as the angle at the circumference is half the angle at the centre, angle $D \hat{P} E$ is $15^\circ$.

Similarly, angle $G \hat{P} H$ is also $15^\circ$.

Triangle $F E P$ is isosceles as line segments $E F$ and $P F$ are both tangent to the same circle, as is triangle $F G P$. This means that angle $E \hat{P} G$ is $90^\circ$.

Therefore, angle $D \hat{P} H$ is $15^\circ + 90^\circ + 15^\circ = 120^\circ$.

Angles $D \hat{Q} E$ and $D \hat{P} E$ a