To start with we can assume each side of the hexagon to be 1 unit without affecting the size of the angle. Now work out the lengths of the lines AG and AH using Pythagoras (having first worked out the horizontal and vertical components of each). It transpires that both AG and AH are equal to sqrt(7)/2 (~1.323). This alone is not of much use in working out the angle in question, however if we now calculate the length of the dashed line GH we discover that this too has a length of sqrt(7)/2. Therefore AGH is an equilateral triangle, and the angle we seek is 60 degrees.

If you’re interested about how I created this puzzle, it is a special case of an interesting theorem I stumbled across: if you have three non-overlapping chords of length x, inside a circle of radius x, and then you connect the ends of those chords to form three new chords (which now don’t necessarily have to be x long), then the midpoints of those new chord form an equilateral triangle.

In my puzzle, the x length chords are AB, CD and FA. The fact that the circumscribed circle has radius x is implicit in the fact that those chords are part of a regular hexagon. Therefore the midpoints A, G and H form an equilateral triangle.