By "rational angles" I mean a rational degree measure (equivalently, angle a rational multiple of $\pi$). Obviously similar triangles should be counted once.
Off the top of my head we have: 30-60-90, equilateral triangles, 45-45-90, and 36-36-72.
The triangles would exist in a regular polygon of 4, 5, 6, sides. So there's also 36-72-72 and 108-36-36 in the pentagon, 30-30-120, 30-60-90 and 60-60-60 in the hexagon, and 45-45-90 in the square. So there's six of them.
The angle at the circumcentre of the triangle for each side, is twice the angle opposite angle. Since the angles make rational angles, the vertices must fall in the a polygon representing the least common denominator.
Since the chords of polygons are solutions to a polynomial of order ½ totient(2n), it's simply a matter of satisfying this for this=2, gives n=4, 5, 6 as the only solutions. 7 and 9 solve cubics, 8, 10, 12, 15 solve biquadratics, 11 solve quintics, 13, 14, 18, 21 solve hexics, and so forth.