A student asked what angles are constructible, considering only angles that measure a whole number of degrees (which I'll call integral angles). The answer is very simple: The only constructible integral angles are those which measure 3n degrees, where n is any natural number.
To see this, first note that angles of measure 60 and 72 (vertex angles in a regular pentagon) are constructible. Since angles may be bisected repeatedly by construction, this means that angles of measure 15 = 60/4 and 9 = 72/8 degrees may be constructed. Since 3 = 9*2 – 15, an angle of 3 degrees may be constructed by constructing two adjacent 9 degree angles, and then a 15 degree angle inside the resulting 18 degree angle. By adding n 3-degree angles together, a 3n degree angle can be constructed.
The proof that only these integral angles can be constructed seems to require the more advanced result that there is at least one integral angle that is not constructible. For example, an angle of 20 degrees is non-constructible. (This is the standard proof, using Galois theory, that demonstrates the impossibility of angle trisection.) If there were an integral angle of measure n that is not a multiple of 3 that is constructible, then since (n,3) = 1, there is a linear combination with integral coefficients of n and 3 that gives 1, and so an angle of 1 degree would be constructible, and hence an angle of 20 degrees would be constructible, which it is not..