I bit late but i i think it is proven there is no solutions, except for the special case 0° and side lengths 1, 1 and 0. Let us consider the triangle with a²+b²=c² and a = c sin(pi q) where q is the angle as a fraction of half a circle. So you are looking for a solution where a, b, c are integer and q is rational. So we first need to find a rational value for q where sin(pi q) is rational. According to https://math.stackexchange.com/questions/87756/when-is-sinx-rational#87768 this happens only for the well known case of 30°, so q=1/6 and a/c=1/2. However, in this case b=c/2 × sqrt(3) which is irrational, so with this angle we can never create integer side length.
this post was submitted on 28 Apr 2024
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Do you mean rational in radians, or in degrees?
I was thinking of degrees, or fractions of a whole (2π radians), though either would be interesting. I would be quite surprised if any such triangles existed with rational angles in radians, given that π is irrational.
It looks like there aren't any, by Niven's Theorem https://www.proofwiki.org/wiki/Niven%27s_Theorem
Thanks!