I am puzzled by this problem encountered. One angle of right triangle with sides $ 3^2 + 4^2 = 5^2$ is
$$ \theta = \arccos(\frac{4}{5}) $$
$$ \exp(i\theta) = \frac{4}{5} + \frac{3}{5} i$$
and
$$\theta = \dfrac{2\pi r}{t} $$
where $r$ and $t$ most likely integers.
Is there a way for finding $r$ and $t$?
On
If $\cos \theta$ and $\sin \theta$ are both rational then either $\tfrac{\theta}{2\pi}=\tfrac{n}{4}$ for some integer $n$ or $\tfrac{\theta}{2\pi}$ is transcendental.
$\tfrac{\theta}{2\pi}$ is either rational or irrational.
Suppose that $\tfrac{\theta}{2\pi}$ is rational. Then $\mathrm{e}^{\mathrm{i}\theta}$ is a root of unity, so its minimal polynomial is a cyclotomic polynomial. But $\mathrm{e}^{\mathrm{i}\theta}$ is also a Gaussian rational, so its minimal polynomial is of degree $1$ or $2$. The only cyclotomic polynomials of degree $1$ or $2$ are those corresponding to $1,2,3,4,6$, and of these only $1,2,4$ correspond to roots of unity where both the real and imaginary parts are rational. Hence $\tfrac{\theta}{2\pi}=\tfrac{n}{4}$ for some integer $n$.
Suppose $\tfrac{\theta}{2\pi}$ is irrational. If $\tfrac{\theta}{2\pi}$ were algebraic, then $\tfrac{2\theta}{\pi}$ would be irrational and algebraic. Since $\mathrm{e}^{\mathrm{i}\pi/2}$ is algebraic, $\mathrm{e}^{\mathrm{i}\theta}=(\mathrm{e}^{\mathrm{i}\pi/2})^{2\theta/\pi}$ would have to be transcendental, by the Gelfond–Schneider theorem. But by hypothesis $\mathrm{e}^{\mathrm{i}\theta}$ is algebraic, so $\tfrac{2\theta}{\pi}$ can't be algebraic. Therefore it must be transcendental, so $\tfrac{\theta}{2\pi}$ is also transcendental.
$$\theta = arccos(\frac{4}{5}) = (0.204832765...)\pi $$
I do not think it is a rational multiple of $\pi$