Inspired by a comment by @QC_QAOA on Question 3458920, which mentioned the ratio between the acute angles in a $3:4:5$ triangle, I would like to know if we can prove that this ratio is irrational.
The symmetries of the arguments of the functions suggested that this could be a fun and potentially tractable problem.
We can prove that the ratio is equal to $\displaystyle \frac{\tan^{-1}\frac{3}{4}}{\tan^{-1}\frac{4}{3}}=\frac{\log\left(\frac{7+24i}{25}\right)}{\log\left(\frac{-7+24i}{25}\right)}=0.69395$, with the complex-logarithm definition of $\tan^{-1}x$. So, it could also be expressed as a solution to $$25^{z}\left(7+24i\right)=25\left(-7+24i\right)^{z}$$
I had then tried to use $x^{p/q}=(x^p)^{1/q}$ but as the numbers are complex, it changed their values. The constant, nor its reciprocal, appear in the OEIS and I can't find it elsewhere.
We know that at least one of the angles (and quite likely both) is irrational, as their sum is $\frac\pi2$.
I believe I've found a proof. Label the two angles, $a=\tan^{-1}\frac34,b=\tan^{-1}\frac43$, and their ratio, $z=\frac{a}{b}$. Then, as $a+b=\frac{\pi}{2}$, we have $z=\frac{\pi}{2b}-1$.
By Niven's theorem, the only values of $0\leq x\leq \frac{\pi}{2}$ that are a rational multiple of $\pi$ and have a rational $\sin$ value are $0,\frac{\pi}{6},\frac{\pi}{2}$. But, as $\sin b = \frac45$, and $b$ is not equal to those values, it cannot be not a rational multiple of $\pi$. Therefore $z$ must be irrational.