Consider a triangle where the three angles must sum to a fixed total. I don't care if they sum to $\pi, \frac{\pi}{2}, 360, 180, 1$ or any other number.
If every angle is irrational and they add up to the total 'degrees' in a triangle, is it possible that the ratios of the angles are also irrational?
Put another way: are there $3$ irrational numbers that sum to $1$, such that the ratios of any pair of these numbers is also irrational?
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Take $A=\sqrt2, B=\sqrt3,C=\sqrt5$. Clearly for any real $k>0$, the three numbers $kA, kB,kC$ are not rational multiples of each other. Choose $k$ such that $kA+kB+kC=\pi$. Now a triangle whose angles (in radians) are $kA, kB, kC$ exists and provide a solution to the question.
EDIT (prompted by the comment of dxiv pointing out a lacuna).
As $\pi$ is transcendental, for our choice of $k$, we will have $kA, kB,kC$ irrational.
Consider the irrational numbers $a,b,c$.
Let $a = \frac{5-\sqrt{2}}{9}, b= \frac{4-\sqrt{2}}{9}, c = \frac{2\sqrt{2}}{9}$
Then $a+b+c = 1$
$$\frac{a}{b} = \frac{5-\sqrt{2}}{4-\sqrt{2}} = \frac{18+\sqrt{2}}{14}$$
$$\frac{a}{c} = \frac{ 5-\sqrt{2}}{2\sqrt{2}} = \frac{-2+5\sqrt{2}}{4}$$
$$\frac{b}{c} = \frac{4-\sqrt{2}}{2\sqrt{2}} = \frac{-1+2\sqrt{2}}{2}$$
Which are all irrational as required.