"Critical" Triangle Problem

[DanteKennedy]

Homework Statement

Let triangle ABC be a right-angled triangle with angle C = 90°. Let a, b, and c denote the lengths of the sides opposite to vertices A, B, and C, respectively, and let alpha and beta denote the measures of the interior angles at A and B.

Relevant Equations

A right triangle is called "critical" if it simultaneously satisfies the following conditions:
1. All side lengths a, b, and c are strictly positive rational numbers

2. The measures of the acute angles alpha and beta, when expressed in pure degrees, are rational numbers (ex. 90°, 68.4°, etc.)

If such a triangle exists, provide an example; if not, prove its non-existence.

No attempts to solve it by myself yet. It would be very nice if someone can help me

 

[.Scott]

This is a big clue.
Don't click this unless you're ready for a spoiler.

 

[DanteKennedy]

Thanks, so it seems like no such triangle exist

 

[PeroK]

Here's my thinking. We need an angle $\theta = q\pi$ (in radians), where $q$ and both $\cos \theta$ and $\sin \theta$ are rational. We can write this using Euler's identity:
$$e^{iq\pi} = e^{i\theta} = \cos \theta + i\sin \theta = a + ib \ \ (q, a,b \in \mathbb Q)$$Now, if $q = k/m$, then $2mq = 2k$ and:
$$(a + ib)^{2m} = e^{i2k\pi} = 1$$I don't know if that helps.

 

[PeroK]

I think you can finish it off by looking at the imaginary part and using the rational root theorem. This would show that either $a = 0$ or $b = 0$.

There's a key factor in this particular problem that means you don't need to use the full power of Niven's theorem. That is that all sides of the triangle have rational lengths.

Or, perhaps not?

 

[.Scott]

According to Wiki, the case against rational tangent of rational angle in degrees is provided in Nivens "Irrational Numbers" - a book that I do not have.
But that is the specific case required by the OP.