How can tan(30°) be irrational and at the same time be a ratio within a triangle?

Question

Dumb question... if tan(30°) is irrational (what I believe it is, it should be $\sqrt 3$), how can it be that, at the same time, is is describing the ratio of two sides within a triangle?

The triangle I have in mind looks like this (sorry for the cheap sketch): So, $a$ is the angle (in this case 30 degrees), and AFAIK it should be possible to describe that angle by the ratio of the opposite and adjacent sides, $A$ and $B$, so: $a = \frac{A}{B}$... and there certainly exists a real-world triangle for $a=30°$, where the other two angles are $90°$ and $60°$... and the two lengths, $A$ and $B$, are just some real values, so their ratio should be a rational number... or not? :-)

It seems I'm missing something here...

Answer 1

the two lengths, $A$ and $B$, are just some real values

That's correct, but you cannot assume that those values must be integers (or rationals).

so their ratio should be a rational number... or not?

Not so, a rational number is defined as the ratio of two integers, not of arbitrary reals. (Of course, every real can be written as the ratio of two reals e.g. $x = \dfrac{x}{1}$, so that doesn't define any additional or interesting property for such $x$.).

Incidentally, the fact that $\,\tan 30^\circ\,$ is irrational is (one) reason why an equilateral triangle cannot be drawn of graph paper (so that the coordinates of all vertices would be integers).

Answer 2

Given a $\Delta{ABC}$ with $\widehat{BAC}=90^\circ$, then $BC$ is the hypotenuse. Let $BC=a;CA=b;AB=c$ then $\tan{ABC}=\dfrac{b}{c}$. If $\widehat{ABC}=30^\circ$ then $\dfrac{b}{c}=\sqrt{3}$.

Obviously, a number can only be rational if it can be expressed in the form of $\dfrac{a}{b}$ with $\operatorname{GCD}(a,b)=1$ and $a,b\in\mathbb{Z}$ and the sides of the triangle do not have to be rational.