Rationality in Triangle 
How can I justify this answer? I think the answer is infinite, but cannot justify it///
 A: As Gerry Myerson pointed out in his comment:

If the ratios of the sides are all rational, there's a similar triangle where the sides themselves are all rational (integral, even). Then by the Law of Cosines, each angle has a rational cosine. If the angle ratios are all rational, then you can show the angles are all rational (measured in degrees, or in multiples of $\pi$ radians). But now you have a rational angle with a rational cosine, and standard irrationality results tell you exactly what the possibilities are.

In When is the (co)sine of a rational angle equal to a rational number? by Jörg Jahnel I read

Let $\alpha$ be a rational angle. Assume that $\cos\alpha$ is a rational number. Then $$\cos\alpha\in\left\{-1,-\tfrac12,0,\tfrac12,1\right\}.$$

So you have $\alpha\in\{-180°,-120°,-90°,-60°,0°,60°,90°,120°,180°\}$. Of these, a non-degenerate triangle can only have $\{60°,90°,120°\}$ as interior angles. And the only combination of three such angles that sums up to $180°$ is $(60°,60°,60°)$. So the only triangles which satisfy your requirements are the equilateral ones. Sure, there are infinitely many equilateral triangles, but this is nevertheless only a very special set of triangles.
