# Can a non-equilateral triangle with integer sides and integer angles (in degrees) exist?

I'm wondering if a triangle with integer sides and integer angles in degrees exists, which is not equilateral. Could someone come up with a proof for why or why not?

Edit: Please do not answer with a triangle which has integer sides only. The triangle must have integer angles in degrees; this means they must actually be integers, not rounded off.

• How about an equilateral triangle with side length 1? – Frpzzd May 5 '17 at 23:27
• @FranklinP.Dyer Wow, how did I forget that. I'll edit the question so it reads differently. – u8y7541 May 5 '17 at 23:27
• Degrees are artificial. Rational multiples of $\pi$ or some other condition on the radian measures would make more sense. – Ethan Bolker May 5 '17 at 23:30
• @EthanBolker Well, I guess you could say it is a radian measure which satisfies the condition that when put into simplest form has a denominator which is a positive divisor of 180, but that's a bit wordy. – u8y7541 May 5 '17 at 23:32
• Euclidean geometry, right? – chepner May 6 '17 at 13:51

Now, when does an integer angle (in degrees) have a rational cosine? We can go one step further: when does a rational angle (in degrees) have a rational cosine? This question is answered with a clever proof on page 2 of When is the (co)sine of a rational angle equal to a rational number? by Jörg Jahnel. The answer is that the rational cosines of the rational angles are just $\pm1,\pm\frac12,0$. (This is known as Niven's Theorem and an alternate proof can be found at ProofWiki.)
Since cosines of $\pm1$ wouldn't correspond to the angle of a genuine triangle, the only potential options are $\cos\theta=0,\pm\frac12$ for angles of $\theta=60^\circ,90^\circ,120^\circ$. But if there is one angle of $90^\circ$ or $120^\circ$, there isn't enough room in $180^\circ$ for two more angles of at least $60^\circ$. And the only triangles whose angles are all $60^\circ$ are equilateral.