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The reason I ask: I was wondering if it was possible to find the angle of a rational triangle by only using the lengths of its sides and knowledge of $\pi$ (that is, no inverse trig functions).

So, that probably isn't possible unless every angle $\theta$ such that $\sin (\theta)$ is a rational number can be represented as $\theta = a \pi$ with $a$ is some rational number.

Clarification edit: A rational triangle is a triangle where every ratio of sides is rational.

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It isn't quite clear what you mean, but there is a special case of an equilateral triangle of side 1 which has sides of rational length and angles which are rational multiples of $\pi$. – Mark Bennet May 12 '13 at 6:08
Don’t you want to talk about cosines rather than sines? – Lubin May 12 '13 at 7:20
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I don't know what you mean by a "rational triangle", but it is extremely unusual for $\sin\theta$ to be rational for $\theta$ a rational multiple of $\pi$; indeed, the only examples are $a=0,1/6,1/2$ and the ones you can obtain from these by symmetries of the sine function.

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A rational triangle is where any ratio of sides of the triangle is rational. Where can I find a proof that those are the only points where it is a rational multiple of Pi? Also, not that it likely matters at this point, but I was also more generally interested in $a\pi + b$ where $a$ and $b$ are rational, so as not to exclude purely rational angles. – James Sydow May 12 '13 at 6:12
$\sin(p\pi/q)$ is an algebraic irrational (except for $q=1$, $q=2$, and $q=6$); the proofs involve Galois Theory and/or Algebraic Number Theory. $\sin(p/q)$ is transcendental (unless $p=0$); the proofs go through Diophantine Approximation. – Gerry Myerson May 12 '13 at 6:42
This result is sometimes called Niven's Theorem. – Erick Wong May 12 '13 at 7:47

Some observations before I try to answer your question. First, a “rational triangle” will always be similar to a triangle with all sides integers, so I’ll speak rather of “integral triangles”. Second, an angle that’s a rational multiple of $\pi$ is one whose degree measure is a rational number of degrees, and I’ll speak of these as “rational angles”. The issue for angles whose radian measure is a rational number involves mathematics much more advanced than I’ll bring up here, as @Gerry and @Erick have suggested.

Third, and most significantly, it looks like you didn’t bring into the question the Law of Cosines, according to which if the angles of a triangle are $\alpha_i$ for $i=1,2,3$, opposite the sides $s_i$ in the usual way, then $$ \cos\alpha_1=\frac{s_2^2+s_3^2-s_1^2}{2s_2s_3}\,. $$ This tells you that in an integral triangle, all the cosines of the angles are rational numbers. It’s not hard to show, conversely, by an analytic-geometry argument, that every angle $\theta$ with $0^\circ<\theta<180^\circ$ and $\cos\theta$ rational appears as vertex angle of an integral triangle. For instance, you see that the $(3,3,2)$ triangle and the $(9,10,11)$ triangle both have an angle with cosine equal to $1/3$.

Therefore, the question you asked in the title boils down to asking which angles $\theta$ with rational cosine are rational (in degrees, remember). The answer is that there are only three of them, $60^\circ$, $90^\circ$, and $120^\circ$. Now let’s see why.

First step is to note that an angle $\theta$ is rational if and only if there is a positive natural number $n$ with $n\theta$ a multiple of $\pi$ (radian measure now), in other words with $\cos(n\theta)=\pm1$.

Second step is to bring in the Čebyšev (Chebyshev) polynomials (of the first kind), $T_n(X)$, which tell you, among many other things, the relation between $\cos(n\theta)$ and $\cos\theta$. When you use the addition formulas for cosine, you see that not only is $\cos(2\theta)=2\cos^2\theta-1$, but that every $\cos(n\theta$) can be expressed as a polynomial in $\cos\theta$, and this polynomial is $T_n$. You can read about these polynomials in Wikipedia, but the important fact for us is that there’s a recursive definition of them, $T_0=1$, $T_1(X)=X$, and $T_{n+1}(X)=2XT_n(X)-T_{n-1}(X)$. You should check that they do what you want for the cosines, at least in the case of $n=2$ and $n=3$. From the recurrence relation, you see that, first, the Čebyšev polynomials have their coefficients in the integers $\mathbb Z$, and, second, that the leading (highest) coefficient of $T_n(X)$ is $2^{n-1}$, for $n>0$.

Now let’s look at an angle whose cosine $\lambda$ is rational, different from $0$ and $\pm1/2$. There are two cases, either the denominator of $\lambda$ is divisible by an odd prime $p$, or it’s divisible by $4$. Let’s look at the first case, that the “$p$-order” of $\lambda$, often written $v_p(\lambda)$, is negative. You get this “order” of $\lambda$ by writing it as a product of primes with exponents, in fact you can write $$ \lambda=\pm\prod_i p_i^{v_i}\,, $$ where for convenience here only I’ve written $v_i$ instead of $v_{p_i}$ As an example, $-105/52=-2^{-2}3^15^17^113^{-1}$. So $v_2(-105/52)=-2$, $v_{11}(-105/54)=0$. The important thing to remember is that when $\lambda$ and $\mu$ are rational numbers with different $p$-orders, then the $p$-order of $\lambda+\mu$ is the smaller of the two $p$-orders.

Now we’re ready to roll. You see that if $v_p(\lambda)=m<0$, then $v_p(T_n(\lambda))=nm<0$. In other words, if there’s $p$ in the denominator of the cosine of your angle $\theta$, there’s at least $p^n$ in the denominator of $\cos(n\theta)$. In particular, it can’t be $\pm1$, so that $\theta$ is not a rational number of degrees.

The other case is that $4$ divides the denominator of $\lambda=\cos\theta$, i.e. $v_2(\lambda)\le-2$. Then you see that the leading term of $T_n(X)$ has the most important contribution to the $2$-order of $T_n(\lambda)$, because $v_2(2^{n-1}\lambda^n)=n-1+nv_2(\lambda)$, which is still negative, because of the condition on $v_2(\lambda)$.

The upshot? If $\theta$ is an angle of an integer triangle, then $n\theta$ is not a multiple of $180^\circ$ no matter what positive $n$ you may take. This with the exception of the angles $60^\circ$, $90^\circ$, and $120^\circ$ only.

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