Prove that the angle $\theta=\arccos(-12/17)$ is constructible using ruler and compass. Could I just do this?
Proof: if we want to show $\arccos\left(\frac{-12}{15}\right)$ is constructible, can't I just say, take $x_0=\cos(\theta)=-\frac{12}{17}$ implies $17x_0+12=0$ which says that $f(x_0)=17x+0+12$ is a polynomial with root $x_0$. But $f(x)$ is irreducible by Eisenstein's criterion with $p=3$. So $f(x)$ is the minimal polynomial with $x_0$ as a root, thus $\left[\mathbb{Q}[x]:\mathbb{Q}\right]=1$ which is equal to $2^m$ for $m=0$.

The way I would do it is with a triangle whose three sides are $51$, $92$, and $133$. Then Law of Cosines.
