= Solution
Use the course convention that a <root system> is finite and is a <reduced root system> satisfying the <crystallographic root system> condition. These hypotheses are essential for the angle restriction. For <roots> $r,s$, the two <Cartan integers>
$$
a=\frac{2(r,s)}{(s,s)},\qquad b=\frac{2(r,s)}{(r,r)}
$$
satisfy $ab=4\cos^2\theta$. If $r,s$ are not parallel, the strict <Cauchy-Schwarz inequality> and integrality give $ab\in\{0,1,2,3\}$. These values give angles $\pi/2$, $\pi/3$ or $2\pi/3$, $\pi/4$ or $3\pi/4$, and $\pi/6$ or $5\pi/6$, respectively. If the roots are parallel, reducedness implies $s=\pm r$, giving $0$ or $\pi$. Therefore
$$
\boxed{\theta\in\{0,\pi/6,\pi/4,\pi/3,\pi/2,2\pi/3,3\pi/4,5\pi/6,\pi\}.}
$$
Here are examples accounting for every angle. In the <A1 root system>, the pairs $(r,r)$ and $(r,-r)$ give $0,\pi$. Two roots from different summands of $A_1\oplus A_1$ give $\pi/2$. The six unit vectors at angles $k\pi/3$ form an <A2 root system>: taking one root at angle zero and another at $\pi/3$ or $2\pi/3$ gives those two angles. In the <B2 root system> $\{\pm e_1,\pm e_2,\pm e_1\pm e_2\}$, pair $e_1$ with $e_1+e_2$ or $-e_1+e_2$ to obtain $\pi/4,3\pi/4$. In a <G2 root system>, take short unit roots at angles $k\pi/3$ and long roots of length $\sqrt3$ at angles $\pi/6+k\pi/3$. A short root at zero paired with long roots at $\pi/6,5\pi/6$ gives the remaining angles.
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