Three, (50 points) If the three sides of a triangle are all rational numbers, and one of its interior angles is also a rational number, find all possible values of this interior angle.

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In $\triangle A B C$, let $B C=a, C A=b, A B=$ $c(a, b, c \in \mathbf{Q})$, and its internal angle
$$
\angle A=\frac{m^{0}}{n_{1}}=\frac{m}{n} 180^{\circ}.
$$

Since a similar transformation of the triangle does not change its internal angles, we can assume $a, b, c \in \mathbf{Z}_{+}$.

According to the cosine rule, $2 \cos A=\frac{b^{2}+c^{2}-a^{2}}{b c}$ is a rational number.
Since $n \angle A=m 180^{\circ}$, then
$\cos n A=\cos \left(m 180^{\circ}\right)=(-1)^{m}$.
Let $x=2 \cos A$. Denote $f_{k}=2 \cos k A\left(k \in \mathbf{N}_{+}\right)$, then $f_{1}=x, f_{2}=x^{2}-2, \cdots \cdots$ and by the sum-to-product formulas of trigonometric functions, we have
$$
\cos (k+2) A+\cos k A=2 \cos A \cdot \cos (k+1) A,
$$

i.e., $f_{k+2}=x f_{k+1}-f_{k}, f_{1}=x, f_{2}=x^{2}-2$.
According to the recursive relation (1), for any $k(k \in \mathbf{N}_{+})$, $f_{k}$ is a polynomial of $x$ with integer coefficients and leading coefficient 1.
Let $f_{n}=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\left(a_{i} \in \mathbf{Z}\right)$.
Notice that $f_{n}=2(-1)^{n}$, so
$$
\begin{array}{l}
x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}-2(-1)^{m} \\
=0.
\end{array}
$$

Since $x=2 \cos A$ is a rational root of equation (2),
let $x=\frac{p}{q}$ (integers $p, q$ are coprime, $q>0$), substituting into equation (2) gives
$$
\sum_{i=0}^{n} a_{i}\left(\frac{p}{q}\right)^{i}-2(-1)^{m}=0\left(a_{n}=1\right),
$$

i.e.,
$$
\begin{array}{l}
p^{n}+a_{n-1} p^{n-1} q+\cdots+a_{1} p q^{n-1}+\left[a_{0}-2(-1)^{m}\right] q^{n} \\
=0.
\end{array}
$$

Therefore, $q \mid p^{n}$. Since $p, q$ are coprime, we have $q=1$.
Thus, $x$ is an integer.
Since $0<\angle A<180^{\circ}, -1<\cos A<1$, we have $-2<x<2$, so the integer $x$ can only take values $1, 0, -1$.
Thus, $\cos A=\frac{1}{2}, 0, -\frac{1}{2}$.
Therefore, $\angle A=60^{\circ}, 90^{\circ}, 120^{\circ}$.
On the other hand, we can find rational-sided triangles $\triangle A B C=(a, b, c)$ such that $\angle A \in\left\{60^{\circ}, 90^{\circ}, 120^{\circ}\right\}$. For example,
$\triangle A B C=(1,1,1)$, then $\angle A=60^{\circ}$;
$\triangle A B C=(5,4,3)$, then $\angle A=90^{\circ}$;
$\triangle A B C=(7,5,3)$, then $\angle A=120^{\circ}$.
Hence, the possible values for the internal angle are only $60^{\circ}, 90^{\circ}, 120^{\circ}$.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)