Which internal angles can a lattice polygon have?

Solution 1:

Any triangle with vertices in lattice points has tangents of the angles expressed by rational numbers. To do this, notice that you can split the angles of the triangles into two or more angles which can be expressed in right triangles using only the definition, and this brings rational tangents for the small angles. Then, using the formula $$ \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ you see that the tangent of the sum is also rational. Therefore the tangents of the angles of lattice polygons are rational numbers.

[edit] Ok. I forgot that the right angles have tangent $\infty$, so if the angles of a polygon can be angles which have rational tangents and right angles.

Here is one proof of mine that we can draw any angle with rational tangents using lattice points. My proof states that any triangle with rational tangents can be embedded in the lattice points after a certain dilatation.

Let $\tan A=\frac{a}{b}=\frac{abcd}{b^2cd},\ \tan B=\frac{c}{d}=\frac{abcd}{abd^2},\ a,b,c,d \in \mathbb{Z}^*,\ (a,b)=(c,d)=1$. Consider the orthogonal coordinate system $xOy$ with origin $D$, $Ox$ being the support of the segment $[DE]$ and points $E(b^2cd+abd^2,0),\ F(b^2cd,abcd)$. From the construction $DEF$ is similar to $ABC$ because $\tan A=\tan D,\ \tan B=\tan E$.