# 20 degrees is not constructible.
We show quickly that $20\degree$ is not a constructible angle.
If one can construct $20\degree$, then with the length 1, we can construct $\cos 20\degree$.
However note the triple angle formula, $$
\cos 3x = 4 \cos^{3} x - 3 \cos x.
$$If $3x = 60\degree$, $x = 20\degree$, then $u = \cos 20\degree$ is a solution to the polynomial equation $$
\frac{1}{2} = 4 u^{3} - 3u
$$or equivalently a root of $$
8 u^{3} - 6 u - 1 = 0
$$Note this polynomial is irreducible over $\mathbf{Q}$. Since if a cubic polynomial is reducible over $\mathbf{Q}$, then it must have a rational root. But by rational root test the only possible candidate of rational roots are $\pm 1 , \pm 1 / 2, \pm 1/ 4, \pm 1 / 8$, to which none of them are roots.
However, recall that every constructible number must be a root of an irreducible polynomial of some degree $2^{m}$, some $m$.
Whence $20 \degree$ is not constructible.