Exercise 10.1.6

Proof. If a $20^{\text{∘}}$ angle was constructible by straightedge and compass, then ${\zeta }_{18}$ would be constructible. The minimal polynomial of ${\zeta }_{18}$ is ${\Phi }_{18}(x)$.

By Exercise 9.1.12, ${\Phi }_{18}(x)={\Phi }_9(-x)$, and ${\Phi }_9(x)={\Phi }_3(x^3)=x^6+x^3+1$, so

Therefore $[\mathbb{Q}({\zeta }_{18}):\mathbb{Q}]=6$. If ${\zeta }_{18}$ was constructible, by Theorem 10.1.6, there exist subfields

with $[F_i:F_{i-1}]=2$ for $1\le i\le n$, and ${\zeta }_{18}\in F_n$. But then by the tower theorem, $6=[\mathbb{Q}({\zeta }_{18}):\mathbb{Q}]$ divides $[F_n:\mathbb{Q}]=2^n$, and $3\mid 2^{n-1}$: this is a contradiction. So ${\zeta }_{18}$ is not constructible, hence it is impossible to trisect a $60^{\text{∘}}$ angle by straightedge and compass. □