Use the fact that $\alpha=$cos$(2\pi/5)$ satisfies the equation $x^2+x-1=0$ to conclude that the regular $5$-gon is constructible by straightedge and compass.
So, the polynomial $x^2+x-1$ is irreducible in $\mathbb{Q}$ by rational root test. So, $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$.
Then is there any proposition from that I can conclude $\alpha$ is constructible? I found a proposition which says if $\alpha\in \mathbb{R}$ is constructible then for a field $F\subset \mathbb{R}$, $[F(\alpha):F]=2^k$ for some integer $k\geq 0 $. So as you can see I cannot use it here, since I need other way around. Can anyone please help?
Using basic geometric constructions we can prove that the sum, difference, product, quotient of constructible numbers is constructible, and that if $a\ge 0$ is constructible then $\sqrt{a}$ is constructible.
So the roots of our quadratic are constructible. Irreducibility is not needed for the proof.