TITLE: The angle $168^\circ$ is constructible
QUESTION [10 upvotes]: Prove that the angle $\theta=168^\circ$ is constructible using a straightedge and a compass.

It is enough to show that the number $\cos\theta$ is constructible, and WolframAlpha gave  $\cos\theta=\frac18 (1-\sqrt5)-\frac14 \sqrt{{\frac32} (5+\sqrt5)}$. I think after that it is immediate because we first add $\sqrt5$ to $\mathbb{Q}$ and then add $\sqrt{{\frac32} (5+\sqrt5)}$, so both extensions are of degree $2$. But then how can we find that representation of $\cos\theta$? And is there a nicer way?
EDIT I found online a cool criterion: can we just state that the order of  $168^\circ$ is $15$, since $168*15$ is divisible by 360, and then since $15=3\times5$ is a product of Fermat primes we are done? And if so, Do we have to check all numbers $1,...,15$ to see $15$ is indeed the order?

REPLY [10 votes]: Yes it is constructible.
The canonical pentagon is indeed constructible, as mentioned by the OP, and so is its angle
$$
\varphi=\frac{3\pi}{5}=\frac{540^\circ}{5}=108^\circ
$$
Also $60^\circ$ is the angle of equilateral triangle, and so it is another constructible angle.
Hence $\vartheta=108^\circ+60^\circ=168^\circ$ is also constructible.