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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?

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    $\begingroup$ You know(?) that you can construct a regular hexagon ($60°$), and a regular pentagon ($72°$). Then you can also construct $12° = 72° - 60°$. Constructing $12°$ is clearly equivalent to constructing $168° = 180° - 12°$. $\endgroup$ – Daniel Fischer Feb 9 '14 at 21:51
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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.

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The 5gon and 6gon are constructible, hence so are angles $72^\circ$ and $60^\circ$ Note that $168^\circ=14\cdot(72^\circ-60^\circ)$.

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    $\begingroup$ It is enough to know how to construct the pentagon. the hexagon is not needed. $\endgroup$ – user127249 Feb 9 '14 at 21:55
  • $\begingroup$ Why $72$ degrees? $\endgroup$ – Emolga Feb 9 '14 at 22:05
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    $\begingroup$ @Leullame if you draw any diagonal in a pentagon you'll get $36$ and $72$ degrees :) $\endgroup$ – qwr Feb 9 '14 at 22:11

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