We want to show, among other things, that if the number $\sin\alpha$ is a "constructible number," then it is possible to construct by ruler and compass an angle of measure $\alpha$. Note that "constructible" and "construct" have different meanings. The term "constructible" is fundamentally algebraic, while "construct" is purely geometric. This link between algebra and geometry is what makes it possible to prove, for example, that the $60^\circ$ angle is not ruler and compass trisectable.
We start with the assumption that we are given two points in the plane, distance $1$ apart. We need to prove the following lemma, which is not hard to prove, but requires a couple of pages of geometric work.
Lemma: Suppose that we can construct with ruler and compass two points that are distance $x$ apart, and also two points that are distance $y$ apart. Then we can construct two points that are (i) distance $x+y$ apart; (ii) distance $|x-y|$ apart; (iii) distance $xy$ apart; (iv) distance $\frac{x}{y}$ apart (if $y\ne 0$); (v) distance $\sqrt{xy}$ apart.
Now suppose that $\sin\alpha$ is a constructible number, where we assume $0\lt\sin\alpha\lt 1$. We want to construct the angle $\alpha$ between $0$ and $\frac{\pi}{2}$.
By the above lemma we can construct by ruler and compass points $P$ and $Q$ such that the distance from $P$ to $Q$ is $\sin\alpha$. Draw a circle with centre $P$ and radius $1$. Draw the line $\ell$ through $Q$ perpendicular to $Q$, and suppose the line $\ell$ meets the circle at point $R$. Then $\angle RPQ=\alpha$.
For the specific question you asked, which is easier, we need to work in the other direction. Suppose that we can construct, by ruler and compass, an angle $\alpha$, that is, a point $P$ and two rays through $P$ such that the angle between the rays is $\alpha$. Then (if $0\lt \alpha\lt \frac{\pi}{2}$) we can draw a right triangle $PQR$ which has $\alpha$ as one of its angles. Then $\sin\alpha=\frac{QR}{PR}$. Now we need to show that $QR$ and $PR$ are constructible numbers. For that we show that any segment $XY$ constructible by ruler and compass starting from a segment of length $1$ has length a constructible number. To do that we do an induction on the length of the construction. There are two facts of analytic geometry required. (a) Let $\ell_1$ be a line that passes through two points whose coordinates are constructible numbers, and let $\ell_2$ also be such a line. Then the coordinates of the intersection point of $\ell_1$ and $\ell_2$ are constructible numbers. (b) Let $\ell$ be a line with equation whose coefficients are constructible numbers, and $C$ a circle whose equation has constructible coefficients, then the points of intersection of $\ell$ and $C$ have constructible coordinates.
Best Answer
Starting with 2 points in the plane separated by a distance $1$, and if $A$ and $B$ are constructible lengths then so are $\sqrt A,A+B,|A-B|,AB,$ and (if $ B\ne 0$ ) so is $A/B$, by some fairly simple geometric constructions. And every positive integer is a constructible length. Using all this and the quadratic formula,if $A,B,C$ are constructible lengths with $A\ne 0$, and the real number $x$ satisfies $ Ax^2+B +C=0$ then $x$ is constructible. Hence $D=\alpha /2=\cos (2\pi /5)$ is. So take points $p,q$ a distance $D$ apart.Draw circle $\Sigma$ centered at $p$, with radius $1$, draw line $l$ thru $q$ , perpendicular to the line thru $p,q$. Then $\Sigma$ meets the line thru $p,q$ at a point $r$ with $q$ between $p$ and $r$ , and $\Sigma$ meets $l$ at points $s_1$ and $s_2$ . The triangle $p ,r, s_1$ is one fifth of a regular pentagon centered at $p$.