Geometrically Constructible Angles Angle 30 can be constructed, through drawing an equilateral triangle, constructing angle 120, bisecting it multiple times and getting angle 30. Is it possible to contruct 3 degrees using geometric theorems and how?
 A: A theoretical answer: Gauss discovered that the angle $$\frac{360^\circ}{n}$$ for a positive integer $n$ can be constructed with compass and straightedge if the prime factorization of $n$ is of the form $n = 2^k p_1 p_2\ldots p_t$ where $k$ is any nonnegative integer, and the $p_i$ are pairwise distinct Fermat primes.
In our case, that is enough since we have $3^\circ=\frac{360^\circ}{120}$ and $120=2^3\cdot 3\cdot 5$, and $3$ and $5$ are unequal Fermat primes.
Wantzel later supplied the full proof that these are the only $n$ for which $\frac{360^\circ}{n}$ can be constructed (Gauss–Wantzel theorem).
A: It is possible to construct a 30 degrees angle and a 45 degrees angle. At the same time, it is possible to construct a 72 degrees angle. Combining the 30 degrees and 45 degrees to form 75 degrees, then constructing the 72 degree angle allows you to form a 3 degrees angle.
A: 
Here is the construction:


*

*Construct a line $l$ and set two points $A$ and $B$.

*Construct a circle $\Gamma$ with the center $B$ and radius $AB$ which intersects $l$ at $C$.

*Construct a circle $\Pi$ with the center $C$ and radius $BC$ which intersects $l$ at $D$ and $\Gamma$ at $E$.

*Construct a circle $\Omega$ with the center $D$ and radius $DE$ which intersects $l$ at $F$. (not between $BC$)

*Construct a line $EF$ which intersects $\Gamma$ at $G$.

*construct a line $GC$ which intersects $\Omega$ at $H$. (not between $GC$)

*Construct a circle $\Sigma$ with the center $A$ and the radius $AH$ which intersects $EF$ at $I$. 

*construct a line $AI$ and the angle $\angle IAB=3^\circ$
A: Hint With a straightedge and compass, yes. A regular pentagon can be so constructed, and the central angle between two successive vertices is $\frac{360^\circ}{5} = 72^\circ$.
