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So far I know that it’s possible to draw angles which are multiples of 15° (ex. 15°, 30°, 45° etc.).

Could anybody please tell me if it's possible to draw other angles which are not multiples of 15° using only a compass and a ruler.

Russell
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    Yes, there is an interesting connection to Fermat numbers in the construction of angles using compass-and-straightedge. Gauss showed how to construct them, and the complete proof that his characterization was correct was later supplied by Wenzel. – hardmath Jul 31 '20 at 14:43
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    The simple answer is infinitely many because any angle you can construct can be bisected as many times as you want. You can also add any two angles that you can construct. – Ross Millikan Jul 31 '20 at 14:48
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    According to this wiki page, such an angle $\pi \cdot \frac pq$ can be drawn in this way if and only if the denominator of the fully reduced multiple is a power of $2$, or the product of a power of $2$ with the product of one or more distinct Fermat primes. – Ben Grossmann Jul 31 '20 at 14:56

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You can construct a regular $n$-gon with straightedge and compass if and only if $n$ is a power of $2$ times a product of Fermat primes - primes of the form $2^{2^j} +1$.

That tells you what fractional angles you can construct. For example, the $17$-gon is constructible, so you can construct an angle of $360/17$ degrees.

an4s
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Ethan Bolker
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Concerning the nice answer of Ethan, I must say that not every angle constructed with rule and compass must form a regular n-gon. This is only a sufficient condition.

Angles with rational trigonometric values

I can give you another sufficient condition to create angles with compass and ruler: You can draw any angle whose cosine, sine, or tangent is a rational number, i.e., $\dfrac{m}{n}\; |\; m,n\in\mathbb{N}$.

The proof is easy. I will give it only for the tangent, similarly can be done for cosine and sine. Just take a certain length as unit length and draw a right rectangle with the horizontal cathetus having length $n$ unit lengths and the vertical one having $m$ unit lengths.

For example, with the lengths $m=2$ for the vertical cathetus and $n=3$ for the horizontal you have an angle $\alpha\approx33.69^{\circ}$

An example for the cosine: If you take $m=11$ for the hypotenuse and $n=4$ for the horizontal cathetus, then you have an angle with cosine $\dfrac{4}{11}$, i.e., $\alpha\approx68.676^{\circ}$

The power of this method, is that you can compute the trigonometric function (cosine, sine, or tangent) of any angle $\alpha$ and then approximate it by a rational number. Once you have the rational number (if the numerator and denominator are not too big) you might construct with ruler and compass a quite good approximation of the angle.

About the regular n-gons

Just for the sake of completeness. The integer sequence of the numbers of edges of regular polygons constructible with ruler and compass is A003401

The first values of the sequence, i.e., the possible n-gons, are following:

$1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, 102, 120, \ldots$

Now, something very interesting happens with the possible n-gons (constructed with ruler and compass) with odd number of edges. The sequence is finite and quite short (A045544)

$3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295$

So, if there are no more Fermat primes (as it seems the case), the greatest number of sides of a ruler+compass odd n-gon is $4294967295$.

loved.by.Jesus
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