There are ways to draw $180^\circ, 90^\circ, 45^\circ, 30^\circ, 60^\circ, \dots$ angles.

But is there a way to draw a $1^\circ$ angle?

In other words how to divide a circle into $360$ equal parts?

  • $\begingroup$ I am not sure, but I would guess it's probably not possible... $\endgroup$
    – 5xum
    Jun 2, 2016 at 9:15
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    $\begingroup$ Different surface question, but same material as Can a circle's circumference be divided into arbitrary number of equal parts using straight edge and compass only? $\endgroup$
    – AakashM
    Jun 2, 2016 at 9:22
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    $\begingroup$ Perhaps less satisfyingly then @MvG 's answer (indeed the proof is probably the same). It is generally known that you can't trisect all angles using ruler and compass. In particular you can't trisect $60^\circ$, and so, can't construct $20^\circ$. But if you could construct $1^\circ$ then just doing it 20 times, we have a contradiction. $\endgroup$
    – James
    Jun 2, 2016 at 19:55

2 Answers 2


Short answer: No.

Proof by contradiction

Suppose you could construct $1°$ in the unit circle. Then you could simply add that angle to itself a number of times to end up with $40°$. You could draw a line through the resulting point perpendicular to one of the legs of the angle. That line would have a distance from the origin which is

$$\cos(40°)=\sqrt[3]{\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}} + \frac{1}{4 \, \sqrt[3]{\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}}}\approx0.766$$

Notice the cubic roots there? That's not something you can construct in general. So this is already a strong indication that the number $\cos(40°)$ isn't constructible, which in turn means your $1°$ can't be constructible either.

But perhaps you think I just chose a stupid representation of that number. So I have to rule out that there could be an easier solution. Consider the irreducible polynomial $X^3-3X+1$. The number $x_1=2\cdot\cos(40°)$ is a root of that polynomial (the others are $x_2\approx0.174$ and $x_3\approx-0.940$). But the Galois group of the field extension $\mathbb Q(x_1)/\mathbb Q$ has order $3$. So you can't solve it using square roots, you absolutely need cubic roots.

Construction can only provide square roots. If you were able to construct $1°$, you could go on and construct $x_1$ as outlined. Since $x_1$ isn't constructible, $1°$ can't have been constructible either.

So an angle of $\bf{1°}$ cannot be constructed.

Note that when I wrote about the constructibility of $1°$ above, I meant as an angle, not a number. The length $1°=\pi/180$ is even more obviously not constructible, since it's not even algebraic but transcendental. But that's not what you were asking, and not what I was referring to.

(I'm not too experienced with Galois theory myself. Perhaps there are better ways to put the core of my argument into terms more readily understandable by others. So feel free to edit my post if you understand what I tried to say and know a better way of saying it.)

How I found this

I arrived at this answer with some educated trial and error, and the help of Sage as my computer algebra system. I knew about the concept of constructible numbers, so I knew that I'd have to investigate whether some length derived from that angle would be expressible with only square roots or not. I started by looking at the minimal polynomial of $1°$:

sage: deg = QQbar.zeta(360)
sage: deg.real().minpoly()
x^48 - 12*x^46 + 135/2*x^44 - 473/2*x^42 + 37023/64*x^40 - 33579/32*x^38 + 5995185/4096*x^36 - 6580089/4096*x^34 + 11535777/8192*x^32 - 509423/512*x^30 + 597177831/1048576*x^28 - 277344081/1048576*x^26 + 834808065/8388608*x^24 - 126664335/4194304*x^22 + 1969138215/268435456*x^20 - 378710755/268435456*x^18 + 908975295/4294967296*x^16 - 26063991/1073741824*x^14 + 142445393/68719476736*x^12 - 8735955/68719476736*x^10 + 2925693/549755813888*x^8 - 38697/274877906944*x^6 + 4527/2199023255552*x^4 - 27/2199023255552*x^2 + 1/281474976710656

That zeta(360) is $\exp(i2\pi/360)=\exp(i\cdot1°)$, i.e. a complex number $1°$ along the unit circle in the set $\bar{\mathbb Q}$ of algebraic numbers. So its real and imaginary parts correspond to the cosine and sine of that angle. The minimal polynomial of the real part is the smallest (monic) polynomial (i.e. lowest degree) in $\mathbb Q[X]$ which has this cosine as one of its roots.

In this case that's a fairly complicated polynomial. So I looked for ways to simplify the situation. For example, bisecting an angle is no problem, but doing so numerically likely involves additional square roots, increasing the degree and complexity of the minimal polynomial. So for example $8°$ is already easier.

sage: (deg^8).real().minpoly()
x^12 - 3*x^10 - 1/8*x^9 + 27/8*x^8 + 9/32*x^7 - 7/4*x^6 - 27/128*x^5 + 105/256*x^4 + 31/512*x^3 - 9/256*x^2 - 3/512*x + 1/4096

Powers of a complex number multiply the angle with the exponent, so deg^8 represents $8°$. This computation suppots my heuristic intuition: If I divide $360°$ into fewer parts I stand a better chance of simplifying the situation. So how many parts do I need? As you are probably aware, trisecting arbitrary angles using ruler and compass is impossible. On the other hand, trisecting the full circle into $120°$ parts is really easy. That made me look at the prime factor decomposition of $360=2^3\cdot3^2\cdot5$. Next candidates would be $2^3=9$ leading to $360°/9=40°$ or $5$ leading to $360°/5=72°$. But you may know that regular pentagons are constructible, or read that fact off the quadratic minimal polynomial:

sage: (deg^72).real().minpoly()
x^2 + 1/2*x - 1/4

So that led me to investigating the $40°$ more closely. The cosine (real part) looked easier than the sine (imaginary part), so I stayed with that.

sage: (deg^40).real().minpoly()
x^3 - 3/4*x + 1/8
sage: (deg^40).imag().minpoly()
x^6 - 3/2*x^4 + 9/16*x^2 - 3/64

The monic polynomial with integer coefficients used in my example would have been easy enough to find manually, but in this case (me not seeing the wood for the trees) came from the following computation:

sage: NumberField((deg^40).real().minpoly().numerator(), "y")
Number Field in y with defining polynomial 8*x^3 - 6*x + 1
sage: _.galois_closure("z", map=True)
(Number Field in z with defining polynomial x^3 - 3*x + 1, Ring morphism:
   From: Number Field in y with defining polynomial 8*x^3 - 6*x + 1
   To:   Number Field in z with defining polynomial x^3 - 3*x + 1
   Defn: y |--> 1/2*z)

So this is the source of that factor $2$ in front of the cosine. Then I verified the degree of the Galois group:

sage: _[0].defining_polynomial()
x^3 - 3*x + 1
sage: _.galois_group()
Transitive group number 1 of degree 3

I also had Sage help me with the radical expression above:

sage: (deg^40).real()
sage: latex(_.radical_expression())
{\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}} + \frac{1}{4 \, {\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}}}

$${\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}} + \frac{1}{4 \, {\left(\frac{1}{16} i \, \sqrt{3} - \frac{1}{16}\right)}^{\frac{1}{3}}}$$

Cross connection

I hadn't looked at this when I first wrote my answer, but the comment by @AakashM points out Can a circle's circumference be divided into arbitrary number of equal parts using straight edge and compass only? That question essentially asks whether one can construct $360°/k$ for arbitrary integers $k$, so you are dealing with the special case $k=360$. The accepted answer there points out that $k=2^\alpha\cdot p_1\cdots p_s$ works if all the $p_s$ are distinct Fermat primes. For $k=360=2^3\cdot3^2\cdot5$ all the prime factors are indeed Fermat primes, but the exponent at $3^2$ is a problem, making this non-constructible. In fact $k=2^3\cdot3\cdot5=120$ leads to $3°$ as the smallest integer number of degrees which can be constructed.

Also see the Wikipedia or MathWorld page on constructible regular polygons.

Which tools

All the above refers to the canonical interpretation of ruler and compass, where the ruler may only be used to connect two points, or to extend a segment. The answer by Viking uses a more physical interpretation of the ruler, and also allows for incremental adjustments (sliding the ruler). These increased capabilities make it possible to construct otherwise unconstructible numbers, but are at odds with the typical understanding of “straight-edge and compass” in the mathematical community.

  • 1
    $\begingroup$ So 3 degrees would be the smallest? $\endgroup$ Jul 3, 2018 at 20:06
  • $\begingroup$ Among angles with integral numbers of degrees, yes. $\endgroup$ Apr 16, 2021 at 21:02

Here is how you construct 1 degree with ruler and compass(used a 10cm long ruler here, but any would work):

  1. Construct a 54 degree angle (not shown for clarity. This is just 45+72/8 degrees, and is trivially constructed.

  2. Construct a 55 degree angle:

    • First construct a circle with a radius 10, and call the horisontal diameter x-axis.

    • Now construct a 15 degree angle, BAC

    • Place the ruler through C and adjust it so one end touches the x-axis and the other end the circle. Mark the point where it touches the circle B'

    • Now the angle BAB' is 55 degrees.

Superimpose over the 54 degree angle, and subtract the difference, 1 degree. Now you can construct any whole degree by addition.


You can easily see quickly (geogebra or with calculator) that this is correct to 20 digits. I leave the proof of this to you.

Hint: Note that both AB', DB' and AC is 10

constructing 55 degrees

  • 3
    $\begingroup$ While this is true, usually "ruler and compass" means a straight edge and a compass; you aren't allowed to measure distance. $\endgroup$
    – Vedvart1
    Feb 8, 2017 at 2:41
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    $\begingroup$ Your construction of the 55 degree angle involves measuring length, which we usually don't allow in these types of problems. "Ruler and compass" usually refers to a magical infinitely long compass with no measurement capabilities, in other words just a straight edge. If we allowed measuring distance, then we could easily create a 1 degree angle by how you just described. $\endgroup$
    – Vedvart1
    Apr 9, 2017 at 2:05
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    $\begingroup$ Your construction of the 55 degree angle needs distance measurement in step 3. By aligning the "ends of said ruler" you are using its length (10 cm). Without that, there are infinitely many ways to draw BD' through C, the circle, and the x axis. $\endgroup$
    – Vedvart1
    Apr 10, 2017 at 5:02
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    $\begingroup$ There is of course an issue with the proof, but it is a little more subtle. The compass is to translate distances, and all that is done in last step is to translate the radius as indicated in step 3. No need for a measurement. $\endgroup$
    – Viking
    Apr 10, 2017 at 5:30
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    $\begingroup$ The construction is correct. In the last step, you slide the compass/ruler along until it intersect. Still only rules and compass, but it can be discussed is this is acceptable. This can be done in many ways. Another way is to construct a helper diagram on a separate paper, and overlay it. Only unmarked ruler and compass. Additional limitation is often placed on constructions to exclude the above use. $\endgroup$
    – Viking
    Apr 10, 2017 at 23:29

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