My friend and I were building paper cones made from circles in which we cut out sectors of the circle and joined the two sides together. We were wondering about the relationship between the angle of the sector of the circle we cut out and the inclined angle of the cone the circle eventually made. While doing the math, when cutting the circle in half, a 180 degree sector, we eventually figured out the slant angle was somewhere around 56.944 degrees using cosine and the radius and slant height of the cone. I happened to notice that number was also very close to 180/pi, or 57.296 degrees. Is there any relationship between these two? Or is it coincidence that they're close in number? Could anyone explain to me what the relationship between these two degrees are?


The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $\theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-\theta R$. Since the circumference of a circle is $2\pi$ times its radius, we then have for the radius of the cone’s base $$r = \left(1-\frac\theta{2\pi}\right)R.$$

The slant height of the cone is the original circle’s radius $R$, so $$\sin \alpha=\frac r R=1-\frac\theta{2\pi},$$ where $\alpha$ is the half-angle at the cone’s vertex.

If you cut out half of the circle, you’ll have $\sin\alpha=\frac12$, or $\alpha=\frac\pi6$, which is $30$ degrees.

  • $\begingroup$ $r,\theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$. $\endgroup$ – Narasimham Jan 12 '18 at 16:52
  • $\begingroup$ @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $\theta$ of the deleted sector, whereas you used the angle of the remaining sector $\gamma = 2\pi-\theta$. $\endgroup$ – amd Jan 12 '18 at 19:04
  • $\begingroup$ Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting. $\endgroup$ – Narasimham Jan 12 '18 at 21:45
  • $\begingroup$ @Narasimham I’ll make some edits to make this clearer. Thanks! $\endgroup$ – amd Jan 12 '18 at 22:11

The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:

$$ \sin \alpha = \frac{r}{L} = \dfrac{\gamma}{2 \pi}$$

$$ \dfrac{\gamma}{\sin \alpha} = 2 \pi $$

is the relation you are asking for. Accordingly the fully developed angle $\gamma $ at center of sector is

$$ 2 \pi \sin \alpha $$

In your case semi-vertex angle is $ \sin^{-1}\dfrac12 = \sin^{-1} \dfrac{\pi}{2 \pi} = 30^{\circ}$ from straight angle $\pi$ to full circle $2 \pi $ ratio $=\dfrac12$.

If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{\circ}$, not $180/\pi $ degrees or a radian.


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