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July 28th, 2018, 01:50 AM   #1
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Making a cone out of a circle's sector

Hey.

Say that we have a circle sector with sectoral angle $\displaystyle \theta$ and radius $\displaystyle r$, then we join the radii of the sector together, forming the cone in a way that would make the cone's base circumference equal to the length of the sector's arc length. Slant height of the cone would be $\displaystyle r$. The question is, would the angle of the cone's vertex (if we take its face at some point which is a triangle) be the same as $\displaystyle \theta$?



credits to some guy from stack exchange ^

Last edited by skipjack; July 29th, 2018 at 12:38 AM.
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July 28th, 2018, 11:29 AM   #2
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Why are you showing the diagram on left?
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July 28th, 2018, 11:41 AM   #3
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Your notation is a bit confusing.

Let your sector have radius $\ell$ and angle $\theta$

The base circumference of the cone will be $\ell \theta$

and the base radius will be

$r = \dfrac{\ell \theta}{2\pi}$

The cone vertex angle will be

$\phi = \arcsin\left(\dfrac{r}{\ell}\right) = \arcsin\left(\dfrac{\theta}{2\pi}\right)$
Thanks from Denis and Integraluser
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July 29th, 2018, 07:44 AM   #4
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Aha. So it's a new angle at last? Alright thanks, this guy @ stack exchange confused me.

@Denis: the picture isn't mine, I got it from some question @ stack exchange. I think he did that to elaborate and relate things more (theta, the slant height and radius)
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