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July 28th, 2018, 02:50 AM  #1 
Newbie Joined: Jun 2018 From: Jordan Posts: 19 Thanks: 0  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 01:38 AM. 
July 28th, 2018, 12:29 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,592 Thanks: 953 
Why are you showing the diagram on left?

July 28th, 2018, 12:41 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152 
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 $\phi = \arcsin\left(\dfrac{r}{\ell}\right) = \arcsin\left(\dfrac{\theta}{2\pi}\right)$ 
July 29th, 2018, 08:44 AM  #4 
Newbie Joined: Jun 2018 From: Jordan Posts: 19 Thanks: 0 
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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circle, cone, making, sector 
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