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How do we measure an angle of a cone? When making the cone flat we obtain a circle with a sector cut out of it. So when calculating the angle of a cone we would actually calculate the angle of a circle with a sector cut out?

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This usually refers to the opening angle of the cone, which is the angle made by its sides along a cross-section through the apex and center of its base. For a right cone, you’ll also see half of this angle—the angle between the cone’s axis and sides—used as well. For most purposes, these angles are much more useful to know than the angle of arc subtended by the base with the cone “rolled out.”

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the angle of the sector differs from the angle of the cone.

the sector's angle is computed using the formula $\theta=\frac{L}{R}$; where $L$ is the sector's arc length and $R$ is the sector's radius. now say $L=R\theta$.

when you make a cone using the sector, its arc length will become the cone's base perimeter. so you can write $2\pi r=R\theta$; where $r$ is the cone's base radius.

now you can find $r$ to be $\frac{R\theta}{2\pi}$.

the cone's lateral side is $R$ (the sector's radius). so we have a right triangle here which its smallest side is $\frac{R\theta}{2\pi}$, its vertical side is the cone's height and its hypotenuse is $R$. (you can find the height using Pythagorean rule but it's not our goal here).

let's call the top vertex of our triangle $\frac{\alpha}{2}$. ($\alpha$ is our cone's angle which we want to calculate).

$\sin(\frac{\alpha}{2})=\frac{\frac{R\theta}{2\pi}}{R}=\frac{\theta}{2\pi}$.

now find $\frac{\alpha}{2}$ using $\frac{\alpha}{2}=\arcsin(\frac{\theta}{2\pi})$.

finally, find $\alpha$ to be $2\arcsin(\frac{\theta}{2\pi})$.

note that $\theta$ is in radians here; if you want to put it in degrees, you should write $180$ instead of $\pi$.

sometimes you don't have the sector's angle. but you have $n=\frac{S_c}{S_s}$ or $n=\frac{P_c}{P_s}$ where $S_c$ and $S_s$ are circle's area and sector's area respectively. Also, $P_c$ and $P_s$ are circle's perimeter and sector's perimeter respectively.

in these cases, we have $\sin(\frac{\alpha}{2})=\frac{1}{n}$ and so on. (because $n=\frac{2\pi}{\theta}$).

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