Angle of sector formed by cutting a cone

Question

A cone has a height of 10cm and circular base with radius 4.it is slit and cut open to form a sector.find the angle formed by the two radii.which is the simplest method to solve this?

Answer 1

Denote $a$ the edge of the cone, $h$ its height and $r$ its radius: by Pythagoras, $\; a=\sqrt{r^2+h^2}$.

The sector obtained has radius $a$. If its angle is $\theta$, its area is $\;\mathcal A=\frac12\theta a^2$.

On the other hand, this area is equal to the lateral area of the cone, which is equal to $\;\pi ra$. So $$\frac12\theta a^2=\pi ra\iff \theta=\frac{2\pi r}{a}.$$ (angles are in radians).

Answer 2

The angle (in radians) of a circular sector is a ratio of the arc length and the sector's radius, $l/R$. The arc lenght is a circumference of your cone's base $$l=2\pi r$$ and the radius is a slant height of the cone, i.e. the distance from the apex to the base edge: $$R=\sqrt{h^2+r^2}$$

So the answer is $$angle = \frac lR = \frac{2\pi r}{\sqrt{h^2+r^2}}=\frac{8\pi\,\text{cm}}{\sqrt{116}\,\text{cm}}\approx 2.33\,\text{rad}\approx 133^\circ$$

Answer 3

The trick is that you know the circumference of the base of the cone, $2\pi r$, which is the same as the arc length of the flattened sector, $\sqrt{r^2+h^2}\,\theta$.

$$\theta=\frac{2\pi}{\sqrt{1+\dfrac{h^2}{r^2}}}.$$