Angle of sector formed by cutting a cone 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?
 A: 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).
A: 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
$$angle = \frac lR = \frac{2\pi r}{\sqrt{h^2+r^2}}=\frac{8\pi\,\text{cm}}{\sqrt{116}\,\text{cm}}\approx 2.33\,\text{rad}\approx 133^\circ$$
A: 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, $\sqrt{r^2+h^2}\,\theta$.
$$\theta=\frac{2\pi}{\sqrt{1+\dfrac{h^2}{r^2}}}.$$

