From a circular piece of paper with radius $BC$, Jeff removes the unshaded sector shown. Using the larger shaded sector, he joins edge $BC$ to edge $BA$ (without overlap) to form a cone of radius 12 centimeters and of volume $432\pi$ cubic centimeters. What is the number of degrees in the measure of angle $ABC$ of the sector that is not used?
I'm no expert in Geometry, but will give this a go:
First Method:
Knowing the volume of the cone and its radius, we can calculate:
Height = 9 cm. And from this we can calculate the original radius before the cut: 12 =sqrt[r^2 - 9^2], solve for r
r = 15 cm
Circumference of the original paper - Circumference of the cone base =Length of arc= 2*15pi - 2*12pi =6pi
Arc length =[x(in degrees) x pi/180] x r
6pi =[x * pi/180] * 12, solve for x
x = 90 degrees.
Second Method:
Knowing the volume of the cone and its radius, we can calculate:
Lateral surface area =180pi
Base of cone SA =144pi
180pi - 144pi = 36pi - this should be the area of the missing piece.
Area of a sector =[x(in degrees) * pi/360] * r^2
36pi =[x *pi/360] * 12^2, solve for x
x = 90 degrees.
There you have it. At least I got the same answer in both methods!!!!.
