A circular sector of central angle theta and 9 cm radius can be fashioned into a paper cone with volume 100 cm^3. Find all possible values of theta.

Answer 1

arclength = rØ, where Ø is the central angle in radians

so the arclength of the sector is 9Ø
but this becomes the circumference of the cone
2πr = 9Ø
or
Ø = 2πr/9
r = 9Ø/(2π)

we need the height of the cone:
r^2 + h^2 = 9^2
r^2 = 81 – h^2
or
h^2 = 81 - (9Ø)^2/(4π^2)
h = √(81 - (9Ø)^2/(4π^2)) . arrrgjjj!

Now volume of cone
= (1/3)πr^2 h
= 100

πr^2 h = 300
π(81 – h^2)h = 300
πh^3 – 81π h + 300 = 0

using my trusty cubic equation solver at
http://www.1728.com/cubic.htm
I got
h = 8.34
h = a negative and
h = 1.2

when h = 8.34
then r^2 = 81 – 8.34^2
r = 3.38

so Ø = 2πr/9 = 2π(3.38)/9 = 2.36 radians ,(appr 135.2°)

when h = 1.2
r^2 = 81 – 1.2^2
r = 8.92
Ø = 6.22 , which would still be a possible cone , (appr. 356.4°)

This was fun !!!!