# Calc height of cone with only volume and angle

1. Mar 10, 2009

### rustys111

hi all,

Ive been sitting up so late trying to work something out.

If anyone could help that would be great.

How do i calculate the height of a cone if the internal angle of the cone at the top vertex is 60degrees and the total volume for the cone is 2.0m3?

this is just a example - if you could guide me in the correct path and give me a quick awser so i can check my workings that would be great.

Thanks all

2. Mar 10, 2009

### matt grime

What are the formulae for volume of a cone? (Forum rule: no help without you showing your working - you say you have some, so what have you done?)

3. Mar 10, 2009

### rustys111

No worries

Given V = 2 internal angle at vertex = 60

volume = 1/3*Pi*r^2*h

Draw cone cross section - dive into two triangles to make a right angled tri

Angles in tri = 90, 60 & 30

S = hyp of right angled tri

therefore
using Angle 60

r = Scos60
h = Ssin60

v = 1/3 * Pi*r^2*h
2 = 1/3 * Pi * (Scos60)^2 * (Ssin60)

(2*3)/pi = (Scos60)^2 * (Ssin60)

Sqr((2*3)/pi) = Scos60 * Sqr(S) * Sqr(Sin60)

(Sqr((2*3)/pi) / sqr(sin60) = Scos60 * Sqr(S)

((Sqr((2*3)/pi) / sqr(sin60)) /cos60 = S * Sqr(S)

(((Sqr((2*3)/pi) / sqr(sin60)) /cos60) ^2 = S^2 * S

(((Sqr((2*3)/pi) / sqr(sin60)) /cos60) ^2 = S^3

((((Sqr((2*3)/pi) / sqr(sin60)) /cos60) ^2) ^ 1/3 = S

solve for S = 2.0662216704972 (Accuracy required)

input back in

r = Scos60
h = Ssin60

Height (h)= 1.78940045526428

Check
Place back into equation for volume
V = 1/3*Pi*r^2 *h
V = 2

Am i correct?
Thats what i need to know

Last edited: Mar 10, 2009
4. Mar 11, 2009

### HallsofIvy

Staff Emeritus
You should state which angle is 60! Dividing the cross section into two right triangles, you have angles of 60 and 30 degrees. I normally think about the vertex angle but you are using a base angle- that confused me for a moment!

Another way to do this is to recognize that since 60 degrees gives an equilateral triangle, dividing into two triangles gives a right triangle with hypotenuse of length S and one leg of length S/2 (the radius of the cone). By the pythagorean theorem, $S^2= h^2+ S^2/4$ so $h^2= (3/4)S^2$ and $h= (\sqrt{3}/2)S$.

Since r= S/2 and $h= (\sqrt{3}/2)S$, dividing the first equation by the second, we have $r= h/\sqrt{3}$

As above $r= (1/\sqrt{3})h$.

$V= (1/3)\pi r^2 h= (1/3)\pi (1/3)h^3= \pi/9 h^3= 2$
$h^3= 18/\pi= 5.73$
[itex]h= 1.79[/tex]
Just what you have.

Last edited: Mar 11, 2009