Calc height of cone with only volume and angle

[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

 

[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?)

 

[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

Radius (r)= 1.03311083738979
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

Is there any other way about this??

 

[HallsofIvy]

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

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, [itex]S^2= h^2+ S^2/4[/itex] so [itex]h^2= (3/4)S^2[/itex] and [itex]h= (\sqrt{3}/2)S[/itex].

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

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

Radius (r)= 1.03311083738979
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

Is there any other way about this??

As above [itex]r= (1/\sqrt{3})h[/itex].

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