cone algebra question

[reef surge]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[naive wasp]

Suppose the radius is r, the angle is θ and S(base)/S(total) = k.
First, find S(lateral) in terms of r and θ.

Oh, sorry. I thought the ratio was base to lateral, let me correct that.

Done.

Still, you need to find the lateral surface area first.

Note that the lateral surface area of a cone is S(lateral) = πrl, where l is the slant height.

So, you need to start by finding l in terms of r and θ.

No.

"Latus" means "side".

So, rougly speaking, it's the side area.

But it's usually called lateral.

As is anything that is on the side of something.

Try looking at the axial cross-section of the cone.

No, that doesn't make sense.

Try drawing the cross-section.

No.

No.
This is the axial cross-section. What is l?

Your l is almost correct, but not quite.

[naive wasp]

Note how we define the angle here.

Wait, hold on.

Ohh.

I see. You defined the angle differently.

Well, alright. Suppose θ is half of the needed angle. Then:
l = r/sin(θ)
S(lateral) = πrl = πr^2/sin(θ)
S(base) = πr^2
So:
S(base)/S(total) = S(base)/(S(base) + S(lateral)) = πr^2/(πr^2 + πr^2/sin(θ)) = 1/(1 + 1/sin(θ))
We know that it equals k. So:
1/(1 + 1/sin(θ)) = k
Now you can solve for θ. Note that 0 < θ < π.

You're welcome!