First off, it'd probably help a lot to refer to this article:

http://en.wikipedia.org/wiki/Tetrahedron

I want to know the angles of a regular tetrahedron that AREN'T 60 degrees. I'm stupid with math though -- REALLY stupid -- so the article confuses me on if and where they're answering my question here. What I could really use is someone taking one of the pictures and MS-painting to show which angles are where.

Also, in an equalateral triangle, if the sides are a whole number, can its height also be a whole number? Hell, is it possible to at least get the sides to be a small, SIMPLE number so that its height will be too? Again, I suck at math here.

Thanks for any help you all can give.

If you imagine looking at a FACE (ignore the third dimension) you're looking at an equilateral triangle, whose internal angles are all 60 degrees apiece. I'll look into the rest later (I'm christmas shopping)

Graph theory to the rescue! I've never proved anything about the angles between walks in a graph but you can prove that in a graph G, [exist]u,v-walk[/exist] | len(u) = len(v) [forall][isin='u,v']V(G)[/isin][/forall]

Might I remind you that I am stupid in math, JG. That looks like a bunch of gibberish to me.

And Martyn -- yeah, that's the part I know already.

I fixed my notation (what the frack was up with the slowness there?).

It reads There exists a u,v-walk such that the length of u = length of v forall u,v in the set of verticies...**** that should be edges. Ugh I gotta lay off the sauce.

That smells like an inductive proof to me.

Wait aren't ALL the angles in a regular tetrahedron exactly 60 degrees? It seems like it should be that way.

As for the height of an equilateral triangle, unfortunately it's not going to be a whole number because it will be sqrt(3)/2 * the length of the side.

The angle between a face (the base) and another face (standing) should be LESS than 60 degrees, though by how much I do not know. The angle between an edge running from base to peak and a (standing) face will likely be different than 60 degrees as well.

Maybe I'll need to MSPaint what I'm talking about...

Oh I see what you're saying; I thought you just wanted angles between the edges.

Look at the first diagram:

The angle between a face and an edge is theta1 (how do you get math symbols?)
The angle between two faces is theta2

Notice both angles are in the plane of the red triangle. That's a triangle of one triangle edge and two triangle heights. As kyle90 says, we know the height of an equilateral triangle to be sqrt(3)/2 of its edge length.

Diagram 2 is that red triangle. I've cut it in half again to get a right angled triangle. That means we can use simple trigonometry to get the following:

cos(theta1) = (e / 2) / (sqrt(3) * e / 2)
= 1 / sqrt(3)
theta1 ~= 54.74 degrees

likewise
sin(theta2 / 2) = 1 / sqrt(3)
theta2 ~= 70.53 degrees


That what you're after?

Another way to work out the angles would be to read the article under the Area and Volume heading :$

Like I said, the article I linked was confusing for stupid math people like me. I'm afraid yours was even moreso confusing, and the diagrams didn't make it very clear which angles we were talking about. So I'm left with the same position I was before. I'm not seeing how the angle between two faces can be GREATER than 60 degrees, since the slope of the faces are less than the slope of the edges, no?

Perhaps I should clarify again -- I am STUPID with math. Giraffe and JediGandalf have only confirmed that I don't understand equations. I understand VISUALS. The angles marked weren't very clear, and there were a LOT of symbols making things confusing. I will trust you that the math works out, I promise! You don't need to throw all the formulas out -- it just confuses me.

Here -- look at the little slice this diagram made below. That slice (being 2 dimensional) has 3 points. Point "a" has an angle between two faces (from the 3-D whole), and point "b" has an angle between a face and an edge (again, from the 3-D whole).

Now, all I want to know are angles "a" and "b" -- that's it. No proofs, no thetas or "e"s or whatever the graph theory....stuff that JG posted. Just fill in the blanks.

angle "a" = ?
angle "b" = ?

Let's see if this works...

Even though Giraffe did in fact give you the numbers you want, i'll solve it using his methods but i'll substitute numbers in so that there's less nasty algebra for you.

1. Assume all the edges are of length 2.
2. Your line bisecting the aqua triangle (on your diagram) splits the triangle into two right-angled trianges with sides of length 1, 2 and unknown.
3. The angles in this triangle are 90, 60 and 30.
4. We can use basic trig rules to work out that the length of the unknown side (the dashed line you've drawn between A and B) is 2sin60
5. Now we look at the triangle formed by your two dashed lines and the other known edge of length 2
6. We know the length of all the sides and we also know that B and the other unknown angle are the same (look at the diagram until you understand why)
7. We can again split the edge of length 2 to give us two more right-angled triangles which each contain half of A.
8. We know the internal angles of these triangles are A/2, B and 90
9. We also know the length of the hypotenuse is 2sin60.
10. We can calculate A/2 using sin(A/2)=1/2sin60
11. Once we know A it's a simple matter to calculate B using the fact that the internal angles of a triangle must equal 180.

so A = 70.52878 degrees and B = 54.73561 degrees

So I just realized that the length from point "a" to the unlabeled point is in fact different than the length of an edge, and that the proportion between that length and the height from base to peak is likely different than the proportion between the length of the base and the height of one of the 2-D faces. This means that angle "a" being ~70 degrees can make a little more sense to me now.

angle "a" ~ 70.52878 degrees
angle "b" ~ 54.73561 degrees

Thanks Detty for being my soundwall in this matter. So... I'm done now. Thanks.