# 3 dimensional triangle?

Discussion in 'Math' started by suzuki, Apr 30, 2012.

1. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
Hi,

In 2D, we can express the sides of a right angle triangle with sines and cosines. Is there a way to extend this idea to 3D? I.E. i have a right angle triangle with sides A and B, with magnitude of $\sqrt(A^2+B^2)$ and angle $\theta$.

Is there a similar analysis that can be applied to something that has a 3rd dimension? i.e. with magnitude of $\sqrt(A^2+B^2+C^2)$

2. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
Yep. And probably even if it has a 4th dimension.

3. ### 1chance Member

Nov 26, 2011
42
184
Triangles are by definition 2D. You are wanting to deal with a pyramid if you use 3D. Could you draw what you are trying to describe?

4. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
well essentially i have something with magnitude $\sqrt(A^2+B^2+C^2$. I want do be able to break this down into terms of sine and cosine, which usually only applies to 2d right angle triangles.

For example, for $\sqrt(A^2+B^2)$ I can easily form a triangle with hypotenuse of that $\sqrt(A^2+B^2)$ and sides with length A and B. Then i can use sin and cosine to describe $A/(\sqrt(A^2+B^2))$ or $B/\sqrt(A^2+B^2)$ quite easily. However, I am wondering if there is a similar approach or method that can be applied to geometries with magnitudes of $\sqrt(A^2+B^2+C^2)$

hope that makes my question more clear

5. ### WBahn Moderator

Mar 31, 2012
17,445
4,698
If you have a rectangular box with sides of length A, B, and C, then the length of the diagonal is $\sqrt{A^2+B^2+C^2}$. You should try to derive this for yourself. Start with any two sides (which line in a plane) and find the hypotenuse. This is the diagonal of that side. Now use that hypotenuse and the thrid side (which again define a plane) and find the hypotenuse of that. This IS the diagonal of the rectangular solid.