3 dimensional triangle?

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

  1. suzuki

    Thread Starter Member

    Aug 10, 2011

    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
    Yep. And probably even if it has a 4th dimension. ;)
  3. 1chance


    Nov 26, 2011
    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
    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


    Mar 31, 2012
    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.