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Angles on a Sphere

  1. Oct 2, 2005 #1
    When I was looking at a tetrahedral molecular structure, a question arose. In that molecule, there are 4 atoms placed around a sphere, where the bond angle between any two is 109.5o. How could I show that? It's not part of the homework, but I'm just wondering how to find the angle if you have 4 equidistant points on a sphere. Spherical trigonometry maybe?

    Thank you :smile:
  2. jcsd
  3. Oct 2, 2005 #2
    Hmm... Well I was unable to find anything related to it in spherical trig., but maybe I'm looking in the wrong place. It seems like an easy problem, but I'd like to see how it's done...
  4. Oct 2, 2005 #3


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    This is really a problem in Euclidean solid 3D geometry, as opposed to spherical trigonometry (the trigonometry of the nonEuclidean geometry on a sphere, a closed 2D surface).

    Realize that you are studying a regular tetraheron, which can be inscribed in a sphere as well as circumscribed about a smaller sphere. Can you relate the edge-lengths of the tetrahedron with the radius of at least of these spheres? If you can, then you will end up with a problem in ordinary trigonometry.
  5. Oct 3, 2005 #4


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    To find the angle, assume a tetrahedron whose sides are 1 unit long. Basic geometry tells you the height of the tetrahedron is [itex]\sqrt {2/3}[/tex]. Since the center of mass is 1/4 the height above any face, we know the distance between the center of mass and any vertex is therefore

    [tex]r = \frac {3}{4} \sqrt \frac {2}{3}[/tex]

    Now use the law of cosines, [itex]2 r^2 (1 - \cos \theta) = 1[/itex] to obtain the angle as [itex]\cos \theta = -\frac {1}{3}[/itex]
  6. Oct 3, 2005 #5
    m = length of side
    n = height of individual side (triangle)
    h = height of tetrahedron

    This is what I came up with:


    [tex]h=\frac{\sqrt{3}}{2}\,n\implies h=\frac{3}{4}\,m[/tex]

    So my result is different than yours. Why is that? Also, how did you find the center of mass for the tetrahedron?

    Thanks for the help :smile:
  7. Oct 3, 2005 #6


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    In your notation, [itex]h^2 = \sqrt {n^2 - (n/3)^2}[/itex].

    For the center of mass, imagine 1 mass unit placed at each vertex. Consider 3 of them to be a base of the tetrahedron (say z = 0) which makes the average height of the masses 1/4.

    By the way, if [itex]\cos \theta = -1/3[/itex] then [itex]\theta = 109.47^o[/itex]. What did you end up with?
  8. Oct 3, 2005 #7
    I certainly understand your other steps (how you came up with 109.47o, I'm just trying to see where your first statement [itex]h^2 = \sqrt {n^2 - (n/3)^2}[/itex] came from... When I draw it out, it looks like the vector (with head at the top vertex and tail halfway between two vertices on the base) projects onto the base with a component of n/2. I know that the magnitude of that vector is n, so I solved for the height and then substituted everything back knowing what n is in terms of m. Am I wrong in doing this? Also, how did you find the height? Sorry to keep bothering you, I just haven't ever dealt with a tetrahedron before (don't ask why... they seemed to have skipped over it in multivar. calc. and linear algebra).
  9. Oct 3, 2005 #8


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    The vector you describe projects onto the base with a component of n/3 and not n/2. If you draw a line from the midpoint between the two vertices on the base and the opposite vertex on the base then the projection of that vector is 1/3 the length of that line. (The center of a triangle is 1/3 the height from the base of the triangle.)

    The vector itself is the hypotenuse of a right triangle with the projection of that vector onto the base being one of the sides so the height of the tetrahedron is the third side of that right triangle.
  10. Oct 3, 2005 #9
    Wow I can't believe I missed that (1/3 instead of 1/2). Thanks a lot for the help Tide, I appreciate it.
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