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Stereometry problem

  1. Oct 16, 2003 #1
    I have 3 planes intersecting and forming a kind of a pyramid. The angles between pyramid's edges from the top point are known. How can I find the angles between side triangles? (Angles between each two planes from their intersection lines)

    Note: Don't want formulaes here. Just understanding.
    Last edited: Oct 16, 2003
  2. jcsd
  3. Oct 16, 2003 #2


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    Understand this, you need formulas to find the angles.
  4. Oct 17, 2003 #3
    Well, I think I solved it. Without using any formulaes except for simple sin, cos and tan


    Angles given: ADB = 60, ACD = 45, BCD = 45.
    Need angle: GHF.
    Assume: DH = 10. So this solution is a special-case scenario and won't get
    full points on a test. However, the teacher demonstrated the "proper"
    solution today and that took about 30 minutes and the whole blackboard so
    this is considerably smaller.

    1. Solve triangle GHD.
    DH = 10
    GDH = 45
    GHD = 90 because GH is perpendicular to DB.
    Need only GH and that is 10 after a quick tan()

    2. Solve triangle FHD.
    DH = 10
    FDH = 30 because ED splits the corner at ADB.
    FHD = 90 because FH is perpendicular to DB.
    FH = 5 after a quick sin()

    3. Solve triangle GFH.
    GFH = 90
    FH = 5 (from p2)
    GH = 10 (from p1)
    GHF = 60 after a quick cos()

    Since there's bilateral symmetry in this "pyramid", the other such edge will
    also be 60 degrees on the other side. Haven't even tried to calculate the top
    one yet.

    The weird thing is that the book says that the answer is arc cos sqrt(3)/3.
    Which could in some weird ass way be 60 degrees. Or which could be wrong. Or
    I could be wrong. No idea.

    If you have time, see if my solution works or not :)
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