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Math Help - Triangles, Planes and 3D geometry

  1. #1
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    Triangles, Planes and 3D geometry

    The Question

    Given a rectangular tetrahedron (i.e. a pyramid where each face is an equilateral triangle), find the cosine of the angle between two of the faces.

    My Workings

    To be honest I have not really got any worth showing for this one, I seem to be struggling with the concept of questions that ask for the angles between 2 planes. I'm just not sure how to approach them.

    Apparently cos(60) is not the answer
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  2. #2
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    Quote Originally Posted by AliceFisher View Post
    The Question

    Given a rectangular tetrahedron (i.e. a pyramid where each face is an equilateral triangle), find the cosine of the angle between two of the faces.

    My Workings

    To be honest I have not really got any worth showing for this one, I seem to be struggling with the concept of questions that ask for the angles between 2 planes. I'm just not sure how to approach them.

    Apparently cos(60) is not the answer
    Hi Alice,

    imagine you can stand all 3 sides of the tetrahedron perpendicular to the base,
    as if the sides were only touching but can be pulled away from each other,
    so that all 3 sides are now standing at right-angles to the base.

    The vertical height "h" of any of these triangular faces in terms of the base edge length "x" is found using

    \frac{x}{2}=\frac{h}{tan60^o}

    x=\frac{2h}{tan60^o}=\frac{2h}{\sqrt{3}}

    h=\frac{\sqrt{3}}{2}x

    Now tilt the sides until they form the tetrahedron again,
    and referring to the attachment,
    if we have a second blue line going to the third apex of the base not shown in the attachment (think of the 3-D picture),
    then we have the triangle that forms the angle between the faces.

    It's dimensions are x,\ \frac{\sqrt{3}}{2}x,\ \frac{\sqrt{3}}{2}x

    Then the cosine rule finds the angle between the faces

    x^2=\left(\frac{\sqrt{3}}{2}\right)^2x^2(2)-2\left(\frac{\sqrt{3}}{2}x\frac{\sqrt{3}}{2}x\righ  t)cos\theta

    x^2=\frac{6}{4}x^2-\frac{3}{2}x^2cos\theta

    3x^2cos\theta=3x^2-2x^2=x^2

    3cos\theta=1

    \theta=cos^{-1}\left(\frac{1}{3}\right)=70.52878^o

    A 3-D picture would be more helpful here,
    hence an oblique 3-D view is also attached,
    where we need the angle between the blue lines identified by the arrows.
    The angle at the base and at the apex is 60 degrees,
    but the blue lines have a smaller length to the edges,
    hence the angle between the faces is greater than 60 degrees.

    Sorry there is a typo in the diagram.

    The length of the blue line is \frac{\sqrt{3}}{2}x
    Attached Thumbnails Attached Thumbnails Triangles, Planes and 3D geometry-regular-tetrahedron.jpg   Triangles, Planes and 3D geometry-oblique-3-d-view.jpg  
    Last edited by Archie Meade; April 26th 2010 at 03:12 AM. Reason: additional data
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  3. #3
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    Thanks for this I'm going to work through your answer now
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  4. #4
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    Okay its been a bit of work but I think I have finally got it! Thanks again

    The blue lines being longer in the middle really helped me understand why the angle would be different.
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  5. #5
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    Hi Alice,

    the attached drawing might be useful.
    It shows that the angle between the pairs of blue lines are the same.
    It's most convenient for us to calculate angle ABC, however.

    The blue lines are on the base and the backface of the tetrahedron.
    Attached Thumbnails Attached Thumbnails Triangles, Planes and 3D geometry-tetrahedron2.jpg  
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