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Math Help - Solving an irregular tetrahedron

  1. #1
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    Solving an irregular tetrahedron

    If I know the lengths of the three edges of face A on an irregular tetrahedron, and I know the three angles formed at the vertex opposite face A (vertex P), how can I determine the other edges and angles of the tetrahedron?

    In the image, the known elements are sides SQ, QR and RS, and angles SPQ, QPR, and RPS. The three angles on face A (SQR, SRQ, and QSR) can be easily found. What I'm after is one of the angles adjacent to face A (such as PQS) and/or the length of one of the missing edges (such as PQ).
    Attached Thumbnails Attached Thumbnails Solving an irregular tetrahedron-tetrahedron.jpg  
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  2. #2
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    Re: Solving an irregular tetrahedron

    Quote Originally Posted by remettub View Post
    If I know the lengths of the three edges of face A on an irregular tetrahedron, and I know the three angles formed at the vertex opposite face A (vertex P), how can I determine the other edges and angles of the tetrahedron?

    In the image, the known elements are sides SQ, QR and RS, and angles SPQ, QPR, and RPS. The three angles on face A (SQR, SRQ, and QSR) can be easily found. What I'm after is one of the angles adjacent to face A (such as PQS) and/or the length of one of the missing edges (such as PQ).
    1. I've modified your sketch a little bit (see attachment).

    2. Since you know already the lengthes of a, b, c and the values of the angles \alpha, \beta, \gamma you can set up a system of equations using Cosine rule:

    \left| \begin{array}{rcl}a^2&=&x^2+y^2-2xy\cdot \cos(\alpha)\\ b^2&=&y^2+z^2-2yz\cdot \cos(\beta)\\ c^2&=&x^2+z^2-2xz\cdot \cos(\gamma) \end{array} \right.

    3. With the lengthes of x, y, z you are able to determine the missing angles using Cosine or Sine rule.
    Attached Thumbnails Attached Thumbnails Solving an irregular tetrahedron-irreg_pyramide.png  
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