# Solving an irregular tetrahedron

• Oct 11th 2011, 12:43 PM
remettub
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).
• Oct 13th 2011, 11:36 PM
earboth
Re: Solving an irregular tetrahedron
Quote:

Originally Posted by remettub
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.