woodworker wanting to know formulas for odd shaped boxes I am a woodworker and want to build boxes that are not at right angles, a pyramid for example.  I am looking for the formulas that will allow me to relate the two miter cuts necessary (one that will create a trapezoid shape, for example, and the other that will allow the edges of the trapezoids to be glued [for a normal box it is 45, for a frame it is either 0 or 90] to the exterior angle of the box. 
 A: My linear algebra is a bit rusty, but I think it provides a means to solve this if you have Cartesian coordinates for all the verticies.
For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it, $\phi_{A,2}$, the internal angle of the face at $P_{A,2}$ is the arccosine of the cross product of the norms of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$
$$
\phi_{A,2}=\cos^{-1}\left(\left\|P_{A,1}-P_{A,2}\right\|\times\left\|P_{A,3}-P_{A,2}\right\|\right)
$$
For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it (any three points on the face if it has more than 3, as long as they are going clockwise), the normal vector $\vec{n}_A$ (the direction the face is facing) is the norm of the dot product of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$.
$$
\vec{n}_A=\left\|\left(P_{A,1}-P_{A,2}\right)\cdot\left(P_{A,3}-P_{A,2}\right)\right\|
$$
The angle $\theta_{AB}$ between faces $A$ and $B$ with normals $\vec{n}_A$ and $\vec{n}_B$ is the supplement of the angle between the normals, which is the the arccosine of their cross product.
$$
\theta_{AB}=\pi-\cos^{-1}\left(\vec{n}_A\times\vec{n}_B\right)
$$
The mitre angle for their shared edge (assuming equal thicknesses of material) is half that $$\frac{\theta_{AB}}{2}$$
