A truncated tetrahedron

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Question from Brad:

If you take a regular tetrahedron and truncate it so you keep the original three 60degree angles around one vertex but the legs originating from it become any three consecutive terms of the Fibonacci series the new base is one triangle of a pentagon.
I want to know the height of the new pyramid relative to its new base and the angles between the base and the other three sides.

Brad,

Some tools to use:

The "box product" or determinant formula for the volume of a tetrahedron implies that if the point E is on the edge AB of the tetrahedron ABCD, vol(AECD) = AE/AB vol(ABCD). (The same thing works for areas of triangles)
You can find the length of the new edges using the cosine law; with 60 degree angles this takes a particularly simple form.

Good Hunting!
RD

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