Finding a particular angle withion Johnson Solid J2

  • Context:
  • Thread starter Thread starter GreyArea
  • Start date Start date
  • Tags Tags
    Angle Solid
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
2 replies · 3K views
GreyArea
Messages
3
Reaction score
0
MY apologies that I don't think this is particularly advanced; my algebra and trigonometry is quite good, but all seems to fall apart once i move from 2D to 3D environments.

Johnson Solid J2 is a pentagonal pyramid consisting of five equilateral triangles on a pentagonal pase, meeting at the apex. It's the shape you get if you slice off the top five triangular faces of an icosahedron.

I can find a lot of info on this shape, height, surface area etc, but I need to know one specific angle that eludes me; the angle formed by the junction of;

1. A vertical line dropped from the apex of the solid to the base
2. a line that bisects one of the equilateral triangles, rising from the mid point of base of the triangle to its apex

If it can be calculated as a function of the height of the pyramid, or the height of the face that's fine, but I get the feeling it SHOULD be a constant value, similar to the dihedral angle of (approx) 138 .2 degrees that is formed by the junction of any pair of triangular faces in this solid.

Thanks for your help!
 
Mathematics news on Phys.org
As is often the case...I get stuck, post for help...then get lucky!

I've found the formula for the "slant height" (a term I did not previously know) which is the length of the bisecting line down the midpoint of the face.

Since I already have the formula for the height of the pyramid, I believe I have everything I need to find the angle as follows;

a = length of edge of face.

Slant height (s)
$$0.5 * \sqrt{3} * a$$

Pyramid height (h)
$$sqrt{((5 - sqrt{5})/10)} * a$$

So if I'm right, s is my hypotenuse, h my adjacent...so I just need to use cosine?
 
I could probably substitute and simplify given time...but when Excel is so easy...the answer should anyone ever need it, is 52.62263 degrees and as I suspected does not vary with the length of the triangular faces.