Dihedral Angle Of The Rhombic Triacontahedron

Figure 1.  We are looking for the dihedral angle,  BIE . Figure 1 shows an r.t. “cap” (in red) over the face of the dodecahedron, and shows that J and G are on AH, the height of the pentagon. F is directly above the mid-face of the pentagon, at G.  I is directly above J on BE.  BJI is right, allowing us to calculate  BIJ or  EIJ, and  BIE, the dihedral angle, is then just twice that.
AH and BE lie on the pentagonal plane ABCDE.
BI is a line along an r.t. face, EI is a line along another of the faces.

Here is our plan of attack:
We know from Rhombic Triacontahedron that  BFA is 63.43494882° This can be rewritten without loss of information as . Triangle BIF is right by construction. BF is the side of the r.t., or rts. Here we must use trigonometry to get BI, by taking the sine of  BFA so we can find BI. BE is bisected by AH, the angle bisector of  BAE and a diagonal of the pentagon.  AJ is known from Pentagon Construction. To get  BIJ, we can take the sine of  BIJ =  BJ / BI. The dihedral angle  BIE is twice angle BIJ.

First let’s find BJ. BJ is one-half the  diagonal of the pentagon with sides equal to the side of the dodecahedron. We must change units in terms of the side of the r.t. Remember that  
So we write  BJ =
 = BI / BF.
BI = BF * .
BI =

 

Now we can calculate the dihedral angle BIE.

Figure 4  The dihedral angle BIE
We can write
 
 BIJ  = 72°.
The dihedral angle BIE is 2 *  BIJ.
Dihedral Angle = 144°.

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