Angle between two faces in this polyhedron? I am wondering as the polyhedron in the following picture, what is the angle between the red regular 5-gon and the yellow square, and what is the angle between the yellow square and blue equilateral triangle, if they do form such a structure? Here for the angle, I mean the angle between two faces.
My idea is to look at the big circle around this structure, but I don't get enough equations to solve the degrees in this way.

 A: This might be a more intuitive approach, if less rigorous. 
If you only had the red pentagons, you would have a dodecahedron. But instead we have a yellow square between the faces. Thus the supplement of the angle between the red and yellow will be half the supplement of the dihedral angle of the dodecahedron. Looking up the dihedral angle in the dodecahedron we see it's $\pi-ArcTan(2)$. Thus the supplement is $ArcTan(2)$, and so the angle between the red and yellow is $\pi-\frac{1}{2}ArcTan(2)$
Similarly, if you only had the blue triangles, you would have an icosahedron. Looking up the dihedral angle we see it's $\pi-ArcCos(\frac{\sqrt{5}}{3})$. So the angle between the blue and yellow is $\pi-\frac{1}{2}ArcCos(\frac{\sqrt{5}}{3})$.
A: What about considering https://bendwavy.org/klitzing/explain/dihedral.htm ?
The polyhedron you consider there is given at https://bendwavy.org/klitzing/incmats/srid.htm . There its dihedrals are given in the form
between {3} and {4}:   arccos(-(1+sqrt(5))/sqrt(12)) = 159.094843°
between {4} and {5}:   arccos(-sqrt[(5+sqrt(5))/10]) = 148.282526°  
--- rk
A: The actual calculations will be rather tedious, but here goes. Start with this: 
$\phi=\pi/5$, $P_1P_0P_2$ are the sides of a regular pentagon, and $P_1P_0P_3$ and $P_2P_0P_4$ are the sides of two squares.
If we start rotating the first square around the axis $P_0P_1$ and the second square around the axis $P_0P_2$, there will be one position where the distance $P_3^\prime P_4^\prime$ is equal to one. The rotations by angle $\alpha$ will map $P_3$ and $P_4$ to $P_3^\prime, P_4^\prime = (\pm \cos \alpha \sin \phi, -\cos \alpha \cos \phi, -\sin \alpha)$. Equating the norm of $P_3^\prime P_4^\prime$ to one, we find $\alpha$, after which the first dihedral angle is $\pi - \alpha$, yielding
$$\pi - \arccos \frac {\csc \phi} 2.$$
The normals to the square faces is $(0,0,1)$ rotated in the same way, which gives $(\pm \sin \alpha \sin \phi, -\sin \alpha \cos \phi, \cos \alpha)$. The normal to the triangular face can be computed as the cross product of $P_0P_4^\prime$ and $P_0P_3^\prime$, which gives $(0, -\sin 2\alpha \sin \phi, \cos^2 \alpha \sin 2 \phi)$. The second dihedral angle is $\pi$ minus the angle between the normals, yielding
$$\pi - \arccos \frac {2 \cos \phi} {\sqrt 3}.$$
