Pentagonal trapezohedron with face perpendicular to side How do I calculate the angles of the kites in a pentagonal trapezohedron (i.e., a d10) such that the edge opposite a face is perpendicular to that face?
I.e., I'm trying to make $\alpha$ be 90 degrees in this picture:

 A: As we work up to an enlargement factor, we can assume that the projections of the 10 "intermediate" points $A_k$ and $B_k$ belong to the unit circle, i.e., we can assume that these points have the following coordinates:
$\begin{cases}A_k=(\cos(2k\tfrac{\pi}{5}),& \sin(2k\tfrac{\pi}{5}),& \ \ a)\\B_k=(\cos((2k+1)\tfrac{\pi}{5}),& \sin((2k+1)\tfrac{\pi}{5}),&-a)\end{cases}$ for $k=0 \cdots 4$ with parameter $a$.
[Points $A_k$ (resp. $B_k$) are the vertices of the upper (resp. lower) pentagon, the lower pentagon being rotated by an angle $\tfrac{\pi}{5}$ with respect to the upper pentagon.] 
Let $S_+=(0,0,b)$ and $S_-=(0,0,-b)$ denote the upper and lower summits of the polyhedron ($b$ being a second parameter).
We have to determine $a$ and $b$.
As all faces have an identical shape, let us consider face $S_-B_4A_0B_1$. 


*

*A first constraint is that it is ... a face, i.e., that all points are coplanar.


This is classically expressed (To find the volume of the region that is bordered by 4 points in 3D space) as the following zero determinant:
$$\tag{1}\begin{vmatrix}x&x'&x''&x'''\\y&y'&y''&y'''\\z&z'&z''&z'''\\1&1&1&1\\\end{vmatrix}=0,$$
where $(x,y,z),(x',y',z'),\cdots$ are resp. coordinates of points $S_-,B_4,A_0,B_1$.
Setting $\begin{cases}c:=\cos(\tfrac{\pi}{5})\\s:=\sin(\tfrac{\pi}{5})\end{cases}$, constraint (1) gives $$\tag{1}\begin{vmatrix}0&c&1&c\\0&-s&0&s\\-b&-a&a&-a\\1&1&1&1\\\end{vmatrix}=0 \ \iff \ 2s(a - b + ac + bc)=0.$$
As $s\neq0$, this is finally equivalent to equation 

$$\tag{1a}a-b+\cos(\tfrac{\pi}{5})(a+b)=0.$$



*

*The second constraint is the desired orthogonality:


As suggested by the dashed segment $A_0S_-$ of your figure, it is equivalent to the orthogonality of $\vec{A_0S_-}$ and $\vec{A_0S_+}$ i.e., $\vec{A_0S_-}.\vec{A_0S_+}=0$ which turns to be :

$$\tag{2}a^2-b^2+1=0.$$

(see Remark below for a different, lengthier, manner).
Essential remark: relationship (2), written under the form $\sqrt{1+a^2}=b$ means that all the 12 points $A_k, B_k, S_, S_-$ must belong to the sphere centered in $O$ with radius $b$. This is quite evident... afterwards : considering diameter $S_+S_-$, we have the classical right angle property with respect to this diameter...
(1a) and (2) generate a system of 2 equations with 2 unknowns that will give the values :

$$\begin{cases}a=\tfrac12 \left(\tfrac{1}{\sqrt{c}} - \sqrt{c} \right)\approx 0.106166\\b=\tfrac12 \left(\tfrac{1}{\sqrt{c}} + \sqrt{c} \right)\approx 1.005620\end{cases} \ \text{where} \  c=\cos(\tfrac{\pi}{5})$$

Knowing $a$ and $b$, angles $\beta$ and $\gamma$ are obtained by using dot products. With the help of @Tordek, here they are : $\beta=\pi/2$ which is rather remarkable, and $\gamma=arccos(\tfrac{\sqrt{5}-1}{2})\approx 52°$.
Remark : 
Orthogonality could have been expressed by considering the normal to the face given by cross product:
$$\vec{N}=\vec{S_-B_4} \times \vec{S_-B_0}$$
which is constrained to have a zero dot product with $\vec{S_+A_0}$:
$$\vec{N}.\vec{S_+A_0}=0 \ \ \iff \ \ \det(\vec{S_-B_4},\vec{S_-B_0},\vec{S_+A_0})=0$$
("mixed product" property : $(U \times V).W=\det(U,V,W)$)
A: This is a roadmap for calculating everything.
Label the vertices like in the picture. Point $M$ is the center of regular pentagon $BDEJI$ and $N$ is the midpoint of $BD$. 

It is clear that $\angle BGM = \angle CAB = \angle NAB$ and $\angle GMB = 90^\circ = \angle AMN$. This means that $\triangle BGM \sim \triangle NAM$. 
If additionally $\alpha = \angle ABG = 90^\circ$ then $\angle ABM = 90^\circ - \angle MBG = \angle BGM$ and since $\angle BMA = 90^\circ = \angle GMB$, it follows that $\triangle ABM \sim \triangle BGM \sim \triangle NAM$ as well. 
Below we will express everything in terms of $BD$.
The values of $BM$ and $MN$ can be calculated from the formulas for circumradius and inradius of regular pentagon of side $BD$.
Using similarities above we find 
$$\frac{BM}{MA}=\frac{MA}{MN} \implies AM^2=BM \cdot MN,$$
so we can calculate $AM$.
Using Pythagorean theorem we can calculate $AB^2=BM^2 + MA^2$ and $AN^2=MN^2+AM^2$.
By similarity $\triangle BGM \sim \triangle ABM$ we can calculate $AC = BG = \frac{AB \cdot BM}{AM}$.
Then $CN = AC - AN$ and $BC^2 = CN^2 - BN^2 = CN^2 - \frac 14 BD^2$.
Now that we know the values of $AB, BC, CA$ we can find (using law of cosines)
$$\beta = \arccos \frac{AB^2+BC^2-AC^2}{2AB\cdot BC}$$
and
$$\gamma = \arccos \frac{2BC^2-BD^2}{2BC^2}.$$
