Find angles of two intersected planes. I'm really bad at math so I'll try to explain as best as I can.
Here's a visual representation of what I need to do. Basically it's a pop-up book.
There is a plane which can be folded on the blue line, so it makes two planes.  I have dimensions of planes, angles of planes on Z axis and angles of the pages on which these planes are positioned.
I need to get the X axis angles for each of two planes to have them appear seamless as a folded one while I turn pages.
 A: Let's start with the OP's illustration given in a comment.

I am going to refer to the different planes and vectors seen on the image as fallows. 
The red right page (with the script "N'T STAND IT!") will be called $r$. The red left page (with the script "BIG FROG CA") will be called $l$. The green left hand side of the frog will be called $gl$ the green right hand side of the frog will be called $gr$. When the pop up book is open then $l$ and $r$ coincide, $gl$ and $gr$ are perpendicular to $l$ and $r$, respectively. $gl$ and $lr$ intersect in a line perpendicular to the open book when $l$ and $r$ coincide.
$a$ $b$, and $c'$ are unit vectors on the common lines of $r$ and $gr$; $l$ and $gl$; and $gl$ and $gr$, respectively. $c'$ is perpendicular to $a$ and $b$ when the book is open, i.e. when the two red pages $l$ and $r$ coincide.
The next figure simplifies the two red and the two green pages. The upper figure depicts the fully open book. 

Let $\alpha$ characterize the two vectors $a$ and $b$ when the book is flat open. Let $\beta$ characterize an in between situation when the two red pages do not coincide. At the beginning $\beta=0$.
Let's fix our coordinate system as shown above. Now,
$$a=\begin{bmatrix}
\cos(\alpha)\cos(\beta)\\
\sin(\alpha)\\
\cos(\alpha)\sin(\beta).
\end{bmatrix}$$
and 
$$b=\begin{bmatrix}
-\cos(\alpha)\\
\ \sin(\alpha)\\
0
\end{bmatrix}.$$
Vector $c$ is perpendicular to the flat red plane when the book is flat open; it is perpendicular to $a$ and $b$.  And, $c$ remains perpendicular to the plane determined by $a$ and $b$ while the book is closing or, in other words while $\beta$ goes from $0$ to $\pi$. This is because $c$ and $a$, and $c$ and be are rigidly connected to each other. See the model 
Let $c$ be a vector pointing in the direction of $c'$. For $0\le \beta \le \pi$
$$c=a\times b=
\begin{vmatrix}
i&j&k\\
\ \cos(\alpha)\cos(\beta)&\sin(\alpha)&\cos(\alpha)\sin(\beta)\\
-\cos(\alpha)&\sin(\alpha)&0
\end{vmatrix}=$$
$$= \begin{bmatrix}
- \sin(\alpha)\cos(\alpha)\sin(\beta)\\
\cos^2(\alpha)\sin(\beta)\\
\sin(\alpha)\cos(\alpha)(1+\cos(\beta))
\end{bmatrix}.$$
My solution will be rather sketchy from this point on.
Watch out: $c$ is not a unit vector and $\lim_{\beta \rightarrow \pi}=0$ which result is misleading. So, let's normalize $c$ and, obviously $c'$, mentioned above, is its normalized version.
Then calculate $d=c'\times a$ and $e=c'\times b$ and let the corresponding unit vectors be denoted by $d'$ and $e'$, respectively.
Now, $c'$ is the unit normal vector of the plane $r$ and $d'$ is the unit normal vector of the plane $gr$; $c'\cdot d'$ is the cosine of the angle between $r$ and $gr$. So the angle of these two planes can be calculated. Similarly, $e'\cdot c'$ equals the cosine of $l$ and $gl$. Also, $e'\cdot k$ is the cosine of the angle between $gl$ and $r$. The angle between $l$ and $r$ is $\pi-\beta$.
