Dihedral angle Finding I meet a wall while some problem solving.
The wall is following question.
There is a triangle ABC, The vertice A touch bottom plane and the distance from B, C to the bottom : BE = b, CD = c .
When ED= l_1 , DA=l_2 , AE = l_3 , What can I find the dihedral angle between plane ABC and plane AED ? 

I want to break this wall. 
Please some help to me !

Let $A = O = (0, 0, 0)$ - is origin. $\overrightarrow{Oz}$ axis directed away from $A$ and $\overrightarrow{Oz} \perp AED$. $\overrightarrow{Ox}$ is codirectional to $\overrightarrow{AE}$, $\Rightarrow E = (l_3, 0, 0)$ and $B = (l_3, 0, b)$. What are $x$ and $y$ coordinates of $D$ and $C$?
$$
\begin{array}{lr}
p = \frac{l_1 + l_2 + l_3}{2} & \\
y = \frac{2}{l_3} \cdot \sqrt{p \cdot (p - l_1) \cdot (p - l_2) \cdot (p - l_3)} &
\text{($y$ is height of the $EAD$ triangle from $D$ to $EA$)} \\
x = \sqrt{l_2^2 - y^2} &
\text{$x$ is $x$-coordinate of mentioned above height}
\end{array}
$$
$\Rightarrow$ $D = (x, y, 0)$ and $C = (x, y, c)$.
It is well-known that normalized normal hyperplane equation for hyperplane supports the simplex spanned on $p_i, i \in \{1, 2, \dots, D\}$ points is:
$$
\sum \limits_{i = 1}^{D}\left(x_i \cdot \frac{n_i}{|\overrightarrow{n}|}\right) - \frac{d}{|\overrightarrow{n}|} = 0
$$
where $\overrightarrow{n} = (n_1, n_2, \dots, n_D)$ — normal to the hyperplane and $\frac{d}{|\overrightarrow{n}|}$ is a distance from the origin to the hyperplane:
$$
n_i =
\begin{vmatrix}
p_1^1  & p_1^2  & \dots  & p_1^{i - 1} & 1      & p_1^{i + 1} & \dots  & p_1^D  \\
p_2^1  & p_2^2  & \dots  & p_2^{i - 1} & 1      & p_2^{i + 1} & \dots  & p_2^D  \\
\vdots & \vdots & \ddots & \vdots      & \vdots & \vdots      & \ddots & \vdots \\
p_D^1  & p_D^2  & \dots  & p_D^{i - 1} & 1      & p_D^{i + 1} & \dots  & p_D^D  \notag
\end{vmatrix} \\
d =
\begin{vmatrix}
p_1^1  & p_1^2  & \dots  & p_1^D  \\
p_2^1  & p_2^2  & \dots  & p_2^D  \\
\vdots & \vdots & \ddots & \vdots \\
p_D^1  & p_D^2  & \dots  & p_D^D  \notag
\end{vmatrix}
$$
In our case $D = 3$ and above huge formulas became much simplier. All we need are just a coordinates of the normal $\overrightarrow{n}$ to each hyperplane (id est to triangles $ABC$ and $AED$).
$$
n_{ABC} =
\left(
\begin{vmatrix}
1 & 0 & 0 \\
1 & 0 & b \\
1 & y & c
\end{vmatrix},
\begin{vmatrix}
0 & 1 & 0 \\
l_3 & 1 & b \\
x & 1 & c
\end{vmatrix},
\begin{vmatrix}
0 & 0 & 1 \\
l_3 & 0 & 1 \\
x & y & 1
\end{vmatrix}
\right) = (-b \cdot y, b \cdot x - l_3 \cdot c, l_3 \cdot y)
$$
$$
n_{AED} =
\left(
\begin{vmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
1 & y & 0
\end{vmatrix},
\begin{vmatrix}
0 & 1 & 0 \\
l_3 & 1 & 0 \\
x & 1 & 0
\end{vmatrix},
\begin{vmatrix}
0 & 0 & 1 \\
l_3 & 0 & 1 \\
x & y & 1
\end{vmatrix}
\right) = (0, 0, l_3 \cdot y)
$$
Finally, dihedral angle between $ABC$ and $AED$ is:
$$
\angle (ABC, AED) = \arccos\left(\frac{(\overrightarrow{n_{ABC}} \cdot \overrightarrow{n_{AED}})}{|\overrightarrow{n_{ABC}}| \cdot |\overrightarrow{n_{AED}}|}\right) = \arccos\left(\frac{l_3 \cdot y}{\sqrt{(b \cdot y)^2 + (b \cdot x - l_3 \cdot c)^2 + (l_3 \cdot y)^2}}\right)
$$
is an $\arccos(\cdot)$ of dot product of normalized normals to planes, supports the triangles.
