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 !
 A: Without loss of generality suppose $b \ge c$.  If $b = c$, then the dihedral angle will be the angle whose tangent is the ratio of this common height $b = c$ to the altitude of $\triangle AED$ from $A$, which we will call $h_1$; i.e., $$\tan \theta = \frac{b}{h_1}.$$  It is easy to observe that $$\frac{h_1 l_1}{2} = |\triangle AED| = \sqrt{s(s-l_1)(s-l_2)(s-l_3)}$$ where $s = (l_1 + l_2 + l_3)/2$, is the semiperimeter, by Heron's formula; thus given the three side lengths of $\triangle AED$ it is straightforward to calculate the desired altitude and therefore the desired dihedral angle.
If $b > c$, then the extension of $BC$ and $ED$ will meet at point $F$, and $AF$ will be the common line of intersection of planes defined by $ABC$ and $AED$.  Then the dihedral angle $\theta$ once again satisfies the relationship $$\tan \theta = \frac{b}{h_E},$$  where now $h_E$ is the altitude of $\triangle AEF$ from $E$.
With this information in mind, I leave it to you to complete the calculation for this case.  You would do well to demonstrate your understanding of the solution.  You will need to use properties of similar triangles, and the Law of Cosines.
A: 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.
