Given two angle and a segment, can you find $h$? While designing another problem I came up with the following question:

Considering the picture below, can you determine $h$ based only on the information of $\alpha$, $\beta$ and $x$ given the fact that the segment $x$ is the continuation of the height?


My guess is that the problem is not solvable (in the sense that one can construct multiple $h$ with the same $\alpha,\beta$ and $x$), but cannot see how to show this.
So far I gave the problem a try by splitting the angle and combining some trigonometric identities but couldn't conclude.

So far I only obtained the following
\begin{align*}
\tan(\alpha)=\frac{(h_1+h_2)(x+y)}{(x+y)^2-h_1h_2}
\end{align*}
and
$$\tan(\beta)=\frac{(h_1+h_2)\cdot y}{y^2-h_1h_2}.$$
Many thanks in advance.
 A: It is not possible to determine $h$ using only $\alpha, \beta, x$. 
For example, if we take $\alpha = \alpha_1, \beta= \beta_1$ (i.e. making a right triangle), we get $$h = \frac{x \tan \alpha \tan \beta}{\tan \beta - \tan \alpha}$$
On the other hand, if we take $\frac{\alpha}{2} = \alpha_1 = \alpha_2, \frac{\beta}{2} = \beta_1 = \beta_2$ (i.e. making an isosceles triangle), then we can cut it in two to get right triangles, so we get  $$h = 2\frac{x \tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})}{\tan(\frac{\beta}{2}) - \tan(\frac{\alpha}{2})}$$
Trying a few simple values for $x, \alpha, \beta$ (I used $1, 30^\circ, 45^\circ$) shows that $h$ takes different values.
A: 
Draw the circumscribed circles 
of $\triangle ABC$ and $\triangle EBC$,
select the point $A_1$ and $E_1$ on these circles
such that $A_1E_1\parallel AE$
and $|A_1E_1|\ne|AE|$.
We know that then
$\angle CA_1B=\alpha$.
$\angle CE_1B=\beta$.
Construct $E_2\in A_1E_1:\ |A_1E_2|=|AE|$.
Draw the line $L_1$ through $E_2$ parallel to $E_1B$
and the line $L_2$ through $E_2$ parallel to $E_1C$.
Point $B_1=L_1 \cap A_1B$,
point $C_1=L_2 \cap A_1C$.
point $D_1=BC \cap A_1E_2$,
point $D_2=B_1C_1 \cap A_1E_2$.
This new construction that includes $\triangle A_1B_1C_1$
and$\triangle E_2B_1C_1$ has the same 
properties as the original one,
namely 
$\angle C_1A_1B_1=\alpha$,
$\angle C_1E_1B_1=\beta$,
$A_1E_1\perp B_1C_1$,
$|A_1E_1|=x$, but 
$|B_1C_1|\ne|BC|$.

Edit
As @pm2595 noted, this construction and conclusion
is valid only if the point $E_1:\ |A_1E_1|\ne x$ exists.
And such a point can always be found 
excluding a special case,
when the circumradius $R_{ABC}$ 
of $\triangle ABC$
and the circumradius $R_{EBC}$ 
of $\triangle EBC$
are the same:
\begin{align} 
R_{ABC}&=R_{EBC}
,\\
\frac{h}{2\sin\alpha}
&=
\frac{h}{2\sin\beta}
,
\end{align}
in other words, unless we are given that
\begin{align} 
\beta&=180^\circ-\alpha
.
\end{align} 

In this special case for any point 
$A_1\in\operatorname{arc}(CAB)$
the distance to
the corresponding point 
$E_1\in\operatorname{arc}(CEB)$ 
will always be the same, $|A_1E_1|=x$
and $E_2\equiv E_1$.
Surprisingly, in such a case
we indeed can find a unique
value for $h$,
since it would be the same 
for all possible constructions,
and we can easily find it
from the configuration, in which
both 
$\triangle ABC$
and $\triangle EBC$
are isosceles: 

In this case we have
\begin{align}
\triangle ADC:\quad
\tfrac12h&=|AD|\tan\tfrac\alpha2
\tag{1}\label{1}
,\\ 
\triangle EDC:\quad
\tfrac12h&=|ED|\tan\tfrac\beta2
=(|AD|-x)\tan\left(\tfrac12(180^\circ-\alpha)\right)
\tag{2}\label{2}
,
\end{align} 
so we can solve the system \eqref{1}-\eqref{2}
for $|AD|$ and $h$.
Thus in case when
$\beta=180^\circ-\alpha$,
we have the unique answer 
\begin{align}
h&=x\tan\alpha
.
\end{align}.
Note that even in this case 
the value of $|AD|$ is still uncertain:
that one we've found from 
\eqref{1}-\eqref{2}
is valid for isosceles configuration only
and it would be different otherwise.
A: I like using the law of sines for problems like this.  Consider the upper leftmost triangle in your diagram, the one whose angles are $\alpha_1$, $\pi - \beta_1$, and $\beta_1 - \alpha_1$.  Let's write $r$ for the length of the side opposite the angle $\pi - \beta_1$.  By the law of sines,
$$
\frac{\sin(\beta_1 - \alpha_1)}{x} = \frac{\sin(\pi-\beta_1)}{r}
$$
Notice that $r$ is determined by the values of $x$, $\alpha_1$ and $\beta_1$.
But $h_1$ is the product $r\, \sin(\alpha_1)$, so $h_1$ too is determined by the values of $x$, $\alpha_1$ and $\beta_1$.
A: 
This may not be the easiest solution but it is a different one which I thought would be interesting to share as well. Looking at those points we know that $\angle ACB=\angle ALB=\beta,$ $\angle AMB=\angle AKB=\alpha$ and by construction $MC$ and $KL$ are prependicular to $AB$.  Moreover, we have $h=\overline{AB}$ and as we can see in the image $\overline{MC}< \overline{KL}.$ So, we have two triangles with the same $h,\alpha,\beta$ but different $x$. 
Let us now think of the homothety centered at $B$ with ratio $\frac{\overline{KL}}{\overline{MC}}$ and let $C',M',A'$ be the images of $C,M,A$ respectively. By the  homothty's properties   we have
\begin{align}
 \angle BC'A'&=\angle BCA=\beta\\
\angle BM'A'&=\angle BMA=\alpha\\
\overline{M'C'}&=\frac{\overline{KL}}{\overline{MC}}\overline{MC}=\overline{KL}\\ \overline{A'B}&=\overline{AB}\frac{\overline{KL}}{\overline{MC}}>\overline{AB}.
\end{align} 
Furthermore, $M'C'$ will be prependicular to $A'B.$
Hence, we have $\angle A'C'B=\angle ALB=\beta,$ $\angle A'M'B=\angle AKB=\alpha$ and $x'=\overline{M'C'}=\overline{KL}=x$ but $h'=\overline{A'B}>\overline{AB}=h,$  therefore having $x$ and both angles is not enough.
A: This is what I would practically do upfront before attempting  a solution.
Take $ 30 ^{\circ}$ and $ 45 ^{\circ}$ set square triangles from a geometry box as a particular example. 
Place two vertices at distance $x$ along $x-$ axis.
No matter  how you rotate the triangles about the fulcrum vertices,line connecting  side intersections do not occur with perpendicularity in the manner you suggest. So drop further general solution attempts.
