Simple 9-th grade geometry problem I have a geometry problem which states that

Find the range of $x$ in following figure.
  

Given that $AD$ and $AC$ are equal, and the values and angles are also given. How to estimate the range of the problem possibly using basic geometry methods? Judging from other problems, this is 8th grade or 9th grade problem. I have used the following methods to estimate the range of the $x$ 

Since $2x-10 > 0 \implies x > 5$ and $2x-10 < 38 \implies x < 24$. Therefore $5<x<24$.

Using the fact that $\displaystyle \frac {\sin x}{x}$ is decreasing gives more tight bound which is $$\displaystyle 2x - 10 > \arcsin \left(\frac9{10} \sin (38^\circ)\right) \implies x > 11.8^\circ $$ But I don't think it is suitable since all other problems are either related to congruency or similarity. What is the simplest (basic geometry) way to estimate the range of $x$?
 A: First set $\theta = 2x-10$ and find the domain of $\theta$. The law of cosines states
$$ \begin{align}
  9^2 & = D^2 + L^2 -2 D L \cos \theta \\
 10^2 & = D^2 + L^2 -2 D L \cos 38°
\end{align} $$
where $D$ is the common edge, and $L$ is the equal outside edge (marked with $\parallel$)
Now consider the ratio $\lambda = \frac{L}{D}$ with the second equation solved for $$ D = \frac{10}{\sqrt{1+\lambda^2-2 \lambda \cos 38°}} $$
which is used in the first equation to yield
$$ 9^2 = 10^2 \frac{1+\lambda^2-2 \lambda \cos \theta}{1+\lambda^2-2 \lambda \cos 38°} $$
The above is solved for $$\cos \theta = \frac{9^2}{10^2} \cos 38° + \frac{10^2-9^2}{10^2} \frac{\lambda^2+1}{2 \lambda} $$
The above has a minimum value for $cos \theta=0.82829$ at $\lambda=1$ which translates to $$x \le \frac{90}{\pi} \frac{\cos^{-1}( 0.82829)}{2}+5 = 22.03836 $$
Now the smaller $x$ and $\theta$ is the larger $\cos\theta$ is. The maximum value for cosine is $\cos \theta=1$ which translates to $$x > \frac{\cos^{-1}(1)}{2}+5 = 5 $$
The table below summarizes the findings.
$$\begin{matrix}
\lambda & \theta & x & L & D \\
\hline 0.2837930 & 0& 5& 3.5662& 12.566\\
 1& 34.077°& 22.03836&  15.358&  15.358 \\
 3.523694&  0& 5& 12.566& 3.5662
\end{matrix}$$
A: I don't follow your argument about the tighter lower bound on the angle. The way I read it, the angle $2x-10$ can indeed become zero, namely when the absolute value of the difference in edge lengths becomes $9$:
$$\lvert AB-AC\rvert=9\implies\angle ACB,\angle CBA\approx71°\pm58.316971°$$
I'm more concerned about the upper bound. For reasons of symmery (and this is no proof yet!), the maximal value for $2x-10$ should correspond to the isosceles case. So start with an isosceles triangle to the right, where by the law of sines you have
\begin{align*}
\angle ACB=\angle CBA&=\frac{180°-38°}{2}=71°\\
AB=AC&=10\frac{\sin71°}{\sin38°}\approx 15.357767
\end{align*}
Then you have all the edge lengths for the left triangle as well, and can use the cosine law to compute the angle in question:
\begin{align*}
\cos\angle BAD &=\frac{AB^2+AC^2-9^2}{2\cdot AB\cdot AC}\approx 0.82828871 \\
\angle BAD &\approx 34.076654° \\
x &\approx 22.038327
\end{align*}
That number $x$ does not appear to be an algebraic number, as far as my numeric experiments indicate. I have no idea how students should come up with this except by using the laws of sines and cosines similar to how I did it.
To conclude, I'd state that the full and precise range is
$$
5°\le x°\le\frac12\left(10°+\arccos\left(1-
\frac12\left(\frac{9\sin38°}{10\sin71°}\right)^2\right)\right)
\approx 22.038327°
$$

