<s>[INST] Two squares and angle can someone answer this please?
I have two equal squares (picture). We know the lengths of three lines inside them. One of them is upside down. Find angle alpha. The picture is not perfect.

Thanks
 [/INST] Following is one boring way of solving the problem using coordinate geometry.
Even though I suspect there are more geometry intuitive and clever solutions,
a solution is a solution and here we go...
Choose a coordinate system such that the lower left corner of the upper square is the origin.
Let $\ell$ be the side and $(x,y)$ be the intersection of the three 3 rays of/in the upper square.
We have
$$
\begin{eqnarray}
x^2 + y^2 &= 2^2 = 4 &\tag{*1a}\\
(x-\ell)^2 + y^2 &= 3^2= 9 &\tag{*1b}\\
x^2 + (y-\ell)^2 &= 1^2 = 1 &\tag{*1c}\\
\end{eqnarray}
$$
"Subtract" $(*1a)$ from $(*1b)$ and $(*1c)$, we get
$$
\begin{cases}
\ell^2 - 2\ell x &= 9-4 = 5\\
\ell^2 - 2\ell y &= 1-4 -3
\end{cases}
\quad\implies\quad
\begin{cases}
x &= \displaystyle\;\frac{\ell^2 - 5}{2\ell}\\
y &= \displaystyle\;\frac{\ell^2 + 3}{2\ell}
\end{cases}
$$
Substitute this back into $(*1a)$, we get
$$
 (\ell^2 - 5)^2 + (\ell^2 + 3)^2 = 4(2\ell)^2
\iff  2 \ell^4 - 20 \ell^2 + 34 = 0
\iff \ell^2 = 5 \pm 2\sqrt{2}
$$
Since $x > 0$, we find $\ell^2 = 5 + 2\sqrt{2}$.
Apply cosine rule to the triangle with side $2,3$ and $\ell$, we find
$$\begin{align}
&\cos\alpha = \frac{2^2 + 3^2 - \ell^2}{2(2)(3)} = \frac{13 - (5 + 2\sqrt{2})}{12} = \frac{4 - \sqrt{2}}{6}\\
\implies & \alpha = \cos^{-1}\left(\frac{4 - \sqrt{2}}{6}\right) \approx  64.47122063449069^\circ
\end{align}
$$
</s>