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

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
$$\begin{align}
&\cos\alpha = \frac{2^2 + 3^2 - \ell^2}{2(2)(3)} = \frac{13 - (5 + 2\sqrt{2})}{12} = \frac{4 - \sqrt{2}}{6}\\
\implies & \alpha = \cos^{-1}\left(\frac{4 - \sqrt{2}}{6}\right) \approx  64.47122063449069^\circ
\end{align}
$$
