<s>[INST] Angles between vectors of center of two incircles I have two two incircle between rectangle and two 
quadrilateral circlein. It's possible to determine exact value of $\phi,$ angles between vectors of center of two circles.

 [/INST] 
Let $|AB|=|CD|=a$,$|BC|=|AD|=b$,
$|O_1E|=r_1$,
$|O_2F|=r_2$,
$\angle O_1AO_2=\phi$,
$\angle O_1AF=\alpha$
$\angle O_2AF=\beta$.
Then
\begin{align}
\tan\alpha&=\frac{b-r_1}{a/2}
\tag{1}\label{1}
,\\
\sin\alpha&=\frac{b-r_1}{b+r_1}
,\\
\tan\alpha&=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}
=\frac{b-r_1}{b+r_1}
\left/
\sqrt{1-\Big( \frac{b-r_1}{b+r_1} \Big)^2} \right.
=\tfrac12\,\frac{b-r_1}{\sqrt{b\,r_1}}
\tag{2}\label{2}
.
\end{align} 
From \eqref{1}$=$\eqref{2} it follows
\begin{align}
a&=4\,\sqrt{b\,r_1}
\tag{3}\label{3}
.
\end{align} 
Similarly,
\begin{align}
\tan\beta&=\frac{2r_2}a
\tag{4}\label{4}
,\\
\sin\beta&=\frac{r_2}{b-r_2}
,\\
\tan\beta&=
\frac{\sin\beta}{\sqrt{1-\sin^2\beta}}
=\frac{r_2}{\sqrt{b\,(b-2\,r_2)}}
\tag{5}\label{5}
.
\end{align}
From \eqref{4}$=$\eqref{5}:
\begin{align}
a&=2\,\sqrt{b\,(b-2\,r_2)}
\tag{6}\label{6}
,
\end{align}
and from \eqref{3}$=$\eqref{6}
we have
\begin{align}
b&=4r_1+2r_2
,\\
a&=4\,\sqrt{r_1\,(4\,r_1+2\,r_2)}
.
\end{align}
\begin{align}
\tan\phi&=\tan(\alpha-\beta)
=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
=\frac{2(3r_1+r_2)\,\sqrt{2\,r_1\,(2\,r_1+r_2)}}{16\,r_1^2+11\,r_1\,r_2+2\,r_2^2}
\end{align}
For $r_1=1$, $r_2=2$ we have
\begin{align}
a&=8\,\sqrt2\approx 11.31370850
,\\
b&=8
,\\
\phi&=\arctan\Big(\frac{10\,\sqrt2}{23} \Big)
\approx 31.586338^\circ
.
\end{align}
</s>