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.

 A: $\qquad\qquad\qquad$

Place the origin $O$ at the lower left-hand corner, and let the positive $x,y$-axes be the the natural choices based on the diagram.

Let $r$ be the radius of the red circles.

Let $P,Q$ be the centers of the circles of radii $2,1$, respectively.

Then we can express the coordinates of $P,Q$ as $P=(x,2)$ and $Q=(x,r-1)$, where $x$ is an unknown.

Computing $|OQ|$ in two ways, we get the equation
$$\sqrt{x^2+(r-1)^2}=r+1$$
hence
$$x^2+(r-1)^2=(r+1)^2\tag{eq1}$$
Computing $|OP|$ in two ways, we get the equation
$$\sqrt{x^2+2^2}=r-2$$
hence
$$x^2+4=(r-2)^2\tag{eq2}$$
From $(\text{eq}1){\,-\,}(\text{eq}2)$, we get $r=8$.

Plugging $r=8$ into $(\text{eq}2)$, we get $x=4\sqrt{2}$.

Using the known values of $r$ and $x$, the distance formula yields
\begin{align*}
|OP|&=6\\[4pt]
|OQ|&=9\\[4pt]
|PQ|&=5\\[4pt]
\end{align*}
hence by the law of cosines, $\cos(\large{\phi})={\large{\frac{23}{27}}}$, so we get $\large{\phi}=\cos^{-1}\bigl({\large{\frac{23}{27}}}\bigr)$.
A: There are constraints that $A$ and $B$ must fullfill, i.e., the following system of equations coming from Pythagoras theorem applied to certain right triangles :
$$\begin{cases}(B-2)^2+2^2=(A/2)^2\\(B-1)^2+(A/2)^2=(B+1)^2\end{cases}$$
giving $A=8 \sqrt{2}$ and $B=8$.
If now we take equations as in the partial solution you gave (a good idea) :
$$\tan(\theta) = \dfrac{2}{A/2} = \dfrac{\sqrt{2}}{4}\tag{1}$$
and 
$$\tan(\phi + \theta) = \dfrac{B-1}{A/2} = \dfrac{7}{4 \sqrt{2}}\tag{2}$$
Equation (2) can also be written :
$$\dfrac{\tan(\phi) + \tan(\theta)}{1-\tan(\phi)\tan(\theta)} = \dfrac{7\sqrt{2}}{8}\tag{3}$$
Let $T:=\tan(\phi)$. (3) is equivalent to :
$$\dfrac{T + \sqrt{2}/4}{1-T\sqrt{2}/4} = \dfrac{7\sqrt{2}}{8}\tag{4}$$
giving $T=\dfrac{10 \sqrt{2}}{23}$. Therefore 


$\phi=\arctan(\dfrac{10 \sqrt{2}}{23})\approx 31.5863 \ \text{degrees}.$


A: Using Pythagoras theorem is not difficult to find that the radius $B$ of red arcs is $8$. It follows that $PQ=6$, $PR=9$ and $QR=5$. Apply now the cosine rule to triangle $PQR$ to find 
$$
\cos\phi={23\over27}.
$$

A: 
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}
