Trivial geometry question about an angle between two parallel blocks (CG) Trivial but not for me.
See the figure below. I'd like to draw a kind of "N" where the thickness of the 3 segments is variable.
From "r" the ratio of the thickness over the side of the square, how to find the value of "α", the angle of the inclined segment?

(Edit: every segment has a thickness of "r")
 A: 
From your figure, the width of central narrow white rectangle $=(1-2r) $
Lengths of red and green line segments when added is 1.
$$r/\sin \alpha + ( 1-2r)  \tan \alpha =1 $$
If $ t= \tan(\alpha/2) $ we have
$$r\frac{1+t^2}{2t}+(1-2r) \frac{1-t^2}{2t} =1 $$
simplifies to a quadratic equation
$$ t^2( 3r-1) -2t +(1-r)=0 $$
which can be solved as
$$ \tan ( \alpha/2)=\frac {\sqrt{3r^2-4r+2}-1}{(1-3r)}$$
with proper sign before sqrt radical. When $r=0$ it should check $ \alpha /2=22.5^{\circ}
\text{  from }  t^2+2t -1 =0, $
which checks alright.
A: Expanding on the comment by @peterwhy, one needs to solve the equation
$$\frac{r}{\sin\alpha}+\frac{1-2r}{\tan\alpha}=1$$
Upon rearrangement we get
$$\sin\alpha-(1-2r)\cos\alpha=r\iff \sin(\alpha-\delta)=\frac{r}{1+(1-2r)^2}$$
where $\sin\delta:=(1-2r)/\sqrt{1+(1-2r)^2}$. This can  now be solved quite easily to yield the unique solution
$$\alpha(r)=\sin^{-1}\frac{r}{\sqrt{1+(1-2r)^2}}+\sin^{-1}\frac{1-2r}{\sqrt{1+(1-2r)^2}}$$
We find that
$$\alpha(1/5)=40.84^\circ~,~ \alpha(2/5)=34.40^\circ$$
and also $\alpha(0)=\pi/4$ which works a sanity check.
