MHB Finding $\theta$ in a Geometric Diagram

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Solve for $\theta$ in the diagram below.

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anemone said:
Solve for $\theta$ in the diagram below.
Let $a,b,O,A,B,M$ as in the following picture ($O$ is the origin).
\begin{tikzpicture}
\def\a{4}
\def\b{4}
\draw[fill=black] (-\a,0) -- node[below] {a} (0,0) node[below] {O} -- node[below] {a} (\a,0);
\draw (-\a,0) node[above right] {$70^\circ$} -- node
{b} +(70:\b) coordinate (A) node[above] {A};
\draw (\a,0) node[above left] {$80^\circ$} -- node
{b} +({180-80}:\b) coordinate (B) node[above] {B};
\draw (A) -- (B);
\draw[fill=black] (0,0) -- +(85:{\b*sin(75)}) node[above] {M};
\draw[thick] (0,0) node[above right] {$\theta$} +(.6,0) arc (0:85:.6);
\end{tikzpicture}

Then the coordinates of A are $(-a+b\cos 70^\circ, b\sin 70^\circ)$.
And the coordinates of B are $(a-b\cos 80^\circ, b\sin 80^\circ)$.
Point M is the average of A and B, which is:
$$M = \left(\frac b2(\cos 70^\circ - \cos 80^\circ), \frac b2(\sin 70^\circ+\sin 80^\circ)\right)$$
Using the sum and difference formulas of sine and cosine, we get:
$$M = b(\sin 75^\circ \sin 5^\circ, \sin 75^\circ \cos 5^\circ)$$
And with the complement rule:
$$M = b\sin 75^\circ (\cos 85^\circ, \sin 85^\circ)$$

Thus $\theta = 85^\circ$.​
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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