What is $\theta$? Suppose that $AB=BC=CD=DE=EA=BG=AG=AF=FE=DF=GC$. Then What is $\theta$?

 A: One way to understand this geometric figure is start from the rhombus $ABCG$, rotate it with respect to $A$ clockwisely for some angle $\phi$ until it coincides with rhombus $AFDE$. Under this rotation, $C$ get moved to $D$ and $G$ get moved to $E$. This means $\angle GAE = \phi = \angle CAD\,\color{blue}{{}^{[1]}}$.
We can compute the last angle $\angle CAD$ by applying cosine rules to 
triangle $CAD$. Since $ABCG$ is a rhombus formed from two equilateral triangles $ABG$ and $BCG$, we have 
$$AD : AB = AC : AB = \sqrt{3} : 1$$
This leads to
$$
\phi 
= \cos^{-1}\left(\frac{AC^2 + AD^2 - CD^2}{2\cdot AC \cdot AD}\right)
= \cos^{-1}\left(\frac{\sqrt{3}^2+\sqrt{3}^2-1^2}{2\sqrt{3}^2}\right)
= \cos^{-1}\left(\frac56\right)
$$
From this, we can deduce
$$\theta = \angle FAG = \angle FAE - \angle GAE = \frac{\pi}{3} - \phi =  \frac{\pi}{3} - \cos^{-1}\left(\frac56\right) \approx 26.44269^\circ$$
Notes


*

*$\color{blue}{[1]}$ Another way to see $\angle CAD = \angle GAE$ goes like this.


*

*By SSS, $\triangle CAG \simeq \triangle CAB \implies \angle CAG = \frac12 \angle BAG = 30^\circ$.  

*By SSS again, $\triangle DAE \simeq \triangle DAF \implies \angle DAE = \frac12\angle FAE = 30^\circ$.


This implies $\angle CAG = \angle DAE$ and hence
$$\angle CAD = \angle CAE - \angle DAE = (\angle CAG + \angle GAE) - \angle DAE\\
= \angle GAE + (\angle CAG - \angle DAE) = \angle GAE$$
A: Let the angles FDC and GCD be $x$
Then for the whole pentagon, we have $$540=8\times 60-\theta+2x$$
$$\Rightarrow2x-\theta=60$$
Let all the lengths indicated as equal have length $1$. Now consider the isosceles triangle ADC, where the side AD has length $\sqrt{3}$.
Therefore angle ADC $=\arccos\frac{1}{2\sqrt{3}}=x+30$
Eliminating $x$, $\theta=26.4$
