I was doing a little trigonometry challenge given by my professor, which is not anything especially complicated. However, it is completely based on getting the area of the green triangle in relation to the side of the square and the triangle, equal to $x$.
Reference image: 
I've already got a bit of data from some trig. relations. What doesn't seem to be doable is finding a relation between $x$ and the sides $L_1$ and $L_2$ or isolating the sine of $\beta$ from the area of that triangle, in relation to $x$ as well.
Anyway, the following is the data gathered so far.
-
Sine of $\alpha$ (both $\alpha_1$ and $\alpha_2$):
$$ \begin{aligned}
h_1^2 &= x^2 + \frac{x^2}{4} = \frac{5x^2}{4} \\
h_1 &= \frac{x \sqrt{5}}{2}
\end{aligned} \\
\begin{aligned}
\sin {\alpha} &= \frac{x}{h_1} = \frac{x}{\frac{x \sqrt{5}}{2}} = \frac{2x}{x \sqrt 5} = \frac{2}{\sqrt 5} = \frac{2 \sqrt 5}{5}
\end{aligned} $$ -
Relation between sides $L_1$ and $L_2$, as well as sine of $\beta$ (through the law of sines):
$$ \begin{aligned}
\frac{\sin \alpha}{L_2} &= \frac{\sin \beta}{\frac{x}{2}} = \frac{\sin 30^\circ}{L_1} \\
\frac{2 \sqrt 5}{5L_2} &= \frac{2 \sin \beta}{x} = \frac{1}{2L_1}
\end{aligned} \\
\begin{aligned}
L_1 &= \frac{L_2 \sqrt 5}{4} = \frac{x}{4 \sin \beta} \\
L_2 &= \frac{4L_1 \sqrt 5}{5} = \frac{x \sqrt 5}{5 \sin \beta}
\end{aligned} $$ -
Relation between $h_2$ and sides $L_1$ and $L_2$:
$$ \begin{aligned}
\sin \alpha = \frac{2 \sqrt 5}{5} &= \frac{h_2}{L_1} \\
\sin 30^\circ = \frac{1}{2} &= \frac{h_2}{L_2}
\end{aligned} \\
\begin{aligned}
h_2 &= \frac{L_2}{2} = \frac{2L_1 \sqrt 5}{5} = L_1 \sin \alpha
\end{aligned} $$ -
Calculating the area with the data so far (1) (using sine of $\beta$):
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot L_1 \sin \alpha}{2} \\
&= \frac{x}{4} \cdot L_1 \sin \alpha \\
&= \frac{x}{4} \cdot \frac{x}{4 \sin \beta} \sin \alpha \\
&= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta}
\end{aligned} $$ -
Calculating the area with the data so far (2) (using the sides' values)
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot \frac{2L_1 \sqrt 5}{5}}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_1}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_2 \sqrt 5}{8} \\
&= x \cdot \frac{L_2}{8}
\end{aligned} $$
Additionally, I've tried assuming (for the sake of an approximation) that $\beta$ is a right angle, since it is roughly equal to 86 degrees, and applied a theorem I've noticed in right triangles while meddling with this challenge.
-
Given a triangle $\Delta ABC$, where $a = BC$, $b = CA$ and $c = AB$, the height $h$ of the triangle, perpendicular to the base (assumed to be AB), is equal to the product of the product of the cathetuss divided by the hypotenuse. That is,
$$ \begin{aligned}
h = \frac{a \cdot b}{c}.
\end{aligned} $$ -
Moreover, with a substitution of these values, it is possible to get the value of the sides from $x$, as well as confirm the angle of $\beta$.
$$ \begin{aligned}
h_2 &= \frac{2L_1L_2}{x} \\
&= \frac{L_2}{2} = L_1 \sin \alpha
\end{aligned} \\
L_1 = \frac{x}{4},\ L_2 = \frac{x \sqrt 5}{5} \\
x = 4L_1 = L_2 \sqrt 5 \\
\sin \beta = \sin 90^\circ = 1
$$ -
And finally, calculate an estimated approximation for the area, either with the value of $\beta$ or the value of the sides in respect to $x$.
$$ \begin{align}
A &= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta} \\
&= \frac{x^2}{16} \cdot \frac{2 \sqrt 5}{5} \\
&= \frac{2x^2 \sqrt 5}{80} = \frac{x^2 \sqrt 5}{40} \\ \newline
A &= x \cdot \frac{L_2}{2} \\
&= x \cdot \frac{ \frac{x \sqrt 5}{5} }{8} \\
&= x \cdot \frac{x \sqrt 5}{40} = \frac{x^2 \sqrt 5}{40} \\
\end{align} $$
How one would come to derive $x$ into the sides' values or the sine of $\beta$, in order to get the area only through $x$? What sort of hindsight is required to do so, even?
Best Answer
You can try this simpler alternative, which is in the spirit of finding the area of $\triangle GHB$ in terms of $x$.
Since we basically have a right triangle $ADE$, we can show that: $$x^2+(2x)^2=DE^2\implies DE=\sqrt{x^2+5x^2}=x\sqrt5 \,\,\text{and:}\\ \sin \angle ADE=\frac{2x}{x\sqrt5}\implies \angle ADE=\arcsin\left(\frac2 {\sqrt{5}} \right)$$ It follows that $\angle CDG=90-\angle ADE$ and $\angle CDG=\angle DEA$.
Now we need to find $DG$: $$\cos \angle CDG=\frac x{DG}\\ \cos\left(90-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)=\frac x{DG}\\ \implies DG=\frac{x\sqrt5}{2}$$
It follows that: $$x^2+CG^2=DG^2=\frac{5x^2}{4}\implies CG=\sqrt{\frac{x^2}{4}}=\frac{x}{2} \,\,\text{implying}:\\ GB=CG=\frac x2$$
Now we need to find $\angle DGC$, since $\angle DGC=\angle BGH$: $$\sin \angle DGC= \frac{2x}{x\sqrt5}\implies \angle DGC=\angle BGH=\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Since $\angle CBA$ is right, and $\angle FBE=60^\circ$, then it follows that $\angle GBH=30^\circ$
Now, since we know $\angle FBE=30^\circ$ and $\angle BEH=\angle DEA=90-\arcsin\left(\frac2 {\sqrt{5}} \right)$, then $\angle BHE$ is: $$\angle BHE=30^\circ+\arcsin\left(\frac2 {\sqrt{5}} \right) \,\, \text{thus:}\\ \angle GHB=150^\circ -\arcsin\left(\frac2 {\sqrt{5}} \right)$$
Now, we need to find $BH$. By the law of sines, we have: $$\frac{\sin\angle GHB}{GB}=\frac{\sin BGH}{HB}\,\,\,\, \text{implying:}\\ HB=\frac{GB\sin\angle BGH}{\sin\angle GHB}=\frac{\frac x2\sin\left(\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}{\sin\left(150^\circ-\arcsin\left(\frac2 {\sqrt{5}} \right)\right)}$$
Keeping in mind that: $\cos(\arcsin(x))=\sqrt{1-x^2}$, we get that: $$HB=\frac{2x}{1+2\sqrt3}$$
Now, finally, recalling that: $$\triangle=\frac12 ab\sin A \,\,\text{where:}\\ $$ For $a:=GB=\frac x2, b:=HB=\frac{2x}{1+2\sqrt3}, A:=\angle GBH=30^\circ$