What's the Angle?

Redrawing the diagram to scale for better visualization. After angle chasing, we note that $\angle BFE = 90^\circ$ . Let $EF=a$ . By Pythagorean theorem $BE^2 = BF^2+EF^2 = 1+a^2$ , $\implies BF = \sqrt{1+a^2}$ .

As $\triangle BCG$ and $\triangle ECH$ are similar, we have:

$\begin{aligned} \frac {BC}{CG} & = \frac {EC}{CH} \\ \frac {\sqrt{1+a^2}+1}{\frac {\sqrt 3}2} & = \frac 1{\frac a2} \\ a(\sqrt{1+a^2} + 1) & = \sqrt 3 \\ a\sqrt{1+a^2} & = \sqrt 3 - a \\ a^2 + a^4 & = 3 - 2\sqrt 3 a + a^2 \\ a^4 + 2\sqrt 3 a - 3 & =0 \end{aligned}$

Solving for the equation (I used numerical method), we get $a \approx 0.766420937$ . Let $\angle DCB = \theta$ , then $m\angle BEF = 30^\circ + \theta = 30^\circ + \sin^{-1} \left(\dfrac a2\right) \approx \boxed{52.5}^\circ$ .

$\triangle{BDF}$ is an equilateral triangle $\implies \angle{BDF} = \angle{DFB} = \angle{DBF} = 60^{\circ}$ and $AB$ is a straight line $\implies \angle{ADF} = 120^{\circ}$

Let $\angle{BAF} = \theta$ since $\triangle{ADF}$ is an isosceles triangle with $AD = DF = 1 \implies \angle{DAF} = \theta$ and vertical angles are congruent $\implies \angle{EFD} = \theta$ .

Let $AF = m$

Using the law of cosines on $\triangle{ADF} \implies m^2 = 2 - 2\cos(120^{\circ}) = 3 \implies m = \sqrt{3}$

Using the law of sines on $\triangle{ADF}$ , where $m = \sqrt{3} \implies$

$\dfrac{1}{\sin(\theta)} = \dfrac{\sqrt{3}}{\sin(120^{\circ})} = 2 \implies \sin(\theta) = \dfrac{1}{2} \implies \theta = 30^{\circ}$

$AE$ is a straight line $\implies \angle{BFE} = 90^{\circ}$ and $BC$ is a straight line $\implies \angle{FEC} = 180^{\circ} - \beta \implies \angle{ECF} = \beta - 30^{\circ}$ .

Let $BE = x$ and $FE = y$ .

Using right $\triangle{BEF} \implies y = \sqrt{x^2 - 1}, \cos(\beta) = \dfrac{y}{x} = \dfrac{\sqrt{x^2 - 1}}{x}$ and $\sin(\beta) = \dfrac{1}{x}$ .

Using the law of sines on $\triangle{FEC} \implies$

$\dfrac{1}{\sin(30^{\circ})} = \dfrac{y}{\sin(\beta - 30^{\circ})} \implies 2 = \dfrac{y}{\sin(\beta - 30^{\circ})} \implies$

$y = \sqrt{3}\sin(\beta) - \cos(\beta) = \sqrt{x^2 - 1} \implies$ $\dfrac{\sqrt{3}}{x} - \dfrac{\sqrt{x^2 - 1}}{x} = \sqrt{x^2 - 1} \implies$

$\sqrt{3} = \sqrt{x^2 - 1}(x + 1) \implies 3 = (x^2 - 1)(x^2 + 2x + 1) = x^4 + 2x^3 - 2x - 1 \implies$

$x^4 + 2x^3 - 2x - 4 = 0$

By inspection $x = -2$ is a root of $x^4 + 2x^3 - 2x - 4 = 0$ and dividing $x^4 + 2x^3 - 2x - 4$ by $x + 2$ we obtain:

$x^4 + 2x^3 - 2x - 4 = (x + 2)(x^3 - 2) = 0 \:\ x > 0 \implies x = 2^{\frac{1}{3}} \implies$ $\sin(\beta) = \dfrac{1}{2^{\frac{1}{3}}}$

$\implies \beta = \boxed{52.532688789^{\circ}}$ .