I've used cosine & sine rule, and after playing around with them, I'll always end up with an equation involving $\sin\theta,\cos\theta$ and other weird (and wonderful expressions). I've approached the question in many different ways: by showing $AD=x$ (since $\angle DAC=\angle DCA=30^\circ$ it forms an isosceles triangle), also tried to show $\sin\theta=1/2$ ... but I never got the correct answer.
This has been bothering me for weeks.
Thanks in advance
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In right $\Delta BAC$ $$\sin\theta=\frac{AB}{BC}=\frac{x}{BC}\implies BC=x\csc\theta$$ $$BD=BC-CD=x\csc\theta-x \quad \angle ADB=\theta+30^\circ$$
Using Sine rule in $\Delta ADB$ as folows $$\frac{\sin(\theta+30^\circ)}{x}=\frac{\sin60^\circ}{x\csc\theta-x}$$ $$2\sin(\theta+30^\circ)(1-\sin\theta)=\sqrt3 \sin\theta$$ Solving above equation for $0<\theta<90^\circ$, we get $\color{blue}{\theta=30^\circ}$
Drop altitude DT = $y$. Then, AT = DT$\>\cot30$ = $\sqrt3 y$. From similar triangles CDT and CBA, we have $\frac{BA}{DT}=\frac{AC}{TC}$, or
$$\frac xy = \frac{\sqrt{x^2-y^2} +\sqrt3y} {\sqrt{x^2-y^2}} =1+\frac {\sqrt3y}{\sqrt{x^2-y^2}} $$
Rearrange to get $(x-2y)(x^3+2y^3)=0$, which yields $x=2y$. Thus, $\sin\theta = \frac12$ and $\theta =30$.