Simple(?) -find the angle- triangle problem For $AD=2DC$, $\angle{ADB}=60^\circ$ and $\angle{DBC}=15^\circ$, find the value of $\angle{A}$.

It seems quite simle, maybe it is. I only manage to find that $\angle{C}=45^\circ$, $\angle{BDC}=120^\circ$. I also noticed that if $E$ is the point of $AC$ such that $AE=ED=DC=x$ then the areas $\triangle ABE=\triangle BED=\triangle BDC$. Further more, using the $\sin$ formula to compute areas, $$BD=\frac{x}{\sin 15^\circ}\frac{\sqrt{2}}{2}, BC=\frac{x}{\sin 15^\circ}\frac{\sqrt{3}}{2}$$ and $$AB\sin A=BC\frac{\sqrt{2}}{2}$$.

hint
Sine rule at triangle $BDC$
$$\frac{BD}{\sin45}=\frac{CD}{\sin15} \Rightarrow \frac{BD}{CD}=\frac{\sin 45}{\sin15}$$
Sine rule at $ABD$
$$\frac{BD}{\sin \alpha}=\frac{2CD}{\sin(120-\alpha)} \Rightarrow \frac{BD}{CD}=\frac{2\sin \alpha}{\sin(120-\alpha)}$$
Where $\alpha=\angle A$ 
