Find measure of angle $x$ Can you help me about the title? I don't know what's appropriate for a geometry problem!
In the following figure find measure of angle $x$.

I wrote sin law two times. One in ABC like 
$$\frac{8+BD}{\sin 120}=\frac{4}{\sin x}$$ 
and one in ABD like this 
$$\frac{BD}{\sin 30}=\frac{4}{\sin (\angle{BDA})}$$ 
Now if you eliminate $BD$ and use $\angle{BDA}=90+x$, a trigonometric equation comes out. Solving gives $x=20$ degrees. 
I feel that there is a neat way around of doing this. Can you help?

Let $I$ be the midpoint of $[DC]$ then $AI=\frac{1}{2} DC=4$ (since $ADC$ is a right triangle)
Now you get that, $ABI$ is an isoscles triangle of vertex $A$ which gives you that $\angle ABI =\angle AIB=2x$
(Because if you draw a circle $(C)$ circumscribed about triangle $ACD$ you'll notice that $\angle AID$ is a central angle which is double $\angle ACD$ the interior angle in $(C)$ facing the same arc $AD$)
Finally by summing up the angles in triangle $ABC$ you'll get that $2x+x+120=180$ which gives you that $x=20^\circ$

