Q:

Solve irregular quadrilateral with all sides and one outer angle

I know four sides of an irregular quadrilateral and one outer angle pictured below.

How can I determine angle $(a, d)$.
My attempts:
With the law of sines I got $$\sin(a, d) = \frac{(b + x) \sin\alpha}{a}$$ 
Then I tried to find the $x$ with the law of cosines trying different triangles that could be formed, but I couldn't figure it out. 

A:

Let $ABCD$ be our quadrilateral, where $AB=a$, $BC=b$, $CD=c$, $AD=d$, 
$BC\cap AD=\{E\},$ $\measuredangle E=\alpha$ and $\measuredangle A=x.$
Thus, by law of sines we obtain:
$$\frac{AE}{\sin(\alpha+x)}=\frac{a}{\sin\alpha},$$ which gives
$$DE=AE-AD=\frac{a\sin(\alpha+x)}{\sin\alpha}-d.$$
By the similar way we obtain:
$$CE=\frac{a\sin{x}}{\sin\alpha}-b$$ and by law of cosines for $\Delta ECD$ we obtain an equation on $x$:
$$\left(\frac{a\sin(\alpha+x)}{\sin\alpha}-d\right)^2+\left(\frac{a\sin{x}}{\sin\alpha}-b\right)^2-2\left(\frac{a\sin(\alpha+x)}{\sin\alpha}-d\right)\left(\frac{a\sin{x}}{\sin\alpha}-b\right)\cos\alpha=c^2.$$

