Answer
$x = \frac{d~sin~\alpha~sin~\beta}{sin~(\beta-\alpha)}$
Work Step by Step
The angle at $B$ in the triangle ABD is $180^{\circ}-\beta$. We can find the angle $D$ in the triangle ABD.
$\alpha+(180^{\circ}-\beta)+D = 180^{\circ}$
$D = 180^{\circ}-\alpha-(180^{\circ}-\beta)$
$D = \beta-\alpha$
We can find an expression for the side $c$ which is opposite the angle $\alpha$:
$\frac{c}{sin~\alpha} = \frac{d}{sin~D}$
$c = \frac{d~sin~\alpha}{sin~(\beta-\alpha)}$
We can find an expression for $x$:
$\frac{x}{sin~\beta} = \frac{c}{sin~90^{\circ}}$
$x = \frac{c~sin~\beta}{1}$
$x = \frac{d~sin~\alpha~sin~\beta}{sin~(\beta-\alpha)}$
