## Trigonometry (11th Edition) Clone

$x = \frac{d~sin~\alpha~sin~\beta}{sin~(\beta-\alpha)}$
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 $x$: $\frac{x}{sin~\beta} = \frac{c}{sin~90^{\circ}}$ $x = \frac{c~sin~\beta}{1}$ $x = \frac{d~sin~\alpha~sin~\beta}{sin~(\beta-\alpha)}$