Do I need to solve the sides?

At first, I drew a height from $D$ to $\overline{AB}$ , intersecting at $F$ so $\angle BDF=\alpha + \beta$ . We have $cos(\alpha +\beta)=\frac{4}{5}=\overline{ED}$ $\rightarrow sin(\alpha)=\frac{\frac{4}{5}}{4} = \frac{1}{5}$

$\angle ABC=90-(\alpha+\beta)$

Appliying sine law at $\triangle ABC$ , $\frac{\sin\alpha}{BC}=\sin{(90-(\alpha+\beta))}{AC}\Rightarrow\frac{\sin\alpha}{1}=\frac{\sin{(90-(\alpha+\beta))}}{4}$ but $\sin (90-(\alpha+\beta))=\cos(90-(\alpha+\beta))$

$\therefore \sin\alpha=\frac{\cos (\alpha+\beta)}{4}=\frac{4}{5}\times \frac{1}{4}=\boxed{\frac{1}{5}}$