Ans: A
In $\Delta BCD$,
In $\Delta BCD$,
$\begin{array}{rcl}
\sin \alpha & = & \dfrac{BD}{BC} \\
BD & = & \ell \sin \alpha
\end{array}$
In $\Delta ABD$,
$\begin{array}{rcl}
\cos \beta & = & \dfrac{BD}{AB} \\
AB & = & \dfrac{BD}{\cos\beta} \\
& = & \dfrac{\ell \sin \alpha}{\cos\beta}
\end{array}$