Ans: C
In $\Delta ABD$,
In $\Delta ABD$,
$\begin{array}{rcl}
\sin 40^\circ & = & \dfrac{AD}{12} \\
AD & = & 12 \sin 40^\circ
\end{array}$
In $\Delta ACD$,
$\begin{array}{rcl}
\cos 20^\circ & = & \dfrac{AD}{x} \\
x & = & \dfrac{12 \sin 40^\circ}{\cos 20^\circ} \\
& = & 8.208~483~44 \\
& \approx & 8.21
\end{array}$
