Ans: A
Add a point $X$ on the figure as shown.
Add a point $X$ on the figure as shown.
In $\Delta ABX$, we have
$\begin{array}{rcl}
\sin \angle XAB & = & \dfrac{10}{20} \\
\angle XAB & = & 30^\circ
\end{array}$
Therefore, the bearing of $B$ from $A$ is $030^\circ$.
