2011-II-15

Ans: A
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X

on the figure as shown.

In $Δ A B X$ , we have

$\begin{array}{rcl} \sin ∠ X A B & = & \frac{10}{20} \\ ∠ X A B & = & 30^{\circ} \end{array}$

Therefore, the bearing of $B$ from $A$ is $030^{\circ}$ .