2006-II-16
Ans: A
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Since the bearing of $Q$ from $A$ is $\text{S}28^\cir
Since the bearing of $Q$ from $A$ is $\text{S}28^\cir
$\begin{array}{rcl}
\alpha + 110^\circ & =
$\begin{array}{rcl}
\angle ABC
$\beg
$\b
$\begin{array}{rcl}
\tan 30
Note that
$\begin{array}{rcl}
AC^2 + BC^2 & =
$\beg
$\beg
Add a point $E$ on $AB$ such that $AC//ED$. Then $AEDC$
Let $\theta$ be the angle of depression of $Q$ from $P$