JAMB 2001 · UME · Q31

Elevation, depression & bearings

The bearings of $P$ and $Q$ from a common point $N$ are $020^\circ$ and $300^\circ$ respectively. If $P$ and $Q$ are also equidistant from $N$ , find the bearing of $P$ from $Q$ .

Worked solution (try it first)

The angle between the bearings $300^\circ$ and $020^\circ$ (through north) is $60^\circ + 20^\circ = 80^\circ$ .
$NP = NQ$ , so the base angles of triangle $NPQ$ are $\frac{180^\circ - 80^\circ}{2} = 50^\circ$ .
The bearing of $N$ from $Q$ is the back bearing $300^\circ - 180^\circ = 120^\circ$ .
$P$ is $50^\circ$ anticlockwise from that direction, so the bearing of $P$ from $Q$ is $120^\circ - 50^\circ = 070^\circ$ , option C.

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