JAMB 1993 · UME · Q40

Elevation, depression & bearings

In the figure, $P$ , $Q$ and $H$ are on level ground with $PQ = 5$ m, and $TH$ is vertical. The angles of elevation of $T$ from $P$ and $Q$ are $30^\circ$ and $45^\circ$ . Calculate $TH$ .

Worked solution (try it first)

Let $TH = h$ .
At $Q$ the angle is $45^\circ$ , so $QH = h$ .
At $P$ : $\tan30^\circ = \frac{h}{PH}$ , so $PH = \frac{h}{\tan30^\circ} = h\sqrt3$ .
$P$ is further from $H$ than $Q$ by $PQ$ : $h\sqrt3 - h = 5$ , so $h(\sqrt3 - 1) = 5$ .
So $TH = \dfrac{5}{\sqrt3 - 1}$ m, option B.

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