WAEC 2012 · Paper 2 · Q8

Elevation, depression & bearings

A point $H$ is $20\text{ m}$ away from the foot $F$ of a tower on the same horizontal ground. From the point $H$ , the angles of elevation of a point $P$ on the tower and the top $T$ of the tower are $30^\circ$ and $50^\circ$ respectively. Calculate, correct to 3 significant figures:

(a)
$|PT|$ ;
(b)
the distance between $H$ and the top of the tower;
(c)
how far $H$ must be from the foot of the tower if the angle of depression of $H$ from the top of the tower is to be $40^\circ$ .

Worked solution (try it first)

Draw the tower $FT$ vertical, with $P$ on it, and $H$ on the ground 20 m from $F$ .
$\angle FHP = 30^\circ$ and $\angle FHT = 50^\circ$ , both measured up from the ground.

(a)

In triangle $HFT$ : $|FT| = 20\tan 50^\circ \approx 23.84$ m.
In triangle $HFP$ : $|FP| = 20\tan 30^\circ \approx 11.55$ m.
So $|PT| = |FT| - |FP|$
$\approx 23.84 - 11.55$
$= 12.29$
$\approx 12.3\text{ m}$ .

(b)

$HT$ is the hypotenuse of triangle $HFT$ : $\cos 50^\circ = \frac{20}{|HT|}$ , so $|HT| = \frac{20}{\cos 50^\circ}$
$\approx 31.1\text{ m}$ .

(c)

An angle of depression of $40^\circ$ from $T$ equals an angle of elevation of $40^\circ$ from the new point (alternate angles).
So $\tan 40^\circ = \frac{|FT|}{d}$ , and $d = \frac{23.84}{\tan 40^\circ}$
$\approx 28.4\text{ m}$ .

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