WAEC 2015 · Paper 2 · Q10

(a)
Without using mathematical tables or calculators, simplify $\dfrac{2\tan60^\circ + \cos30^\circ}{\sin60^\circ}$ .
(b)
From an aeroplane in the air, at a horizontal distance of $1050\text{ m}$ from a control tower, the angles of depression of the top and base of the tower are $36^\circ$ and $41^\circ$ respectively. Calculate, correct to the nearest metre, the: (i) height of the control tower; (ii) shortest distance between the aeroplane and the base of the control tower.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Use the exact values $\tan 60^\circ = \sqrt3$ , $\cos 30^\circ = \frac{\sqrt3}{2}$ and $\sin 60^\circ = \frac{\sqrt3}{2}$ : $\frac{2\sqrt3 + \frac{\sqrt3}{2}}{\frac{\sqrt3}{2}}$ .
Multiply the top and bottom by 2: $\frac{4\sqrt3 + \sqrt3}{\sqrt3} = \frac{5\sqrt3}{\sqrt3}$
$= 5$ .

Draw the aeroplane $A$ above and 1050 m horizontally from the tower.
The angles of depression, $36^\circ$ to the top and $41^\circ$ to the base, are measured down from the horizontal through $A$ , and equal the angles of elevation from the ground level lines (alternate angles).

The drop from the aeroplane's height to the base of the tower is $1050\tan 41^\circ \approx 912.75$ m, and to the top of the tower is $1050\tan 36^\circ \approx 762.87$ m.
The tower is the difference: $912.75 - 762.87 \approx 150$ m.

The shortest distance from the aeroplane to the base is the straight line, the hypotenuse of the bigger triangle: $\cos 41^\circ = \frac{1050}{d}$ , so $d = \frac{1050}{0.7547} \approx 1391$ m.