WAEC 2010 · Paper 2 · Q2

(a)
The angle of depression of a boat from the mid-point of a vertical cliff is $35^\circ$ . If the boat is $120\text{ m}$ from the foot of the cliff, calculate the height of the cliff.
(b)
Towns $P$ and $Q$ are $x\text{ km}$ apart. Two motorists set out at the same time from $P$ to $Q$ at steady speeds of $60\text{ km/h}$ and $80\text{ km/h}$ . The faster motorist got to $Q$ 30 minutes earlier than the other. Find the value of $x$ .

Worked solution (try it first)

Draw the cliff with its mid-point $M$ at height $\frac h2$ above the foot $F$ , and the boat $B$ $120\text{ m}$ from $F$ .
The angle of depression at $M$ equals the angle of elevation at $B$ : $\angle MBF = 35^\circ$ .
In the right-angled triangle $MFB$ : $\tan 35^\circ = \dfrac{h/2}{120}$ .
So $\frac h2 = 120 \tan 35^\circ$
$= 120 \times 0.7002$
$= 84.02\text{ m}$ .
Double it for the whole cliff: $h = 2 \times 84.02 = 168.0\text{ m}$ .
The cliff is about $168\text{ m}$ high.

Time is distance over speed: the slower motorist takes $\frac{x}{60}$ hours and the faster one $\frac{x}{80}$ hours.
30 minutes is $\frac12$ hour, so $\dfrac{x}{60} - \dfrac{x}{80} = \dfrac12$ .
Multiply by 240 (the LCM of 60, 80 and 2): $4x - 3x = 120$ .
So $x = 120$ : the towns are $120\text{ km}$ apart.