WAEC 2021 · Paper 2 · Q10

(a)
A cottage is on a bearing of $200^\circ$ and $110^\circ$ from Dogbe's and Mamu's farms respectively. If Dogbe walked $5\text{ km}$ and Mamu $3\text{ km}$ from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms; (ii) the nearest degree, the bearing of Mamu's farm from Dogbe's.
Separate values with commas, e.g. 3, −2
(b)
A ladder $10\text{ m}$ long leaned against a vertical wall $x\text{ m}$ high. The distance between the wall and the foot of the ladder is $2\text{ m}$ longer than the height of the wall. Calculate the value of $x$ .

Worked solution (try it first)

The bearings given are from the farms to the cottage, so turn them round: from the cottage $C$ , Dogbe's farm $D$ is on $200^\circ - 180^\circ = 020^\circ$ (5 km) and Mamu's farm $M$ is on $110^\circ + 180^\circ = 290^\circ$ (3 km).
The angle between these at $C$ is $360^\circ - 290^\circ + 20^\circ = 90^\circ$ .

At $D$ : $\tan\angle CDM = \frac35$ , so $\angle CDM \approx 31.0^\circ$ .
At $D$ , the cottage is on $200^\circ$ and Mamu's farm is $31.0^\circ$ further round clockwise: $200^\circ + 31.0^\circ \approx 231^\circ$ .

The wall ( $x$ m), the ground ( $x + 2$ m) and the ladder (10 m) make a right-angled triangle: $x^2 + (x + 2)^2 = 10^2$ .
Expand: $2x^2 + 4x + 4 = 100$ , so $x^2 + 2x - 48 = 0$ .
Factorise: $(x - 6)(x + 8) = 0$ .
A height can't be negative, so $x = 6$ .