WAEC 2023 · Paper 2 · Q8

(a)
A tree is $8\text{ km}$ due south of a building. Musa is standing $8\text{ km}$ due west of the tree. (i) Illustrate the information on a diagram. (ii) How far is Musa from the building? (iii) Find the bearing of Musa from the building.
Separate values with commas, e.g. 3, −2
(b)
If a man rides a bicycle from his house at $5\text{ km/h}$ , he will get to the office 45 minutes later than when he rides at $6\text{ km/h}$ . Calculate the distance between his house and the office.

Worked solution (try it first)

Draw the building $B$ , the tree $T$ 8 km due south of it, and Musa $M$ 8 km due west of the tree, so the angle at $T$ is a right angle.

By Pythagoras, $|BM| = \sqrt{8^2 + 8^2}$
$= \sqrt{128}$
$\approx 11.31\text{ km}$ .

Triangle $BTM$ is right-angled and isosceles, so $\angle TBM = 45^\circ$ .
From $B$ , Musa is $45^\circ$ west of due south.
Bearings are measured clockwise from north: $180^\circ + 45^\circ = 225^\circ$ .

Let the distance be $D$ km.
At 5 km/h the ride takes $\frac D5$ hours.
At 6 km/h it takes $\frac D6$ hours.
The slower ride takes 45 minutes $= \frac34$ hour longer: $\frac D5 - \frac D6 = \frac34$ .
Multiply by 60: $12D - 10D = 45$ .
So $2D = 45$ and $D = 22.5$ .
The distance is 22.5 km.