WAEC 2016 · Paper 2 · Q8

An eagle flies from point $X$ to $Y$ , a distance of $320\text{ km}$ , on a bearing of N $35^\circ$ W. It then changes direction and flies $550\text{ km}$ to point $Z$ on a bearing of S $55^\circ$ W. It then returns to its starting point $X$ from $Z$ .

(a)
Illustrate the information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw a north line at each point before measuring a bearing. From $X$ , $Y$ is 320 km on N $35^\circ$ W ( $35^\circ$ west of north). From $Y$ , $Z$ is 550 km on S $55^\circ$ W ( $55^\circ$ west of south). The angle at $Y$ is $35^\circ + 55^\circ = 90^\circ$ , which makes the calculation easy. Join $Z$ back to $X$ .
(b)
Calculate the: (i) total distance covered, correct to 3 significant figures; (ii) bearing of $X$ from $Z$ , correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

N $35^\circ$ W is the bearing $325^\circ$ , and S $55^\circ$ W is $235^\circ$ .
Draw north at $X$ and draw $XY$ (320 km) on $325^\circ$ .
Draw north at $Y$ and draw $YZ$ (550 km) on $235^\circ$ .
Join $Z$ to $X$ .

(b)(i)

At $Y$ , the direction back to $X$ is $325^\circ - 180^\circ = 145^\circ$ and the direction to $Z$ is $235^\circ$ , so $\angle XYZ = 235^\circ - 145^\circ$
$= 90^\circ$ .
Pythagoras: $|XZ| = \sqrt{320^2 + 550^2}$
$= \sqrt{404\,900}$
$\approx 636.3$ km.
The total distance flown is $320 + 550 + 636.3 = 1506.3 \approx 1510$ km (3 significant figures).

(ii)

In the right-angled triangle, at $Z$ : $\tan\angle YZX = \frac{320}{550}$ , so $\angle YZX \approx 30.2^\circ$ .
At $Z$ , the direction to $Y$ is the back bearing of $235^\circ$ , which is $055^\circ$ .
$X$ is $30.2^\circ$ further round clockwise: $055^\circ + 30.2^\circ \approx 085^\circ$ (N $85^\circ$ E).

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