KJSEA 2025 Maths Q37 — Bearings and Scale Drawing

KJSEA 2025 Grade 9 Geometry

Published

The Question

“From point A, a learner walked 50 m on a bearing of 120° to reach B. From B, she walked 70 m on a bearing of 210° to reach C. Using a scale of 1 cm to represent 10 m, show the positions of A, B and C, then find the distance and the bearing of C from A.”

1

Set up the scale

First turn the real distances into drawing lengths using the given scale. Every 10 metres on the ground becomes 1 centimetre on paper, so divide each distance by ten. This keeps the picture small enough to draw yet exactly in proportion to the real journey.

50 m÷10=5 cm(leg AB)50 \text{ m} \div 10 = 5 \text{ cm} \quad (\text{leg } AB)
70 m÷10=7 cm(leg BC)70 \text{ m} \div 10 = 7 \text{ cm} \quad (\text{leg } BC)

A bearing is always measured clockwise from the north line, and always written with three figures.

2

Draw the first leg to B

Mark point A near the top of your page and draw a north line straight up from it. Place your protractor centre on A, measure 120° clockwise from north, and rule a line 5 cm long. The end of that line is B. Because 120° is past east, this leg heads down and to the right.

Bearing of B from A=120∘\text{Bearing of } B \text{ from } A = 120^\circ
3

Draw the second leg to C

At B draw a fresh north line parallel to the one at A. Measure 210° clockwise from this new north line and rule a line 7 cm long to reach C. A new north line is needed at every turning point because bearings are always taken from north wherever you are standing.

Bearing of C from B=210∘\text{Bearing of } C \text{ from } B = 210^\circ
4

Join A to C and measure

Rule a straight line from A directly to C. Measure its length with a ruler and measure its bearing by drawing a north line at A and reading the clockwise angle to AC with a protractor. Then convert the drawing length back to real metres by multiplying by ten.

AC≈8.6 cmAC \approx 8.6 \text{ cm}
8.6 cm×10=86 m8.6 \text{ cm} \times 10 = 86 \text{ m}
Bearing of C from A≈174∘\text{Bearing of } C \text{ from } A \approx 174^\circ

Final Result

The line AC measures about 8.6 cm, so the real distance of C from A is about 86 metres, and the bearing of C from A is about 174°.

Why this method works

The scale drawing works because dividing every real length by ten shrinks the whole journey without changing any of its angles, so the paper triangle is an exact copy of the real one. Since bearing 120° and bearing 210° differ by exactly 90°, the path actually turns through a right angle at B, which is why the two legs meet neatly and the drawn triangle is easy to measure accurately.

✓

Because the turn at B is a right angle, Pythagoras confirms the drawing: AC = √(50² + 70²) = √7400 ≈ 86 m, matching the measured value.