Two-dimensional routes and right-triangle models
Build the correct right triangle for bearings, diagonals and perpendicular offsets.
Section Paper 1 · Core · Non-calculator
Core and Extended: 2D right-triangle problems and bearings. Extended also explicitly requires knowing that perpendicular distance is the shortest point-to-line distance. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Read three-figure bearings.
- Use Pythagoras and trigonometric ratios.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which directions form perpendicular legs on a map?
What you will learn
- Convert a described route into perpendicular components.
- Use bearings and right triangles without confusing the starting point.
Choose the triangle before choosing the formula
A route going 6 km east and 8 km north creates perpendicular legs, so the direct distance is √(6² + 8²) = 10 km. Its bearing from the start is measured clockwise from north, giving tan θ = east/north = 6/8 and θ ≈ 36.9°, written as a bearing about 037°. The return bearing differs by 180°.
For a 10 km journey on bearing 060°, the north component is 10 cos 60° = 5 km and east component is 10 sin 60° = 5√3 km. The reference angle is from north, so the north leg is adjacent. A direction west or south needs the appropriate signed component; a negative coordinate is direction, not negative physical distance.
Combine steps without losing the geometry
A rectangle's diagonal may be found first and then used as a side in another right triangle. Label intermediate points so you know which segment a result describes. A drawing can be schematic, but each claimed right angle must follow from the context or stated geometry, such as north being perpendicular to east.
Extended: the perpendicular from a point to a line is its shortest distance. Any sloping connection forms a right triangle with the perpendicular and an extra along-line leg, so its length is at least the perpendicular length by Pythagoras. This prevents using a slant measurement when a problem asks for distance to a road or boundary.
Pause and explain
A 10 km route has bearing 060°. Find its northward component.
Worked example
A walker goes 6 km east and 8 km north. Find direct distance and the approximate three-figure bearing from the starting point.
Show the worked solution
- The direct distance is √(36 + 64) = 10 km.
- The bearing angle from north is tan⁻¹(6/8) ≈ 36.9°.
- For a whole-degree three-figure bearing, write 037°; the direction is northeast.
Answer 10 km; bearing approximately 037°.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A route is 9 km east then 12 km north. Find direct distance.
Give me a hint
Use the two perpendicular legs.
Compare my reasoning
- The squared direct length is 81 + 144 = 225.
- The positive root is 15 km.
- It is shorter than the travelled 21 km route, as expected.
15 km.
Look for these in your work
- Used cardinal perpendicularity.
- Distinguished direct and travelled distance.
A journey of 20 km has bearing 030°. Find north and east components.
Give me a hint
The bearing is measured from north.
Compare my reasoning
- North component is 20 cos 30° = 10√3 km.
- East component is 20 sin 30° = 10 km.
- Both components are positive in the northeast direction.
North 10√3 km; east 10 km.
Look for these in your work
- Matched adjacent with north.
- Applied the chosen reference angle consistently.
Extended: a point is 5 m perpendicularly from a straight road. Another road point is 12 m along the road from the perpendicular foot. Find the sloping distance and explain which is shortest.
Give me a hint
Use a 5–12 right triangle.
Compare my reasoning
- The sloping length is √(25 + 144) = 13 m.
- The perpendicular distance remains 5 m.
- Any extra along-road displacement adds a non-negative square, so it cannot shorten the connection.
Sloping distance 13 m; shortest distance 5 m (Extended reasoning).
Look for these in your work
- Identified the perpendicular foot.
- Explained the shortest-distance condition.
Common mistakes
- Using east as the bearing reference.
- Confusing distance travelled with direct distance.
- Calling a sloping connection the perpendicular distance.
What can you explain now?
Which component is adjacent to a bearing measured from north?
Compare with the explanation
The north–south component.
A bearing's reference line is north, so the adjacent leg follows that axis rather than the east–west axis.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, sketch a bearing route and label its north and east components before calculating.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.