Two-dimensional routes and right-triangle models
Build the correct right triangle for bearings, diagonals and perpendicular offsets.
Core and Extended: 2D right-triangle problems and bearings. Extended also explicitly requires knowing that perpendicular distance is the shortest point-to-line distance.
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°.
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources