High School Maths · Year 2 · Unit 11 · Lesson 5 of 5

Bearings and 3D Problems: Navigate Without a Right Angle

About 15 minutes · north lines and back bearings · the angle inside the triangle at a turn · the cosine and sine rules with bearings · finding a bearing at the end of a journey · triangles inside cuboids and pyramids · angles of elevation from two directions · calculator allowed · exit quiz

Hook

A ship sails 12 km from port P on a bearing of 050° to Q. Then it turns and sails 18 km on a bearing of 130° to R.

How far is the ship from port now? The two legs do not meet at a right angle.

About 23.3 km. At Q, the bearing back to P is 050° + 180° = 230°, so the angle inside the triangle at Q is 230° − 130° = 100°. Two sides and the angle between them: the cosine rule gives PR² = 12² + 18² − 2 × 12 × 18 × cos 100° = 543.0…, so PR = 23.30… km. In Year 1 you solved bearings problems with right-angled triangles. Now any turn will do.

Learn

Master the essentials

Back bearing at B: 040° + 180° = 220°. Angle ABC = 220° − 150° = 70°.

North lines

Draw a north line at every point. A bearing is measured clockwise from north and has three figures.

A bearing of 65° is written 065°.

Back bearings

The bearing back is 180° more, or 180° less if that gives more than 360°.

065° there, 245° back

Angle at a turn

At the turning point, the angle inside the triangle is the back bearing minus the new bearing.

040° then 150°: 220° − 150° = 70°

Parallel north lines

All north lines are parallel, so co-interior angles between them add up to 180°.

Use this to check an angle at a turn.

Then choose a rule

Two distances and the angle between them: the cosine rule. A side with its opposite angle: the sine rule.

The same rules, in a new setting.

A bearing at the end

Find an angle inside the triangle, then add it to, or take it from, a known bearing.

A sketch shows which way to turn.

Triangles in 3D

Pick out the triangle you need and draw it flat, with its true lengths.

Find each length first, often with Pythagoras.

Two steps in 3D

Pythagoras gives lengths in the faces. The cosine rule then gives an angle in a sloping triangle.

Face diagonals of a cuboid make a triangle with no right angle.

Worked examples

See it done

1. A ship sails 15 km from port P on a bearing of 060° to Q. It then sails 22 km on a bearing of 145° to R. How far is R from P? Give your answer correct to 3 significant figures.

  1. At Q, the bearing back to P is 060° + 180° = 240°.
  2. Angle PQR = 240° − 145° = 95°.
  3. Two sides and the angle between them: PR² = 15² + 22² − 2 × 15 × 22 × cos 95°.
  4. PR² = 225 + 484 + 57.52… = 766.5….
  5. PR = 27.68… = 27.7 km.

27.7 km

2. A boat sails 20 km from P on a bearing of 075° to Q, then 13 km on a bearing of 200° to R. Find the bearing of R from P, correct to 1 decimal place.

  1. At Q, the back bearing to P is 255°, so angle PQR = 255° − 200° = 55°.
  2. Cosine rule: PR² = 20² + 13² − 2 × 20 × 13 × cos 55°, so PR = 16.45… km.
  3. Sine rule: sin QPR = 13 × sin 55° ÷ 16.45…, so angle QPR = 40.3°. It is acute, because 13 km is not the longest side.
  4. R is clockwise from Q, seen from P: bearing = 075° + 40.3° = 115.3°.

115.3°

3. A cuboid ABCDEFGH has base ABCD, with E, F, G and H directly above A, B, C and D. AB = 9 cm, BC = 4 cm and AE = 6 cm. Find angle FAH, correct to 1 decimal place.

  1. Each side of triangle AFH is a face diagonal. Use Pythagoras on each face.
  2. AF² = 9² + 6² = 117, AH² = 4² + 6² = 52 and FH² = 9² + 4² = 97.
  3. Cosine rule: cos FAH = (117 + 52 − 97) ÷ (2 × √117 × √52).
  4. √117 × √52 = √6,084 = 78, so cos FAH = 72 ÷ 156 = 6 ÷ 13.
  5. Angle FAH = cos⁻¹(6 ÷ 13) = 62.5°.

62.5°

Common mistake: Using a bearing as if it were the angle in the triangle. A bearing is measured clockwise from north, not along a side of the triangle. Draw a north line at each point and work out the angle inside the triangle (back bearing minus new bearing) before you use any rule.

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed: check it is in degrees mode. Draw a sketch with a north line at each point, or pick out the flat triangle inside the solid. Give lengths correct to 3 significant figures and angles and bearings correct to 1 decimal place. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.The bearing of B from A is 072°. Find the bearing of A from B.

Not quite. This is 180° − 72°. The back bearing is 180° MORE than 072°.

Solution

  1. To face back along the same line, turn through 180°.
  2. 072° + 180° = 252°.

Answer: B. 252°.

Correct.

Solution

  1. To face back along the same line, turn through 180°.
  2. 072° + 180° = 252°.

Not quite. This is 360° − 72°. Add 180° instead.

Solution

  1. To face back along the same line, turn through 180°.
  2. 072° + 180° = 252°.

Answer: B. 252°.

Not quite. This adds 90°. Facing back means turning through 180°.

Solution

  1. To face back along the same line, turn through 180°.
  2. 072° + 180° = 252°.

Answer: B. 252°.

2.A path goes from A on a bearing of 035° to B, then turns onto a bearing of 120° towards C. Find angle ABC, the angle inside the triangle at B.

Not quite. This is 120° − 35°. At B, use the back bearing to A: 215° − 120°.

Solution

  1. At B, the bearing back to A is 035° + 180° = 215°.
  2. Angle ABC = 215° − 120° = 95°.
  3. Check: 145° + 120° + 95° = 360° around B ✓.

Answer: C. 95°.

Not quite. This adds the two bearings.

Solution

  1. At B, the bearing back to A is 035° + 180° = 215°.
  2. Angle ABC = 215° − 120° = 95°.
  3. Check: 145° + 120° + 95° = 360° around B ✓.

Answer: C. 95°.

Correct.

Solution

  1. At B, the bearing back to A is 035° + 180° = 215°.
  2. Angle ABC = 215° − 120° = 95°.
  3. Check: 145° + 120° + 95° = 360° around B ✓.

Not quite. This is only the angle between the north line at B and the path back to A.

Solution

  1. At B, the bearing back to A is 035° + 180° = 215°.
  2. Angle ABC = 215° − 120° = 95°.
  3. Check: 145° + 120° + 95° = 360° around B ✓.

Answer: C. 95°.

3.Point B is 6.5 km from A on a bearing of 030°. Point C is 4.2 km from A on a bearing of 130°. Which working shows that BC = 8.33 km, correct to 3 significant figures?

Not quite. This adds the bearings. The angle at A is 130° − 30° = 100°.

Solution

  1. Both bearings are measured from the same north line at A.
  2. Angle BAC = 130° − 30° = 100°.
  3. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100° = 69.37….
  4. BC = 8.328… = 8.33 km ✓.

Answer: C. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100°.

Not quite. This is Pythagoras. The angle at A is not 90°.

Solution

  1. Both bearings are measured from the same north line at A.
  2. Angle BAC = 130° − 30° = 100°.
  3. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100° = 69.37….
  4. BC = 8.328… = 8.33 km ✓.

Answer: C. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100°.

Correct.

Solution

  1. Both bearings are measured from the same north line at A.
  2. Angle BAC = 130° − 30° = 100°.
  3. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100° = 69.37….
  4. BC = 8.328… = 8.33 km ✓.

Not quite. This gives BC², not BC. You still need the square root.

Solution

  1. Both bearings are measured from the same north line at A.
  2. Angle BAC = 130° − 30° = 100°.
  3. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100° = 69.37….
  4. BC = 8.328… = 8.33 km ✓.

Answer: C. BC² = 6.5² + 4.2² − 2 × 6.5 × 4.2 × cos 100°.

Exam standard (Several steps, like real exam questions.)

4.A ship sails 18 km from harbour H on a bearing of 064° to A. It then sails 25 km on a bearing of 158° to B. How far is B from H? Give your answer correct to 3 significant figures.

Not quite. This uses 158° − 64° = 94°. At A, use the back bearing: 244° − 158° = 86°.

Solution

  1. At A, the bearing back to H is 064° + 180° = 244°.
  2. Angle HAB = 244° − 158° = 86°.
  3. HB² = 18² + 25² − 2 × 18 × 25 × cos 86° = 886.2….
  4. HB = 29.76… = 29.8 km.

Answer: D. 29.8 km.

Not quite. This uses Pythagoras. The angle at A is 86°, not 90°.

Solution

  1. At A, the bearing back to H is 064° + 180° = 244°.
  2. Angle HAB = 244° − 158° = 86°.
  3. HB² = 18² + 25² − 2 × 18 × 25 × cos 86° = 886.2….
  4. HB = 29.76… = 29.8 km.

Answer: D. 29.8 km.

Not quite. This is HB². Take the square root.

Solution

  1. At A, the bearing back to H is 064° + 180° = 244°.
  2. Angle HAB = 244° − 158° = 86°.
  3. HB² = 18² + 25² − 2 × 18 × 25 × cos 86° = 886.2….
  4. HB = 29.76… = 29.8 km.

Answer: D. 29.8 km.

Correct.

Solution

  1. At A, the bearing back to H is 064° + 180° = 244°.
  2. Angle HAB = 244° − 158° = 86°.
  3. HB² = 18² + 25² − 2 × 18 × 25 × cos 86° = 886.2….
  4. HB = 29.76… = 29.8 km.

5.The table shows three walking routes. Each route has two straight legs. For route Y, how far is the end of the walk from the start? Give your answer correct to 3 significant figures.

RouteFirst legSecond leg
X13 km on 080°8 km on 200°
Y16 km on 020°11 km on 135°
Z21 km on 310°15 km on 030°

Each route has two straight legs. Bearings are measured clockwise from north.

Correct.

Solution

  1. Route Y: 16 km on 020°, then 11 km on 135°.
  2. At the turn, the back bearing is 200°, so the angle is 200° − 135° = 65°.
  3. d² = 16² + 11² − 2 × 16 × 11 × cos 65° = 228.2….
  4. d = 15.10… = 15.1 km.

Not quite. This uses route X. Read the row for route Y.

Solution

  1. Route Y: 16 km on 020°, then 11 km on 135°.
  2. At the turn, the back bearing is 200°, so the angle is 200° − 135° = 65°.
  3. d² = 16² + 11² − 2 × 16 × 11 × cos 65° = 228.2….
  4. d = 15.10… = 15.1 km.

Answer: A. 15.1 km.

Not quite. This uses 135° − 20° = 115°. At the turn, the angle is 200° − 135° = 65°.

Solution

  1. Route Y: 16 km on 020°, then 11 km on 135°.
  2. At the turn, the back bearing is 200°, so the angle is 200° − 135° = 65°.
  3. d² = 16² + 11² − 2 × 16 × 11 × cos 65° = 228.2….
  4. d = 15.10… = 15.1 km.

Answer: A. 15.1 km.

Not quite. This uses Pythagoras. The angle at the turn is 65°, not 90°.

Solution

  1. Route Y: 16 km on 020°, then 11 km on 135°.
  2. At the turn, the back bearing is 200°, so the angle is 200° − 135° = 65°.
  3. d² = 16² + 11² − 2 × 16 × 11 × cos 65° = 228.2….
  4. d = 15.10… = 15.1 km.

Answer: A. 15.1 km.

6.A boat sails 9.5 km from A on a bearing of 042° to B. It then sails 13 km on a bearing of 128° to C. Find the bearing of C from A, correct to 1 decimal place.

Correct.

Solution

  1. At B, the back bearing is 222°, so angle ABC = 222° − 128° = 94°.
  2. Cosine rule: AC² = 9.5² + 13² − 2 × 9.5 × 13 × cos 94°, so AC = 16.62… km.
  3. Sine rule: sin BAC = 13 × sin 94° ÷ 16.62…, so angle BAC = 51.3°.
  4. Bearing of C from A = 042° + 51.3° = 093.3°.

Not quite. This is angle BAC. Add it to the bearing 042° of B from A.

Solution

  1. At B, the back bearing is 222°, so angle ABC = 222° − 128° = 94°.
  2. Cosine rule: AC² = 9.5² + 13² − 2 × 9.5 × 13 × cos 94°, so AC = 16.62… km.
  3. Sine rule: sin BAC = 13 × sin 94° ÷ 16.62…, so angle BAC = 51.3°.
  4. Bearing of C from A = 042° + 51.3° = 093.3°.

Answer: A. 093.3°.

Not quite. C is clockwise from B, seen from A. Add angle BAC, do not subtract it.

Solution

  1. At B, the back bearing is 222°, so angle ABC = 222° − 128° = 94°.
  2. Cosine rule: AC² = 9.5² + 13² − 2 × 9.5 × 13 × cos 94°, so AC = 16.62… km.
  3. Sine rule: sin BAC = 13 × sin 94° ÷ 16.62…, so angle BAC = 51.3°.
  4. Bearing of C from A = 042° + 51.3° = 093.3°.

Answer: A. 093.3°.

Not quite. This is the bearing of A from C, the back bearing.

Solution

  1. At B, the back bearing is 222°, so angle ABC = 222° − 128° = 94°.
  2. Cosine rule: AC² = 9.5² + 13² − 2 × 9.5 × 13 × cos 94°, so AC = 16.62… km.
  3. Sine rule: sin BAC = 13 × sin 94° ÷ 16.62…, so angle BAC = 51.3°.
  4. Bearing of C from A = 042° + 51.3° = 093.3°.

Answer: A. 093.3°.

7.A cuboid ABCDEFGH has base ABCD, with E, F, G and H directly above A, B, C and D. AB = 8 cm, BC = 5 cm and AE = 4 cm. Find angle FAH, correct to 1 decimal place.

Not quite. This is angle AHF. Angle FAH is opposite FH, so subtract FH².

Solution

  1. AF² = 8² + 4² = 80, AH² = 5² + 4² = 41 and FH² = 8² + 5² = 89.
  2. Angle FAH is opposite FH.
  3. cos FAH = (80 + 41 − 89) ÷ (2 × √80 × √41) = 0.2793….
  4. Angle FAH = 73.8°.

Answer: D. 73.8°.

Not quite. This is cos FAH. Use cos⁻¹ to find the angle.

Solution

  1. AF² = 8² + 4² = 80, AH² = 5² + 4² = 41 and FH² = 8² + 5² = 89.
  2. Angle FAH is opposite FH.
  3. cos FAH = (80 + 41 − 89) ÷ (2 × √80 × √41) = 0.2793….
  4. Angle FAH = 73.8°.

Answer: D. 73.8°.

Not quite. This takes FH² as 8² + 4² = 80. FH is the top diagonal: 8² + 5² = 89.

Solution

  1. AF² = 8² + 4² = 80, AH² = 5² + 4² = 41 and FH² = 8² + 5² = 89.
  2. Angle FAH is opposite FH.
  3. cos FAH = (80 + 41 − 89) ÷ (2 × √80 × √41) = 0.2793….
  4. Angle FAH = 73.8°.

Answer: D. 73.8°.

Correct.

Solution

  1. AF² = 8² + 4² = 80, AH² = 5² + 4² = 41 and FH² = 8² + 5² = 89.
  2. Angle FAH is opposite FH.
  3. cos FAH = (80 + 41 − 89) ÷ (2 × √80 × √41) = 0.2793….
  4. Angle FAH = 73.8°.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A square-based pyramid VABCD has a base with sides of 7 cm. Each sloping edge VA, VB, VC and VD is 10 cm. Find angle AVC, between two opposite sloping edges, correct to 1 decimal place.

Not quite. This is angle AVB, between two edges next to each other. A and C are opposite corners.

Solution

  1. AC is the diagonal of the base: AC² = 7² + 7² = 98.
  2. Triangle AVC has VA = VC = 10 cm, with AC opposite angle V.
  3. cos AVC = (100 + 100 − 98) ÷ (2 × 10 × 10) = 102 ÷ 200 = 0.51.
  4. Angle AVC = cos⁻¹(0.51) = 59.3°.

Answer: B. 59.3°.

Correct.

Solution

  1. AC is the diagonal of the base: AC² = 7² + 7² = 98.
  2. Triangle AVC has VA = VC = 10 cm, with AC opposite angle V.
  3. cos AVC = (100 + 100 − 98) ÷ (2 × 10 × 10) = 102 ÷ 200 = 0.51.
  4. Angle AVC = cos⁻¹(0.51) = 59.3°.

Not quite. This takes AC = 7 + 7 = 14 cm. The diagonal of the square is √(7² + 7²).

Solution

  1. AC is the diagonal of the base: AC² = 7² + 7² = 98.
  2. Triangle AVC has VA = VC = 10 cm, with AC opposite angle V.
  3. cos AVC = (100 + 100 − 98) ÷ (2 × 10 × 10) = 102 ÷ 200 = 0.51.
  4. Angle AVC = cos⁻¹(0.51) = 59.3°.

Answer: B. 59.3°.

Not quite. This is cos AVC. Use cos⁻¹ to find the angle.

Solution

  1. AC is the diagonal of the base: AC² = 7² + 7² = 98.
  2. Triangle AVC has VA = VC = 10 cm, with AC opposite angle V.
  3. cos AVC = (100 + 100 − 98) ÷ (2 × 10 × 10) = 102 ÷ 200 = 0.51.
  4. Angle AVC = cos⁻¹(0.51) = 59.3°.

Answer: B. 59.3°.

9.A yacht must reach a buoy R, which is 14 km from port P on a bearing of 095°. It first sails 9 km on a bearing of 050° to a point Q. On what bearing must it now sail from Q to reach R? Give your answer correct to 1 decimal place.

Not quite. This is the bearing of Q from R, the back bearing.

Solution

  1. Angle QPR = 095° − 050° = 45°.
  2. Cosine rule: QR² = 9² + 14² − 2 × 9 × 14 × cos 45°, so QR = 9.940… km.
  3. Cosine rule again: cos PQR = (9² + QR² − 14²) ÷ (2 × 9 × QR) = −0.09048…, so angle PQR = 95.2°.
  4. From Q, P is on a bearing of 230°. So the bearing of R from Q = 230° − 95.2° = 134.8°.

Answer: C. 134.8°.

Not quite. This uses the acute angle 84.8° at Q. The cosine of angle PQR is negative, so the angle is obtuse.

Solution

  1. Angle QPR = 095° − 050° = 45°.
  2. Cosine rule: QR² = 9² + 14² − 2 × 9 × 14 × cos 45°, so QR = 9.940… km.
  3. Cosine rule again: cos PQR = (9² + QR² − 14²) ÷ (2 × 9 × QR) = −0.09048…, so angle PQR = 95.2°.
  4. From Q, P is on a bearing of 230°. So the bearing of R from Q = 230° − 95.2° = 134.8°.

Answer: C. 134.8°.

Correct.

Solution

  1. Angle QPR = 095° − 050° = 45°.
  2. Cosine rule: QR² = 9² + 14² − 2 × 9 × 14 × cos 45°, so QR = 9.940… km.
  3. Cosine rule again: cos PQR = (9² + QR² − 14²) ÷ (2 × 9 × QR) = −0.09048…, so angle PQR = 95.2°.
  4. From Q, P is on a bearing of 230°. So the bearing of R from Q = 230° − 95.2° = 134.8°.

Not quite. This turns the wrong way at Q. From the bearing 230° of P, R is anticlockwise: subtract angle PQR.

Solution

  1. Angle QPR = 095° − 050° = 45°.
  2. Cosine rule: QR² = 9² + 14² − 2 × 9 × 14 × cos 45°, so QR = 9.940… km.
  3. Cosine rule again: cos PQR = (9² + QR² − 14²) ÷ (2 × 9 × QR) = −0.09048…, so angle PQR = 95.2°.
  4. From Q, P is on a bearing of 230°. So the bearing of R from Q = 230° − 95.2° = 134.8°.

Answer: C. 134.8°.

10.A mast FT, 40 m tall, stands on level ground with its foot at F. Point A is on a bearing of 210° from F and point B is on a bearing of 320° from F. The angle of elevation of the top T is 32° from A and 25° from B. Find the distance AB, correct to 3 significant figures.

Not quite. This uses Pythagoras. Angle AFB is 320° − 210° = 110°, not 90°.

Solution

  1. FA = 40 ÷ tan 32° = 64.01… m and FB = 40 ÷ tan 25° = 85.78… m.
  2. On the ground, angle AFB = 320° − 210° = 110°.
  3. AB² = FA² + FB² − 2 × FA × FB × cos 110° = 15,210….
  4. AB = 123.3… = 123 m.

Answer: D. 123 m.

Not quite. This uses 40 × tan 32° for FA. FA is adjacent to the 32° angle: FA = 40 ÷ tan 32°.

Solution

  1. FA = 40 ÷ tan 32° = 64.01… m and FB = 40 ÷ tan 25° = 85.78… m.
  2. On the ground, angle AFB = 320° − 210° = 110°.
  3. AB² = FA² + FB² − 2 × FA × FB × cos 110° = 15,210….
  4. AB = 123.3… = 123 m.

Answer: D. 123 m.

Not quite. This uses 180° − 110° = 70°. The angle AFB between the two directions is 110°.

Solution

  1. FA = 40 ÷ tan 32° = 64.01… m and FB = 40 ÷ tan 25° = 85.78… m.
  2. On the ground, angle AFB = 320° − 210° = 110°.
  3. AB² = FA² + FB² − 2 × FA × FB × cos 110° = 15,210….
  4. AB = 123.3… = 123 m.

Answer: D. 123 m.

Correct.

Solution

  1. FA = 40 ÷ tan 32° = 64.01… m and FB = 40 ÷ tan 25° = 85.78… m.
  2. On the ground, angle AFB = 320° − 210° = 110°.
  3. AB² = FA² + FB² − 2 × FA × FB × cos 110° = 15,210….
  4. AB = 123.3… = 123 m.

Apply

Flying back to base

A rescue helicopter flies 40 km from its base B on a bearing of 300° to a point P. It then flies 55 km on a bearing of 045° to a point Q. A calculator is allowed.

1. Find angle BPQ.

2. Find the distance from Q straight back to B, correct to 3 significant figures.

3. Find the bearing on which the helicopter must fly from Q back to B, correct to 1 decimal place.

  1. At P, the bearing back to B is 300° − 180° = 120°. Angle BPQ = 120° − 45° = 75°.
  2. Cosine rule: BQ² = 40² + 55² − 2 × 40 × 55 × cos 75° = 3,486…, so BQ = 59.04… = 59.0 km.
  3. Sine rule: sin BQP = 40 × sin 75° ÷ 59.04…, so angle BQP = 40.9° (acute: 40 km is shorter than BQ). From Q, P is on a bearing of 045° + 180° = 225°, and B is anticlockwise from P. Bearing of B from Q = 225° − 40.9° = 184.1°.

Next: the Lesson 5 exit quiz, then the unit test. The unit test has 17 multiple-choice questions (no calculator) and five structured problems (calculator allowed), 43 marks in all. Allow 70 minutes.

PreviousNext