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GCSE Maths · EdexcelChapter 22 · Constructions, loci, bearings and vectors
GCSE Maths · Loci, Bearings and Vectors · Higher · Lesson 22.7

GCSE Maths: Bearings problems: journeys, Pythagoras and trigonometry (Higher)

Part of the GCSE Maths loci, bearings and vectors chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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22.7

Bearings problems: journeys, Pythagoras and trigonometry

Edexcel 1MA1: G15, G20, G21
Grade 4-8

Many bearings questions describe a journey in two parts: a ship sails from A to B on one bearing, then from B to C on another. You are then asked how far C is from A, or what the bearing of C from A is. To solve these, link the bearings to triangles and use the tools you already know: angles in parallel lines, Pythagoras’ theorem and trigonometry (Chapter 20).

  1. Sketch the journey. Draw the points and the lines, and put a north line at every point where a bearing is given.
  2. Find the angles you need. Use parallel north lines (co-interior or alternate angles), the back bearing (± 180°), and angles in a triangle.
  3. Use a triangle. If there is a right angle, use Pythagoras for a length and sin, cos or tan for an angle. In Higher problems with no right angle, use the sine rule and cosine rule.
  4. Turn the angle back into a bearing. Add it to, or take it away from, a known bearing. Check your answer against the sketch: is it in the right quarter of the compass?

The example below follows these steps. A ship sails 8 km on a bearing of 070° from A to B, then 6 km on a bearing of 160° from B to C. The two bearings differ by 90°, so the path turns through a right angle at B, and triangle ABC has a right angle at B.

N070°N160°8 km6 kmABC
Bearings 070° and 160° differ by 90°, so angle ABC is a right angle. AC is the hypotenuse of a 6, 8, 10 triangle.

Using coordinates. If a place is a number of kilometres east and north of another, make a right-angled triangle from those steps. The angle from the north-south line is found with tan, then you convert it to a bearing: for the north-east quarter the bearing is the angle itself; for the south-east quarter it is 180° minus the angle; for the south-west quarter it is 180° plus the angle; for the north-west quarter it is 360° minus the angle.

Isosceles triangles. If two distances are equal (a ship sails the same distance on two legs), the triangle is isosceles. The base angles are equal, and the angle between the equal sides is the difference between the two bearings.

Higher tier: no right angle. If the path does not turn through exactly 90°, the angle at B in the triangle is 180° minus the turn. Use the cosine rule to find the third side from two sides and the angle between them, then the sine rule to find another angle.

Key ideas

  • Draw a north line at every point. North lines are parallel.
  • Back bearing: ± 180°. Angles between bearings from the same point are found by subtracting.
  • Right angle: Pythagoras for lengths, tan, sin, cos for angles. No right angle (Higher): cosine rule for a length, sine rule for an angle.
  • Turn an angle into a bearing by adding or subtracting from a known bearing. Check the quarter of the compass.
  • Give bearings to the nearest degree with three figures.
Worked example 22.7.1
A ship sails from A to B, 8 km on a bearing of 070°. It then sails from B to C, 6 km on a bearing of 160°. Work out (a) the distance AC, (b) the bearing of C from A, (c) the bearing of A from C.
N070°N160°8 km6 kmABC

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Draw the journey and mark the right angle where the ship turns. Use Pythagoras for a distance and trigonometry for an angle, then turn the angle into a bearing.

  1. Draw a sketch and put a north line at A and at B. A bearing is always measured from north, so we need a north line at every point we measure from.
    A to B: 8 km on a bearing of 070°
    B to C: 6 km on a bearing of 160°
    N070°N160°8 km6 kmABC
  2. Find angle ABC. The ship changes direction from 070° to 160°. The bearings differ by 90°, so the ship turns right through a quarter turn and the path makes a right angle at B.
    160° − 070° = 90°
    Angle ABC = 90°. Triangle ABC is right-angled.
  3. Use Pythagoras to find AC. AC is opposite the right angle, so it is the hypotenuse: add the squares, then square root.
    AC2 = 82 + 62 = 64 + 36 = 100
    AC = √100 = 10.0 km
  4. Find angle BAC with trigonometry. In the right-angled triangle we know the side opposite the angle (BC = 6) and the side next to it (AB = 8). Opposite and adjacent means tan.
    tan(angle BAC) = 6 ÷ 8
    angle BAC = tan−1(6 ÷ 8) = 36.9°
  5. Add to the bearing of B from A. C is further round clockwise from north than B is, so add the angle to 070°.
    70° + 36.9° = 106.9°
    Bearing of C from A = 107° (to the nearest degree)
  6. Find the back bearing. The bearing of A from C is the opposite direction to the bearing of C from A, so add or subtract 180°.
    107° + 180° = 287°
    Bearing of A from C = 287°

Check. Using coordinates from A (east, north): B = (7.52, 2.74) and C = (9.57, −2.90), so C is 10.0 km from A on a bearing of 107°. ✓

Watch out. The angle found with trigonometry is inside the triangle. You must add it to (or subtract it from) a bearing before you answer.

Worked example 22.7.2
A plane flies 9 km from P to Q on a bearing of 310°. It then flies 5 km from Q to R on a bearing of 220°. Work out (a) the distance PR, (b) the bearing of R from P.
N310°N220°9 km5 kmABC

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Draw the journey and mark the right angle where the ship turns. Use Pythagoras for a distance and trigonometry for an angle, then turn the angle into a bearing.

  1. Draw a sketch and put a north line at A and at B. A bearing is always measured from north, so we need a north line at every point we measure from.
    P to B: 9 km on a bearing of 310°
    B to C: 5 km on a bearing of 220°
    N310°N220°9 km5 kmABC
  2. Find angle ABC. The ship changes direction from 310° to 220°. The bearings differ by 90°, so the ship turns left through a quarter turn and the path makes a right angle at B.
    310° − 220° = 90°
    Angle ABC = 90°. Triangle ABC is right-angled.
  3. Use Pythagoras to find AC. AC is opposite the right angle, so it is the hypotenuse: add the squares, then square root.
    AC2 = 92 + 52 = 81 + 25 = 106
    AC = √106 = 10.3 km
  4. Find angle BAC with trigonometry. In the right-angled triangle we know the side opposite the angle (BC = 5) and the side next to it (AB = 9). Opposite and adjacent means tan.
    tan(angle BAC) = 5 ÷ 9
    angle BAC = tan−1(5 ÷ 9) = 29.1°
  5. Take away from the bearing of B from A. C is anticlockwise from B as seen from A, so subtract the angle from 310°.
    310° − 29.1° = 280.9°
    Bearing of C from A = 281° (to the nearest degree)

Check. Using coordinates from A (east, north): B = (−6.89, 5.79) and C = (−10.11, 1.95), so C is 10.3 km from A on a bearing of 281°. ✓

Watch out. The angle found with trigonometry is inside the triangle. You must add it to (or subtract it from) a bearing before you answer.

Worked example 22.7.3
A walker finishes 5 km east and 3 km south of her starting point A. Work out the bearing of her finishing point B from A.
N?5 km3 kmAB

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Draw the east and north steps as a right-angled triangle, find the angle with tan, then turn it into a bearing measured clockwise from north.

  1. Find the steps east and north. Draw a right-angled triangle from A to B: one side goes east or west, the other goes north or south.
    East/west step: 5 km east
    North/south step: 3 km south
  2. Find the angle from the north-south line. The east/west step is opposite the angle and the north/south step is next to it, so use tan.
    tan(angle) = 5 ÷ 3
    angle = tan−1(5 ÷ 3) = 59.0°
  3. Turn the angle into a bearing. Bearings are measured clockwise from north. Decide which quarter of the compass the point is in.
    Bearing = 180° − 59.0° = 121.0°
    To the nearest degree, bearing = 121°

Check. 121° points into the east half of the compass, which matches the picture. ✓

Watch out. Measure the angle from the north-south line, not from the east-west line.

Worked example 22.7.4
A, B and C are three buoys. AB = AC = 8 km. The bearing of B from A is 040° and the bearing of C from A is 100°. Work out the bearing of C from B.
N040°100°8 km8 kmABC

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. The triangle is isosceles, so find the base angles. Combine them with the back bearing of the first leg to get the new bearing.

  1. Draw a sketch with north lines. The north lines at different points are parallel.
    Bearing of B from A = 040°, bearing of C from A = 100°
    AB = AC = 8 km
  2. Find angle BAC. It is the difference between the two bearings measured from the same point.
    100° − 40° = 60°
  3. Find the base angles. AB = AC, so triangle ABC is isosceles. Its base angles are equal and the three angles add up to 180°.
    (180° − 60°) ÷ 2 = 120° ÷ 2 = 60°
  4. Find the bearing of A from B. This is the back bearing of the first bearing: add or subtract 180°.
    40° + 180° = 220°
  5. Turn from BA towards BC. At B the line BC is 60° anticlockwise from BA (the base angle), so subtract it.
    220° − 60° = 160°
    Bearing of C from B = 160°

Check. With coordinates: B = (5.14, 6.13) and C = (7.88, −1.39). The bearing of C from B is 160°. ✓

Watch out. Base angles of an isosceles triangle are the two angles beside the unequal side, not the angle between the equal sides.

Worked example 22.7.5
A boat sails 10 km on a bearing of 035° from A to B. It then sails 7 km on a bearing of 110° from B to C. Work out (a) the distance AC, (b) the bearing of C from A, (c) the bearing of A from C.
N035°N110°10 km7 kmABC

Key words. A bearing is an angle that gives a direction. It is measured from north, clockwise, and written with three figures (for example 045° or 270°). A back bearing is the bearing of the way back along the same line.

Method. Find the angle inside the triangle at the turning point from the two bearings. Then use the cosine rule for the missing side and the sine rule for an angle. Finish by turning the angle into a bearing.

  1. Draw a sketch with a north line at A and B. We measure every bearing from a north line.
    A to B: 10 km, bearing 035°
    B to C: 7 km, bearing 110°
  2. Find the bearing of A from B. This back bearing gives the direction of the line B to A. Add or subtract 180°.
    35° + 180° = 215°
  3. Find angle ABC. It is the difference between the bearings of A and C as seen from B (take the smaller angle round).
    215° − 110° = 105°
    Angle ABC = 105°
  4. Use the cosine rule to find AC. We know two sides (10 and 7) and the angle between them (105°), so use a2 = b2 + c2 − 2bc cos A.
    AC2 = 102 + 72 − 2 × 10 × 7 × cos 105°
    AC2 = 185.23
    AC = 13.6 km
  5. Use the sine rule to find angle BAC. We now know a side and the angle opposite it (AC and angle B), and we want the angle opposite BC (7 km).
    sin A ÷ 7 = sin 105° ÷ 13.610
    sin A = 7 × sin 105° ÷ 13.610 = 0.4968
    A = sin−1(0.4968) = 29.8°
  6. Add to (or subtract from) the bearing AB. C is clockwise from B as seen from A, because the ship turned right at B.
    35° + 29.8° = 64.8°
    Bearing of C from A = 065° to the nearest degree
  7. Find the back bearing. The bearing of A from C is half a turn from the bearing of C from A.
    65° + 180° = 245°
    Bearing of A from C = 245°

Check. Using coordinates from A: B = (5.74, 8.19) and C = (12.31, 5.80), so C is 13.6 km from A on a bearing of 065°. ✓

Watch out. Use the angle inside the triangle in the cosine rule, not the turn angle. They add up to 180°.

Watch out

The angle you find with trigonometry is an angle in the triangle, not the bearing. To get the bearing you must add it to a known bearing (or subtract it). Also, make sure you use the right back bearing: the bearing of A from C is not the same as the bearing of C from A.

Exercise 22H  Bearings and right-angled triangles

Calculator allowed
1
FA ship sails 12 km due north and then 5 km due east. Work out (a) how far the ship is from where it started, (b) the bearing of the ship from its start.
N000°N090°12 km5 kmABC
2
FFor each journey, work out (a) the distance AC and (b) the bearing of C from A. Each journey is two legs at right angles: from A to B, then B to C.
(a)A to B: 12 km on 000°. B to C: 5 km on 090°.
N000°N090°12 km5 kmABC
(b)A to B: 20 km on 090°. B to C: 15 km on 180°.
N090°N180°20 km15 kmABC
(c)A to B: 8 km on 180°. B to C: 6 km on 270°.
N180°N270°8 km6 kmABC
(d)A to B: 24 km on 270°. B to C: 10 km on 000°.
N270°N000°24 km10 kmABC
3
FWork out the distance AC and the bearing of C from A for each journey. The ship turns left at B.
(a)A to B: 9 km on 000°. B to C: 12 km on 270°.
N000°N270°9 km12 kmABC
(b)A to B: 15 km on 180°. B to C: 8 km on 090°.
N180°N090°15 km8 kmABC
4
PSA boat sails 10 km from A to B on a bearing of 050°. It then sails 7.5 km from B to C on a bearing of 140°. Work out the distance AC and the bearing of C from A.
N050°N140°10 km7.5 kmABC
5
PSA plane flies 16 km from X to Y on a bearing of 120°, then 12 km from Y to Z on a bearing of 030°. Work out (a) the distance XZ, (b) the bearing of Z from X, (c) the bearing of X from Z.
N120°N030°16 km12 kmABC
6
FWork out the bearing of B from A in each case. Give your answers to the nearest degree. All the distances are in kilometres.
(a)B is 8 km east and 6 km north of A.
(b)B is 7 km east and 4 km south of A.
(c)B is 5 km west and 12 km south of A.
(d)B is 3 km west and 9 km north of A.
7
PSAn aeroplane flies 300 km from an airport A due east to B. It then turns and flies 400 km due south to C. Work out (a) how far C is from the airport, (b) the bearing on which the plane must fly to return from C directly to A.
N090°N180°300 km400 kmABC
8
PSA, B and C are three trees. AB = AC = 6 m. The bearing of B from A is 030° and the bearing of C from A is 110°. Work out the bearing of C from B.
N030°110°6 km6 kmABC
9
PSThree markers A, B and C are such that AB = AC = 5 km. The bearing of B from A is 300° and the bearing of C from A is 250°. Work out the bearing of C from B.
N300°250°5 km5 kmABC
10
RA ship sails 5 km due north and then 12 km due east. Show that the ship is 13 km from its starting point, on a bearing of 067° to the nearest degree.
N?5 km12 kmABC
11
RLeo works out that angle BAC = 36.9° in a journey where the bearing of B from A is 070° and the ship then turns right at B. He says: “The bearing of C from A is 036.9°.” Explain his mistake and give the correct bearing.
12
RChallenge. Two ships leave a port P at the same time. Ship A sails on a bearing of 045° at 12 km/h. Ship B sails on a bearing of 135° at 16 km/h. After 2 hours, work out (a) how far apart the ships are, (b) the bearing of B from A, to the nearest degree.
N045°135°24 km32 kmPAB

Exercise 22I  Harder bearings problems

Calculator allowed
1
PSA boat sails 6 km from A to B on a bearing of 060°, then 4 km from B to C on a bearing of 150°. (a) Make a scale drawing with 1 cm to 1 km. (b) Use your drawing to find AC and the bearing of C from A. (c) Check your answers by calculation.
N060°N150°6 km4 kmABC
2
PSA, B and C are the corners of an isosceles triangle with AB = AC = 9 km. B is on a bearing of 020° from A and C is on a bearing of 120° from A. Work out the bearing of C from B.
N020°120°9 km9 kmABC
3
PSA coastguard station C sees a ship on a bearing of 060°. A lighthouse L is due east of C. The ship S is 5 km from C and the angle at S is 90° (angle CSL = 90°). Work out the bearing of L from S.
N060°5 kmCSL
4
PSTwo paths leave a gate G. Path 1 goes 3 km on a bearing of 040° to a hut H. Path 2 goes 4 km on a bearing of 130° to a lake L. Work out (a) the distance HL, (b) the bearing of L from H, to the nearest degree.
N040°130°3 km4 kmGHL
5
FA boat sails 10 km from A to B on a bearing of 035°, then 7 km from B to C on a bearing of 110°. Work out the distance AC and the bearing of C from A.
N035°N110°10 km7 kmABC
6
PSA helicopter flies 15 km from H to J on a bearing of 200°, then 9 km from J to K on a bearing of 260°. Work out (a) the distance HK, (b) the bearing of K from H, (c) the bearing of H from K.
N200°N260°15 km9 kmABC
7
PSA ship sails 12 km from P to Q on a bearing of 310°, then 8 km from Q to R on a bearing of 020°. Work out the distance PR and the bearing of R from P.
N310°N020°12 km8 kmABC
8
PSA boat sails 10 km from A to B on a bearing of 035° and then 7 km from B to C on a bearing of 110°. Work out the area of triangle ABC, correct to 1 decimal place.
9
RA ship leaves port P and sails 20 km on a bearing of 050° to A. A second ship sails 20 km from P on a bearing of 110° to B. Work out the bearing of B from A and the distance AB.
N050°110°20 km20 kmPAB

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