Back Bearings

The way back is a half turn from the way out.

Two ends, two bearings

A walker goes from A to B on a bearing of 060°. That angle is measured at A: it is the clockwise turn from north at A to the line toward B.

Now the walker wants to go back. The bearing of A from B is measured at the other end, from north at B. It is a different angle, and it is called the back bearing.

North is the same direction everywhere on a map, so the north line drawn at B points exactly the same way as the north line at A. The two north lines are parallel.

NN060°AB

The walk from A to B on a bearing of 060°, with a north line at A and another at B.

Turning round at B

To face A again, the walker at B starts facing north and turns clockwise until facing back along the line to A. That turn goes past east and past south, so it is more than 180°.

NN060°AB

At B the arc turns clockwise from B’s own north round to the line back to A.

Why the difference is 180°

Think of the north lines at A and B as full lines, running north and south through each point. They are parallel, and the line AB crosses both of them, so AB is a transversal and the angle facts for parallel lines apply.

Continue the north line at B downward, so that it points south. The angle at A between north and AB is 60°. The angle at B between south and BA lies on the other side of AB and between the two parallel lines, so the two are alternate angles, and alternate angles are equal: it is 60° too.

So the turn at B is 180° from north round to south, and then 60° more to the line BA. The bearing of A from B is 180 + 60 = 240°.

Co-interior angles give the same answer. The angle at B between north and BA, on the west side, and the 60° at A are co-interior, so they add to 180°: it is 180 − 60 = 120°. The clockwise turn from north at B goes the long way round, 360 − 120 = 240°.

Add or subtract 180°

The argument works for any bearing, so the back bearing always differs from the bearing out by exactly 180°, a half turn. The only question is whether to add 180° or subtract it, and the answer must stay a bearing: at least 000° and less than 360°.

If the bearing out is less than 180°, add 180°. For 060°, 60 + 180 = 240°. Subtracting would give 60 − 180 = −120, which is not a bearing.

If the bearing out is more than 180°, subtract 180°. A journey from P to Q on a bearing of 300° has a back bearing of 300 − 180 = 120°. Adding would give 480°, more than a full turn.

The P to Q journey can be argued in the same way. At P, the line PQ is 360 − 300 = 60° counterclockwise from north. At Q, the alternate angle puts the line QP 60° on the other side of south, counterclockwise from south, so its bearing is 180 − 60 = 120°.

For a bearing out of exactly 180°, due south, the back bearing is 180 − 180 = 000°, due north.

NN300°120°PQ

From P to Q on a bearing of 300°. 300° is more than 180°, so the bearing of P from Q is 300 − 180 = 120°.

NNABbearing of B from A = 120°bearing of A from B = 300°

the bearing of B from A is 120°, clockwise from north; the bearing of A from B is 120° + 180° = 300°, because the two north arrows are parallel and the angles at A and B are co-interior

Drag B to a bearing of 300° from A and read both bearings

Drag B round A. The gold arc at A is the bearing of B from A, and the green arc at B is the bearing of A from B. Wherever B goes, the two differ by 180°: added while B is on the east side of A, subtracted once it is on the west side. Drag B to 300° and read the bearing back.

Check with the compass points

The way back always points in the opposite direction. The bearing 060° lies between north and east, so its back bearing, 240°, must lie between south and west, and it does. A journey out on 135°, southeast, comes back on 135 + 180 = 315°, northwest.

Two quick checks catch most slips: the larger of the two bearings minus the smaller is 180, and the back bearing is between 000° and 359°.

The usual mistakes

Giving the bearing out again. The walk back runs along the same line, but in the opposite direction, so its bearing is 180° different: 240°, not 060°.

Taking the bearing out away from 360°. 360 − 60 = 300° measures the same line AB counterclockwise from north at A. It is not the way back from B.

Using one rule for every bearing. Adding 180° to 300° gives 480°, which is not a bearing: over 180°, subtract.

Worked example: A Trawler Leaving Harbor and the Course Back In

Question A trawler leaves harbor H and sails on a bearing of 064° for 18 km to the fishing ground F. A lighthouse L stands on the same straight course, 7 km beyond the fishing ground. (a) What is the bearing of the harbor from the fishing ground? (b) What is the bearing of the lighthouse from the harbor, and how far is the lighthouse from the harbor?

  1. 1.Mark the harbor H and draw the north line at H. The course of 064° is measured clockwise from that north line, and the fishing ground F lies 18 km along it.

    NHF18 km064 degbearing of F from H: 064 deg
    NHF18 km064 degbearing of F from H: 064 deg
    The course of 064° is measured clockwise from the north line at the harbor.

  2. 2.The bearing of the harbor from the fishing ground is measured at F, so draw a second north line at F. The two north lines are parallel, because north is the same direction everywhere on this chart.

    NNHF18 km064 degbearing of F from H: 064 degthe north line at F is parallel to the one at H
    NNHF18 km064 degbearing of F from H: 064 degthe north line at F is parallel to the one at H
    The bearing home is measured at F, from a second north line parallel to the first.

  3. 3.Between two parallel north lines the way out and the way back differ by a half turn. The course out is 064°, which is less than 180°, so add the half turn. (a) The bearing of the harbor from the fishing ground is 064 + 180 = 244°.

    NNHF18 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 deg
    NNHF18 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 deg
    (a) The course out is under 180°, so add the half turn: 064 + 180 = 244°.

  4. 4.The lighthouse L lies on the same straight course from H, beyond F, so the ray from H to L is the ray from H to F continued. A point further along one straight course is on the same bearing. (b) The bearing of the lighthouse from the harbor is 064°.

    NNLHF18 km7 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 degL is further along the same course: 064 deg
    NNLHF18 km7 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 degL is further along the same course: 064 deg
    (b) The lighthouse is further along the same straight course, so its bearing from the harbor is 064°.

  5. 5.The lighthouse is 7 km beyond the fishing ground, so its distance from the harbor is 18 + 7 = 25 km. Check: 244 − 180 = 064, the course the trawler set out on, so the two bearings really are a half turn apart.

    NNLHF18 km7 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 degL is further along the same course: 064 degH to L: 18 + 7 = 25 km
    NNLHF18 km7 km064 deg244 degbearing of F from H: 064 degthe north line at F is parallel to the one at H064 is under 180, so add the half turnbearing of H from F: 064 + 180 = 244 degL is further along the same course: 064 degH to L: 18 + 7 = 25 km
    The lighthouse is 18 + 7 = 25 km from the harbor.

Answer: (a) 244°; (b) 064°, and the lighthouse is 25 km from the harbor

Common mistakes

  • Giving the bearing of the harbor from the fishing ground as 064° again, because that is the line the trawler sailed along. A bearing names a direction, not a line, and the direction home is the opposite of the direction out.
  • Writing the answer as 064 − 180 = −116°. A bearing is measured clockwise from north and always lies between 000° and 360°, so a course under 180° has its half turn added, not taken away.

More compass directions and bearings problems, worked step by step →

Worked example: A Walker Retracing a Two-Leg Route in the Rain

Question A walker leaves the car park P on a bearing of 105° and walks 2 km to a gate G. From the gate she walks on a bearing of 205° for 3 km to a lake K. Rain sets in and she returns along the same two legs. (a) On what bearing must she walk from the lake back to the gate? (b) On what bearing must she then walk from the gate back to the car park?

  1. 1.Draw the route: a north line at the car park with the first leg on 105°, then a north line at the gate with the second leg on 205°. Draw a north line at the lake as well, because the first bearing back is measured there.

    NNNPGK2 km3 km105 deg205 degout: P to G on 105 deg, G to K on 205 deg
    NNNPGK2 km3 km105 deg205 degout: P to G on 105 deg, G to K on 205 deg
    The route out: 105° from the car park to the gate, then 205° from the gate to the lake.

  2. 2.The walk from the lake to the gate reverses the second leg, the one on 205°. That bearing is more than 180°, so subtract the half turn: (a) 205 − 180 = 25, written with three figures as 025°.

    NNNPGK2 km3 km105 deg205 deg025 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg
    NNNPGK2 km3 km105 deg205 deg025 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg
    (a) Going back reverses the second leg. 205° is over 180°, so subtract: 205 − 180 = 025°.

  3. 3.The walk from the gate to the car park reverses the first leg, the one on 105°. That bearing is less than 180°, so add the half turn: (b) 105 + 180 = 285°.

    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 deg
    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 deg
    (b) Going back reverses the first leg. 105° is under 180°, so add: 105 + 180 = 285°.

  4. 4.Check each answer by reversing it again: 025 + 180 = 205 and 285 − 180 = 105, the two bearings she walked out on.

    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 degcheck: 025 + 180 = 205 and 285 − 180 = 105
    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 degcheck: 025 + 180 = 205 and 285 − 180 = 105
    Check: 025 + 180 = 205 and 285 − 180 = 105, the two bearings she walked out on.

  5. 5.Notice that the two legs were treated differently. Adding the half turn to 205° would give 385°, which is past a full turn, and subtracting it from 105° would give −75°; neither is a bearing, so in each case only one of the two is allowed.

    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 degcheck: 025 + 180 = 205 and 285 − 180 = 105only one of add or subtract leaves a bearing
    NNNPGK2 km3 km105 deg205 deg025 deg285 degout: P to G on 105 deg, G to K on 205 deg205 is over 180, so take off the half turnK back to G: 205 − 180 = 025 deg105 is under 180, so add the half turnG back to P: 105 + 180 = 285 degcheck: 025 + 180 = 205 and 285 − 180 = 105only one of add or subtract leaves a bearing
    Adding to 205° would give 385° and subtracting from 105° would give −75°; neither is a bearing.

Answer: (a) 025°; (b) 285°

Common mistakes

  • Using one rule for both legs, for instance adding 180° every time, which turns 205° into 385°. A bearing is never 360° or more, so that answer has to be brought back by a full turn or found by subtracting instead.
  • Giving the bearing from the lake to the car park as the answer to part (a). She retraces her steps, so she walks first to the gate along the second leg reversed; the car park is not reached in a straight line from the lake.

More compass directions and bearings problems, worked step by step →

Practice Back Bearings in the app