Applications: Compass Directions and Bearings

A bearing is measured clockwise from north, at the place you are standing. Sailing home reverses the direction of the course, and reversing a direction is a half turn, so the bearing home and the bearing out differ by 180^°. A bearing must stay between 000^° and 360^°, so add the half turn to a bearing under 180^° and subtract it from a bearing over 180^°.

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.
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.
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^°.
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^°.
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.

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

techniqueBack Bearings · Three-Figure Bearings

examsO-Level · GCSE Higher

Two bearings on one object fix its position: it lies where the two lines cross. To use the bearings in a triangle they must first be turned into angles of that triangle, and each one is measured against a line whose bearing is already known, here the coast itself.

1. Draw the north line at the station and the north line at the lighthouse, and draw the two sightings. The yacht lies where the two sightings cross.
2. The coast runs east and west, so from the station the lighthouse is due east, a bearing of 090^°, and from the lighthouse the station is due west, a bearing of 270^°.
3. The angle of the triangle at the station is the angle between the coast and the sighting: 090 − 045 = 45^°. The angle at the lighthouse is 315 − 270 = 45^°. The three angles add to 180^°, so the angle at the yacht is 180 − 45 − 45 = 90^°.
4. Drop a perpendicular from the yacht to the coast, meeting it at M. The two base angles are equal, so the triangle is isosceles and M is the midpoint of CL, giving CM = 12 ÷ 2 = 6 km. In the right-angled triangle CMY the angle at C is 45^°, so the two short sides are equal. (a) The yacht is 6 km north of the coast.
5. The bearing of the yacht from the station is 045^°, so the bearing of the station from the yacht is a half turn from it: (b) 045 + 180 = 225^°. Check: the bearing of the lighthouse from the yacht is 315 − 180 = 135^°, and 225 − 135 = 90^°, the angle at the yacht found in step 3.

answer(a) 6 km; (b) 225^°

techniqueThree-Figure Bearings · Back Bearings

examsO-Level · GCSE Higher

The eight points of the compass are equally spaced round a full turn, so each one is 45^° from the next. Naming them as bearings, counted clockwise from north, turns a compass direction into a number and a number back into a compass direction.

1. A full turn is 360^° and the compass has eight points, so one step from a point to the next is 360 ÷ 8 = 45^°.
2. Counting clockwise from north, the eight points are north 000^°, north-east 045^°, east 090^°, south-east 135^°, south 180^°, south-west 225^°, west 270^° and north-west 315^°.
3. The course is 225^°, the sixth of those bearings. (a) The airplane is flying south-west.
4. The return course is a half turn from the course out. The course out is 225^°, which is more than 180^°, so subtract the half turn: 225 − 180 = 45, written with three figures as 045^°.
5. (b) The return course is 045^°, which is north-east. Check: north-east and south-west are opposite points of the compass, four steps apart, and 4 × 45 = 180^°, a half turn.

answer(a) south-west; (b) 045^°, which is north-east

techniqueThe Eight-Point Compass · Three-Figure Bearings · Back Bearings

examsGCSE Higher

A wind is named by the direction it comes from, but anything carried by it travels the opposite way. The two directions are a half turn apart, so the two bearings differ by 180^°.

1. North-west lies halfway between north and west. Counting clockwise from north, west is three quarter-turns, a bearing of 270^°, and half of a quarter-turn is 45^°.
2. (a) The wind comes from a bearing of 270 + 45 = 315^°.
3. The balloon travels the way the wind blows, which is the opposite of the way it comes from, and opposite is a half turn. The bearing 315^° is more than 180^°, so subtract the half turn: 315 − 180 = 135^°.
4. (b) The balloon is carried on a bearing of 135^°, and 135^° is south-east. Check: north-west and south-east are opposite points of the compass, so a north-west wind blows toward the south-east.

answer(a) 315^°; (b) 135^°, which is south-east

techniqueThe Eight-Point Compass · Back Bearings · Three-Figure Bearings

examsGCSE Higher

A turn is a difference of two bearings. Going clockwise, the bearing increases; if the new bearing is the smaller number the ferry has passed north on the way, so a full turn of 360^° is added before the subtraction. The turn the other way round is whatever is left of the full turn.

1. North-east is one step of 45^° clockwise from north, so (a) the new course is a bearing of 045^°.
2. Turning clockwise takes the ferry from 340^° past north and on to 045^°. Because 045 is the smaller number, add a full turn before subtracting: the clockwise turn is 045 + 360 − 340 = 65^°.
3. The two turns together make one full turn, so the counterclockwise turn is 360 − 65 = 295^°.
4. (b) The clockwise turn of 65^° is the shorter one, and it is shorter by 295 − 65 = 230^°. Check: 65 + 295 = 360^°, one full turn, as it must be.

answer(a) 045^°, a turn of 65^° clockwise; (b) the clockwise turn is shorter, by 230^°, since the counterclockwise turn is 295^°

techniqueThe Eight-Point Compass · Three-Figure Bearings

examsGCSE Higher

Compass directions are bearings too: due east is 090^° and due south is 180^°. Written as bearings they can be subtracted, and the difference is the angle the route turns through, which is what makes Pythagoras available.

1. Due east is a bearing of 090^° and due south is a bearing of 180^°, so at the well the route turns through 180 − 090 = 90^°. The angle OWR is a right angle.
2. Triangle OWR is right-angled at W, with OR the side opposite the right angle. By Pythagoras, OR² = OW² + WR² = 120² + 160² = 14400 + 25600 = 40000.
3. (a) OR = √40000 = 200 m, so the chest lies 200 m from the oak.
4. The bearing of the well from the oak is 090^°, due east. The bearing of the oak from the well is a half turn from it, and 090^° is less than 180^°, so add the half turn: (b) 090 + 180 = 270^°.
5. Check: 270^° is due west, and walking back from the well to the oak does indeed go due west. Check on part (a): 200² = 40000, which is 14400 + 25600.

answer(a) 200 m; (b) 270^°

techniqueThree-Figure Bearings · Back Bearings

examsO-Level · GCSE Higher

A bearing is always written with three figures, so that a reader can tell at a glance where the degrees end. Leading zeros do not change the size of the angle; they only fix its length on the page.

1. A bearing runs from 000^° to 360^°, so three figures are always enough and always used. A bearing under 100^° needs one leading zero and a bearing under 10^° needs two.
2. The first vessel lies 8 degrees clockwise from north, so (a) its bearing is written 008^°.
3. The second vessel lies 80 degrees clockwise from north, so its bearing is written 080^°. Written as "80" it could be read as 800, which is not a bearing at all, because a bearing is never 360^° or more.
4. The bearing of the first vessel from the station is 008^°. The bearing of the station from that vessel is a half turn from it, and 008^° is less than 180^°, so add the half turn: (b) 008 + 180 = 188^°.
5. Check: 188 − 180 = 008, the bearing the lifeboat went out on, and 188^° is just past due south, which is where the station must be from a vessel almost due north of it.

answer(a) 008^° and 080^°; (b) 188^°

techniqueThree-Figure Bearings · Back Bearings

examsO-Level · GCSE Higher

Each leg is reversed on its own. Reversing a direction is a half turn, so each bearing back differs from its bearing out by 180^°; whether that half turn is added or subtracted is decided by keeping the answer between 000^° and 360^°.

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

answer(a) 025^°; (b) 285^°

techniqueBack Bearings · Three-Figure Bearings

examsO-Level · GCSE Higher

Walking a closed route brings the hiker back facing the way she started, so the turns she makes add up to one full turn of 360^°. Each turn is the difference between the bearing she was on and the bearing she takes next, measured clockwise.

1. Draw a north line at each of the three corners. The first leg leaves A on 070^° and the second leaves B on 190^°.
2. At the cairn the hiker changes from the course 070^° to the course 190^°. Both are measured clockwise from north, so (a) she turns 190 − 070 = 120^° clockwise.
3. All three legs are 4 km, so the route is an equilateral triangle and the turn at every corner is the same 120^°.
4. At the bothy she turns another 120^° clockwise, so (b) the last leg runs on a bearing of 190 + 120 = 310^°.
5. Check: one more turn of 120^° gives 310 + 120 = 430, and taking off the full turn, 430 − 360 = 070^°, the bearing she set out on. The three turns are 120 + 120 + 120 = 360^°, exactly one full turn, which is what closing the route requires.

answer(a) 120^° clockwise; (b) 310^°

techniqueThree-Figure Bearings · Back Bearings

examsO-Level · GCSE Higher

An eighth-turn is the step from one point of the compass to the next. Counting in eighth-turns and counting in degrees are the same count, because each eighth-turn is 45^°: a clockwise step adds 45^° to the bearing and an counterclockwise step takes 45^° off it.

1. A full turn is 360^° and there are eight points on the compass, so one eighth-turn is 360 ÷ 8 = 45^°. North-east is one eighth-turn clockwise from north, a bearing of 045^°.
2. Three eighth-turns clockwise add 3 × 45 = 135^° to the bearing: 045 + 135 = 180^°.
3. (a) A bearing of 180^° is due south, so after the first move the aerial faces south.
4. Two eighth-turns counterclockwise take 2 × 45 = 90^° off the bearing: 180 − 90 = 90, written with three figures as 090^°.
5. (b) A bearing of 090^° is due east, so the aerial finishes facing east. Check: three steps clockwise and two back leaves one step clockwise from north-east, and one step clockwise from 045^° is 045 + 45 = 090^°, due east.

answer(a) south, 180^°; (b) east, 090^°

techniqueThe Eight-Point Compass · Three-Figure Bearings

examsGCSE Higher