Hard GCSE Bearings and scale drawings Questions

Challenging, exam-style GCSE Bearings and scale drawings questions with worked solutions. Stretch yourself on the hardest three-figure bearings, measuring clockwise from north, back bearings, reverse bearing problems.

three-figure bearingsmeasuring clockwise from northback bearingsreverse bearingparallel north linesangle reasoning
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
The bearing of BB from AA is 310310^\circ. The bearing of CC from AA is 085085^\circ. Which of these is the size of angle BACBAC?
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Worked solution

  1. Recall what a three-figure bearing is.

    clockwise from north, three figures\text{clockwise from north, three figures}

    A bearing is the angle turned CLOCKWISE from the north line, written with three figures — so an angle of 88{}^\circ becomes 008008^\circ.

  2. Note that both bearings are measured at A from the same north line.

    310 and 085 at A310^\circ \text{ and } 085^\circ \text{ at } A

    Both directions are measured from the north line at AA, so angle BACBAC is simply the difference between the two bearings.

  3. Subtract the smaller bearing from the larger one.

    310085=225310 - 085 = 225

    The two directions are 225225^\circ apart one way round the north line at AA.

  4. Take the non-reflex angle.

    225 and 135225^\circ \text{ and } 135^\circ

    The full turn at AA splits into 225225^\circ and 135135^\circ. Angle BACBAC is the one that is not reflex, so it is 135135^\circ.

  5. Check what the answer is measured in.

    135 (degrees)135^\circ \text{ (degrees)}

    The answer is the SIZE of an angle, so it is written 135135^\circ and NOT as a three-figure bearing. Only directions get three figures.

  6. Check the size of the angle is sensible.

    0<135<1800^\circ < 135^\circ < 180^\circ

    An angle inside a triangle is always between 00^\circ and 180180^\circ, and 135135^\circ is.

  7. Check the angle asked for is the one that has been found.

    135 and 225135^\circ \text{ and } 225^\circ

    Two angles meet at that point: 135135^\circ and the reflex angle 225225^\circ. The question asks for the angle of the triangle, which is the smaller one, 135135^\circ.

  8. Check the angle and its reflex angle add to a full turn.

    135+225=360135 + 225 = 360

    The angle 135135^\circ and the reflex angle around the same point add up to a full turn of 360360^\circ, which they do.

  9. Do not write this answer with three figures.

    135135 (a bearing)135^\circ \ne 135^\circ \text{ (a bearing)}

    Three-figure notation is for BEARINGS, which are directions. This answer is the size of an angle, so it is just 135135^\circ.

  10. Check that the two north lines are parallel.

    NANBN_A \parallel N_B

    North is the same direction everywhere on the page, so the north line at one point is parallel to the north line at the other. That is what lets you use the parallel-line angle facts.

  11. Check the angle was measured from north and not from east.

    clockwise, starting at north\text{clockwise, starting at north}

    A bearing always starts at the north line and turns clockwise. Starting anywhere else, or turning the other way, gives the wrong bearing.

  12. Check the angle on an accurate drawing.

    protractor135\text{protractor} \rightarrow 135^\circ

    On a drawing made to scale, measuring the angle with a protractor should give 135135^\circ. That is a useful check but not a proof — the working is the proof.

  13. Note that a bearing does not depend on how far apart the points are.

    direction only, not distance\text{direction only, not distance}

    A bearing records a DIRECTION. Moving the second point twice as far away along the same line does not change the bearing at all.

  14. Check the answer against a rough sketch.

    sketchcheck\text{sketch} \rightarrow \text{check}

    Drawing the north line and the direction roughly to scale is the quickest way to catch an answer that points the wrong way. The sketch does not need to be accurate to do that job.

  15. State the correct option.

    angle BAC=135\text{angle } BAC = 135^\circ

    Angle BAC=135BAC = 135^\circ.

Answer
135135^\circ
Question 2
5 markschallenging
A map is drawn to a scale of 1:500001 : 50000. Two towns are 66 cm apart on the map. Which calculation gives the real distance between the towns, in kilometres?
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Worked solution

  1. Say what the scale means.

    1 cm on the map=50000 cm in real life1\text{ cm on the map} = 50000\text{ cm in real life}

    A scale of 1:500001 : 50000 means every length in real life is 5000050000 times the matching length on the map, with both measured in the same unit.

  2. Turn the map distance into a real distance in centimetres.

    6×500006 \times 50000

    The real distance is 5000050000 times the map distance, so multiply: 6×500006 \times 50000 cm.

  3. Recall how centimetres and kilometres are linked.

    1 km=1000 m=100000 cm1\text{ km} = 1000\text{ m} = 100\,000\text{ cm}

    There are 100100 cm in a metre and 10001000 m in a kilometre, so there are 100×1000=100000100 \times 1000 = 100\,000 cm in a kilometre.

  4. Turn the real distance in centimetres into kilometres.

    ÷100000\div 100000

    Dividing by 100000100000 converts centimetres into kilometres, so the whole calculation is 6×50000÷1000006 \times 50000 \div 100000.

  5. Check the value the correct calculation gives.

    6×50000÷100000=36 \times 50000 \div 100000 = 3

    The correct calculation gives 33 km, which is a sensible distance between two towns. The other options give values that are far too big or far too small.

  6. Check the units of the answer.

    3 km3\text{ km}

    The question asks for the real distance in km, so the answer is 3 km3\text{ km}.

  7. Recall how centimetres and kilometres are linked.

    1 km=1000 m=100000 cm1\text{ km} = 1000\text{ m} = 100\,000\text{ cm}

    There are 100100 cm in a metre and 10001000 m in a kilometre, so there are 100×1000=100000100 \times 1000 = 100\,000 cm in a kilometre.

  8. Check the two lengths were in the same unit before the scale was used.

    cm:cm\text{cm} : \text{cm}

    A scale such as 1:250001 : 25\,000 only compares like with like. Both lengths must be turned into the same unit before the multiplication, and only then is the answer converted to the unit asked for.

  9. Check the size of the answer is sensible.

    3 km>03\text{ km} > 0

    The real distance must be a positive length, and 33 km is a realistic size for it.

  10. Check the answer by working backwards.

    3 kmback to the drawing3\text{ km} \rightarrow \text{back to the drawing}

    Reversing every step — dividing where you multiplied, and multiplying where you divided — must take 33 km back to the length the question started from.

  11. Note the commonest mistake with scales.

    × and ÷ the wrong way round\times \text{ and } \div \text{ the wrong way round}

    Multiplying when you should divide turns a real distance into an impossible one. Ask first whether the answer should be bigger or smaller than the length you were given.

  12. Check the conversion factor was applied only once.

    100000 used once100\,000 \text{ used once}

    It is easy to divide by 100100 and then by 10001000 and then by 100000100\,000 again. Each conversion between units happens exactly once.

  13. Check the answer is written to a sensible accuracy.

    3 km3\text{ km}

    The numbers in the question are exact, so 33 km is exact too and needs no rounding.

  14. State the rule this question uses.

    real=scale factor×map\text{real} = \text{scale factor} \times \text{map}

    Every map-scale question is this one rule, used forwards to get a real distance and backwards to get a map distance.

  15. State the correct calculation.

    6×50000÷1000006 \times 50000 \div 100000

    The real distance is 6×50000÷100000=36 \times 50000 \div 100000 = 3 km.

Answer
6×50000÷1000006 \times 50000 \div 100000
Question 3
6 markschallenging
The bearing of QQ from PP is 283283^\circ. Which statement is correct?
Show worked solution

Worked solution

  1. Recall what a three-figure bearing is.

    clockwise from north, three figures\text{clockwise from north, three figures}

    A bearing is the angle turned CLOCKWISE from the north line, written with three figures — so an angle of 88{}^\circ becomes 008008^\circ.

  2. Note that the statement asked about is the back bearing.

    bearing of P from Q\text{bearing of } P \text{ from } Q

    The bearing given is measured at PP. The statements are all about the bearing measured at QQ, looking back at PP.

  3. State the rule for a back bearing.

    back bearing=bearing±180\text{back bearing} = \text{bearing} \pm 180{}^\circ

    The north lines at the two points are parallel, so the two bearings differ by exactly 180180^\circ: add 180180^\circ if the bearing is less than 180180^\circ, and subtract 180180^\circ if it is more.

  4. Apply the back bearing rule.

    283180=103283 - 180 = 103

    283180=103283 - 180 = 103, so the bearing of PP from QQ is 103103^\circ.

  5. Reject the statements that give any other bearing.

    103 only103^\circ \text{ only}

    Only one statement gives 103103^\circ; every other statement names a different direction and so is false.

  6. Write the answer as a three-figure bearing.

    103103^\circ

    Bearings are always written with three figures, putting in zeros at the front if they are needed. So the answer is written 103103^\circ.

  7. Check the answer is a possible bearing.

    0103<3600 \le 103 < 360

    A bearing is measured clockwise from north and comes all the way back to north after a full turn, so every bearing lies between 000000^\circ and 360360^\circ. 103103^\circ does.

  8. Check which part of the compass the direction points into.

    090<103<180090^\circ < 103^\circ < 180^\circ

    103103^\circ lies between 090090^\circ (east) and 180180^\circ (south), so the direction points somewhere between east and south. A sketch should agree with that.

  9. Check the answer by reversing it.

    103+180=283283103 + 180 = 283 \Rightarrow 283

    Turning through a further 180180^\circ from 103103^\circ leads back to 283283^\circ, which is where the reverse journey points. The two bearings differ by 180180^\circ, as they must.

  10. Note the commonest mistake in bearings questions.

    360103=257103360 - 103 = 257 \ne 103

    Bearings are measured CLOCKWISE from north. Measuring anticlockwise instead would give 257257^\circ, which is a different direction entirely.

  11. Check the units of the answer.

    103 (degrees)103^\circ \text{ (degrees)}

    A bearing is an angle, so it is measured in degrees. The answer is 103103^\circ, not 103103 of anything else.

  12. Compare the answer with the nearest compass point.

    103090103^\circ \approx 090^\circ

    The nearest compass point to 103103^\circ is east (090090^\circ), so the direction should look roughly east on a sketch.

  13. Compare the bearing with the four main compass directions.

    000, 090, 180, 270000^\circ,\ 090^\circ,\ 180^\circ,\ 270^\circ

    North is 000000^\circ, east is 090090^\circ, south is 180180^\circ and west is 270270^\circ. Placing 103103^\circ among these is a fast check that the direction is roughly right.

  14. Check that the two north lines are parallel.

    NANBN_A \parallel N_B

    North is the same direction everywhere on the page, so the north line at one point is parallel to the north line at the other. That is what lets you use the parallel-line angle facts.

  15. State the correct statement.

    bearing of P from Q=103\text{bearing of } P \text{ from } Q = 103^\circ

    The bearing of PP from QQ is 103103^\circ.

Answer
The bearing ofPfromQis103.\text{The bearing of} P \text{from} Q \text{is} 103^\circ \text{.}
Question 4
5 markschallenging
The bearing of BB from AA is 172172^\circ. Work out the bearing of AA from BB.
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Worked solution

  1. Write down the bearing that is given, and note that the north lines at A and B are parallel.

    bearing of B from A=172\text{bearing of } B \text{ from } A = 172^\circ

    Standing at AA and facing north, you turn 172172^\circ clockwise to look at BB. North points the same way at BB, so the two north lines are parallel and the two bearings must differ by 180180^\circ.

  2. Add or subtract 180 degrees to reverse the bearing.

    172+180=352172 + 180 = 352

    172172^\circ is less than 180180^\circ, so add 180180^\circ: 172+180=352172 + 180 = 352.

  3. Recall what a three-figure bearing is.

    clockwise from north, three figures\text{clockwise from north, three figures}

    A bearing is the angle turned CLOCKWISE from the north line, written with three figures — so an angle of 88{}^\circ becomes 008008^\circ.

  4. State the rule for a back bearing.

    back bearing=bearing±180\text{back bearing} = \text{bearing} \pm 180{}^\circ

    The north lines at the two points are parallel, so the two bearings differ by exactly 180180^\circ: add 180180^\circ if the bearing is less than 180180^\circ, and subtract 180180^\circ if it is more.

  5. Check the choice of adding or subtracting was the right one.

    172<180172 < 180

    Adding 180180^\circ to a bearing above 180180^\circ would push the answer past a full turn, and subtracting from a bearing below 180180^\circ would push it below zero. Here 172172^\circ is below 180180^\circ, so adding is right.

  6. Write the answer as a three-figure bearing.

    352352^\circ

    Bearings are always written with three figures, putting in zeros at the front if they are needed. So the answer is written 352352^\circ.

  7. Check the answer is a possible bearing.

    0352<3600 \le 352 < 360

    A bearing is measured clockwise from north and comes all the way back to north after a full turn, so every bearing lies between 000000^\circ and 360360^\circ. 352352^\circ does.

  8. Check which part of the compass the direction points into.

    270<352<360270^\circ < 352^\circ < 360^\circ

    352352^\circ lies between 270270^\circ (west) and 000000^\circ (north), so the direction points somewhere between west and north. A sketch should agree with that.

  9. Check the answer by reversing it.

    352+180=532172352 + 180 = 532 \Rightarrow 172

    Turning through a further 180180^\circ from 352352^\circ leads back to 172172^\circ, which is where the reverse journey points. The two bearings differ by 180180^\circ, as they must.

  10. Note the commonest mistake in bearings questions.

    360352=008352360 - 352 = 008 \ne 352

    Bearings are measured CLOCKWISE from north. Measuring anticlockwise instead would give 008008^\circ, which is a different direction entirely.

  11. Check the units of the answer.

    352 (degrees)352^\circ \text{ (degrees)}

    A bearing is an angle, so it is measured in degrees. The answer is 352352^\circ, not 352352 of anything else.

  12. Compare the answer with the nearest compass point.

    352000352^\circ \approx 000^\circ

    The nearest compass point to 352352^\circ is north (000000^\circ), so the direction should look roughly north on a sketch.

  13. Compare the bearing with the four main compass directions.

    000, 090, 180, 270000^\circ,\ 090^\circ,\ 180^\circ,\ 270^\circ

    North is 000000^\circ, east is 090090^\circ, south is 180180^\circ and west is 270270^\circ. Placing 352352^\circ among these is a fast check that the direction is roughly right.

  14. Check that the two north lines are parallel.

    NANBN_A \parallel N_B

    North is the same direction everywhere on the page, so the north line at one point is parallel to the north line at the other. That is what lets you use the parallel-line angle facts.

  15. State the bearing of A from B.

    bearing of A from B=352\text{bearing of } A \text{ from } B = 352^\circ

    The bearing of AA from BB is 352352^\circ.

Answer
bearing of A from B=352\text{bearing of } A \text{ from } B = 352^\circ
Question 5
5 markschallenging
The bearing of BB from AA is 295295^\circ. The bearing of CC from AA is 140140^\circ. Work out the size of angle BACBAC.
Show worked solution

Worked solution

  1. Mark both bearings from the same north line at A, and subtract them.

    295140=155295 - 140 = 155

    Both bearings are measured at AA from the same north line, so the two directions are 155155^\circ apart going one way round that north line.

  2. Take the angle that is not reflex.

    155 and 205155155^\circ \text{ and } 205^\circ \Rightarrow 155^\circ

    The two directions cut the full turn at AA into 155155^\circ and 205205^\circ. Angle BACBAC is the angle of the triangle, so it is the one that is not reflex: 155155^\circ.

  3. Recall what a three-figure bearing is.

    clockwise from north, three figures\text{clockwise from north, three figures}

    A bearing is the angle turned CLOCKWISE from the north line, written with three figures — so an angle of 88{}^\circ becomes 008008^\circ.

  4. Say why the difference of the bearings is the angle.

    295 and 140 both start at north at A295^\circ \text{ and } 140^\circ \text{ both start at north at } A

    Both bearings start from the SAME north line at AA, so subtracting one from the other removes the north line entirely and leaves the angle between ABAB and ACAC.

  5. Check the subtraction.

    140+155=295140 + 155 = 295

    Adding the difference back onto the smaller bearing returns the larger one, so 155155^\circ is right.

  6. Check what the answer is measured in.

    155 (degrees)155^\circ \text{ (degrees)}

    The answer is the SIZE of an angle, so it is written 155155^\circ and NOT as a three-figure bearing. Only directions get three figures.

  7. Check the size of the angle is sensible.

    0<155<1800^\circ < 155^\circ < 180^\circ

    An angle inside a triangle is always between 00^\circ and 180180^\circ, and 155155^\circ is.

  8. Check the angle asked for is the one that has been found.

    155 and 205155^\circ \text{ and } 205^\circ

    Two angles meet at that point: 155155^\circ and the reflex angle 205205^\circ. The question asks for the angle of the triangle, which is the smaller one, 155155^\circ.

  9. Check the angle and its reflex angle add to a full turn.

    155+205=360155 + 205 = 360

    The angle 155155^\circ and the reflex angle around the same point add up to a full turn of 360360^\circ, which they do.

  10. Do not write this answer with three figures.

    155155 (a bearing)155^\circ \ne 155^\circ \text{ (a bearing)}

    Three-figure notation is for BEARINGS, which are directions. This answer is the size of an angle, so it is just 155155^\circ.

  11. Check that the two north lines are parallel.

    NANBN_A \parallel N_B

    North is the same direction everywhere on the page, so the north line at one point is parallel to the north line at the other. That is what lets you use the parallel-line angle facts.

  12. Check the angle was measured from north and not from east.

    clockwise, starting at north\text{clockwise, starting at north}

    A bearing always starts at the north line and turns clockwise. Starting anywhere else, or turning the other way, gives the wrong bearing.

  13. Check the angle on an accurate drawing.

    protractor155\text{protractor} \rightarrow 155^\circ

    On a drawing made to scale, measuring the angle with a protractor should give 155155^\circ. That is a useful check but not a proof — the working is the proof.

  14. Note that a bearing does not depend on how far apart the points are.

    direction only, not distance\text{direction only, not distance}

    A bearing records a DIRECTION. Moving the second point twice as far away along the same line does not change the bearing at all.

  15. State the size of angle BAC.

    angle BAC=155\text{angle } BAC = 155^\circ

    Angle BAC=155BAC = 155^\circ.

Answer
angle BAC=155\text{angle } BAC = 155^\circ

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