Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Geometry and measures
Practise Bearings and scale drawings. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Bearings give directions as angles measured clockwise from north, and scale drawings and maps use a scale to represent real distances. You measure and draw bearings and lengths, convert using scales, and calculate bearings using angle facts and trigonometry.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Use a scale to find real distancese.g. on a 1 : 50 000 map, 4 cm represents 200 000 cm = 2 km.
3
Measure and draw lines and angles accuratelyMeasure lengths to the nearest millimetre and angles to the nearest degree.
4
Measure and draw bearingsMeasure clockwise from north at the starting point, and write the bearing with three figures.
4
Make a scale drawing from real measurementsDivide each real length by the scale, e.g. at 1 cm to 5 m, 20 m is drawn as 4 cm.
5
Find a back bearingAdd or subtract 180°: if B is on a bearing of 070° from A, then A is on a bearing of 250° from B.
5
Calculate bearings and angles using angle factsUse parallel north lines with co-interior and alternate angles.
5
Use right-angled trigonometry in bearing problemsUse Pythagoras and SOH CAH TOA when the journey makes a right-angled triangle.
Notes
Bearings
A bearing is an angle measured clockwise from north, written with three figures: 45° is written as 045°.
'The bearing of B from A' means stand at A, face north and turn clockwise until you face B. So you measure at A.
North lines at different points are parallel, so you can use alternate, corresponding and co-interior angles.
Back bearing: the bearing of A from B is the bearing of B from A plus 180° (if it is less than 180°) or minus 180° (if it is more).
For a bearing over 180° with a semicircular protractor, measure the anticlockwise angle from north and take it away from 360°.
Scales
A scale such as 1 cm to 4 m, or a ratio such as 1 : 400, links lengths on the drawing to real lengths.
Drawing to real life: multiply by the scale. Real life to drawing: divide.
A ratio scale has no units: 1 : 25 000 means 1 cm represents 25 000 cm, which is 250 m.
Convert units carefully: 100 cm = 1 m and 1000 m = 1 km, so 100 000 cm = 1 km.
Scale drawings
Use a sharp pencil, a ruler (to the nearest millimetre) and a protractor (to the nearest degree).
Work out all the drawing lengths first, then draw.
Draw a north line at each point where you need to measure or draw a bearing.
Cheatsheet
Bearing: from north, clockwise, three figures
'Bearing of B from A': measure at A
Back bearing = bearing ± 180°
North lines are parallel: co-interior angles add up to 180°
Scale 1 : n → 1 cm represents n cm
100 cm = 1 m; 1000 m = 1 km; 100 000 cm = 1 km
How to answer each type of question
Use a map scale and draw a bearing
4 marks4
Convert with the scale: multiply for real distances, divide for map distances, then change units.
Draw a north line at the starting point.
Measure the bearing clockwise from north and mark the distance along that line.
Example. On a map with scale 1 : 20 000, a church C is 6.5 cm from a school S, on a bearing of 115° from S. (a) Work out the real distance from S to C. Give your answer in km. (2 marks) (b) Describe how to mark C accurately on the map. (2 marks)
Show the model answer
(a) 6.5 × 20 000 = 130 000 cm (M1) = 1.3 km (A1) (b) Draw a north line at S and measure 115° clockwise from it (B1). Draw a line in that direction and mark C 6.5 cm from S (B1)
Back bearings and angles from bearings
3 to 4 marks5
Draw north lines at each point and mark the bearings you know.
Use ±180° for a back bearing.
Use straight-line, parallel-line and triangle facts to find the angle you want.
Example. (a) The bearing of a lighthouse L from a boat B is 062°. Work out the bearing of B from L. (1 mark) (b) Town Q is due east of town P. Town R is on a bearing of 040° from P and on a bearing of 320° from Q. Work out the size of angle PRQ. (3 marks)
Show the model answer
(a) 062° + 180° = 242° (B1) (b) PQ is due east (090°), so angle RPQ = 90° − 40° = 50° (M1) From Q, P is due west (270°), so angle PQR = 320° − 270° = 50° (M1) Angle PRQ = 180° − 50° − 50° = 80° (A1)
Calculate a bearing using trigonometry
3 marks5
Sketch the journey with north lines and find the right-angled triangle.
Use SOH CAH TOA to find the angle at the starting point.
Turn the angle into a bearing (measured clockwise from north, three figures).
Example. A ship sails 8 km due north from a harbour H to a point A. It then sails 5 km due east to a point B. Work out the bearing of B from H. Give your answer to the nearest degree.
Show the model answer
tan θ = 5 ÷ 8 (M1) θ = tan⁻¹(5 ÷ 8) = 32.0° (M1) The angle is measured clockwise from north at H, so the bearing is 032° (A1)
Shortcuts and memory tricks
The word 'from' tells you where to put the protractor: the bearing of B from A is measured at A.
Estimate first: north-east is 045°, east 090°, south 180°, west 270°.
A bearing and its back bearing always differ by exactly 180°.
For a 1 : n map scale, 1 km is 100 000 cm, so dividing by 100 000 changes cm to km.
Where marks are lost
Measuring anticlockwise, or measuring at the wrong point.
Writing a bearing with fewer than three figures, e.g. 45° instead of 045°.
Reading the wrong scale on a semicircular protractor.
Converting units wrongly with ratio scales (cm to m to km).
Measuring a line from the end of the ruler instead of from 0.
Exam technique
Draw a north line at every point you use, then mark the angles on the diagram.
Measure carefully: drawn bearings and lengths usually only score if they are within a degree or two and a couple of millimetres.
Give real distances in the units the question asks for, with the unit written.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
The bearing of \(B\) from \(A\) is 070°. Tom says, 'The bearing of \(A\) from \(B\) is 110°.'
(a) Explain why Tom is wrong.[1]
(b) Work out the bearing of \(A\) from \(B\).[1]
Show the answer and mark scheme
(a)Answer: The bearing back is in the opposite direction, so you add 180°: \(70 + 180 = 250\). Tom worked out \(180 - 70\).
C1 for explaining that the back bearing differs by 180° (add 180°), not \(180 - 70\)
Worked solution: Going from \(B\) back to \(A\) is the opposite direction, which is 180° different from 070°.
(b)Answer: 250°
B1 for 250
Worked solution: \(070 + 180 = 250^\circ\)
Question 2Medium3 marks
The map shows the positions of Ashcombe (\(P\)) and Brindley (\(Q\)). The map is accurately drawn. Use the scale bar: 1 cm on the map represents 2 km. A radio mast, \(M\), is on a bearing of 120° from \(P\) and on a bearing of 235° from \(Q\).
[object Object]
(a) On the map, mark the position of \(M\) with a cross (×). Label it \(M\).[2]
(b) Use the map to find the distance of \(M\) from \(P\). Give your answer in km.[1]
Show the answer and mark scheme
(a)Answer: Lines drawn on a bearing of 120° from \(P\) and 235° from \(Q\); \(M\) marked where they cross (3.2 cm from \(P\) on the map).
M1 for a line from \(P\) on a bearing of 120° or from \(Q\) on a bearing of 235°, within ±2°
A1 for both lines drawn within ±2° and \(M\) marked at their intersection
Worked solution: Place the protractor on the north line at \(P\) and measure 120° clockwise; draw a long line. Do the same at \(Q\) for 235°. \(M\) is where the lines cross.
(b)Answer: 6.3 km (accept 5.8 to 7 km) km
B1 for 5.8 to 7 km (ft their \(M\): their length in cm × 2)
Worked solution: \(PM\) measures about 3.2 cm on the map, which represents \(3.2 \times 2 \approx 6.3\) km.