Chapter 22 · Foundation tier

GCSE Maths Loci, Bearings, Constructions and Vectors (Foundation): worked examples and questions with answers

By the end of this chapter you will be able to

Use a scale such as 1 : 50 000 to change between distances on a map or drawing and real distances
Construct perpendicular bisectors, angle bisectors, perpendiculars and a 60° angle with a ruler and compasses, and construct triangles from SSS, SAS and ASA
Describe and draw loci, and shade the region that obeys several rules at once
Measure, draw and calculate with three-figure bearings, including back bearings and journeys in two legs
Write translations as column vectors and add, subtract, scale and find the magnitude of vectors

You should already know

Ratio and units of length (Chapters 15 and 16)
Angles on parallel lines and in triangles (Chapter 17)
Pythagoras’ theorem and trigonometry (Chapter 20)
Coordinates and translations (Chapters 12 and 21)
Congruent triangles (Chapter 21)

Where this leads

Circle theorems (Chapter 26, Higher)
Vector proofs in the Higher exam, grades 8 and 9
Plans, maps and navigation in everyday life

Edexcel GCSE (9–1) Mathematics 1MA1, Foundation tier. Content references: G2, G15, G24, G25, R2.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

22.1 Scale drawings and maps: Worksheet (PDF) · Answers (PDF) · lesson page
22.2 Ruler and compass constructions: Worksheet (PDF) · Answers (PDF) · lesson page
22.3 Constructing triangles: Worksheet (PDF) · Answers (PDF) · lesson page
22.4 Loci: the four basic rules: Worksheet (PDF) · Answers (PDF) · lesson page
22.5 Combined loci and region problems: Worksheet (PDF) · Answers (PDF) · lesson page
22.6 Bearings: measuring, drawing and back bearings: Worksheet (PDF) · Answers (PDF) · lesson page
22.7 Bearings problems: journeys, Pythagoras and trigonometry: Worksheet (PDF) · Answers (PDF) · lesson page
22.8 Vectors: column vectors, addition and scalar multiples: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FChange 3.5 km into centimetres.

Done

FTwo straight lines are parallel. A third line crosses both. One co-interior angle is 65°. Work out the other co-interior angle.

Done

FA right-angled triangle has shorter sides of 6 cm and 8 cm. Work out the length of the hypotenuse.

Done

FIn a right-angled triangle the side opposite the angle x is 3 cm and the side next to it (adjacent) is 4 cm. Work out x to 1 decimal place.

Done

FThe point A is (2, 5). It is moved 3 units right and 4 units down. Write down the new coordinates.

Done

22.1

Scale drawings and maps

Edexcel 1MA1: R2, G15
Grade 1-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Scale drawings and mapsComing soon

A map or a scale drawing is a smaller picture of something real. Every length in the picture is the real length divided by the same number. That number is the scale. Architects draw houses on plans, model-makers build toy cars, and walkers follow maps. All of them use scales.

The most common way to write a scale is as a ratio, for example 1 : 50 000. Both numbers are in the same units, so it means: 1 cm on the map is 50 000 cm in real life (or 1 mm is 50 000 mm, or 1 inch is 50 000 inches). You can read any ratio scale this way. A scale of 1 : 100 means that 1 cm on the drawing is 100 cm = 1 m in real life.

Sometimes a scale is written in words, such as “1 cm represents 2 km”. To turn this into a ratio, change both lengths to the same unit (centimetres) and write the map length first: 1 cm : 2 km = 1 cm : 200 000 cm = 1 : 200 000.

A simple map. The scale bar shows that 2 cm on the map is 1 km in real life, because 1 cm is 50 000 cm = 0.5 km.

Changing units. Lengths on a map are measured in centimetres, but real distances are given in metres or kilometres. Remember: 100 cm = 1 m, 1000 m = 1 km, so 100 000 cm = 1 km. To go from centimetres to metres divide by 100. To go from centimetres to kilometres divide by 100 000.

The two rules. To go from the map to real life, multiply by the scale number (real life is bigger). To go from real life to the map, divide by the scale number (the map is smaller). Do the calculation in centimetres, then change the answer into a sensible unit. A bigger second number in the scale means that the map shows more land in less space: 1 : 50 000 shows a larger area than 1 : 25 000, so it has less detail.

Key ideas

A scale 1 : n means 1 cm on the drawing is n cm in real life.
Map to real life: multiply by n. Real life to map: divide by n.
100 cm = 1 m and 100 000 cm = 1 km.
To write “1 cm to 2 km” as a ratio, change both to cm: 1 : 200 000.
Always give the unit in your answer.

Worked example 22.1.1

The scale of a map is 1 : 50 000. Two villages are 4.5 cm apart on the map. How far apart are they in real life? Give your answer in kilometres.

Key words. A scale compares a length on a map or drawing with the real length. A scale of 1 : 50 000 means that 1 cm on the map is 50 000 cm in real life. A ratio compares two amounts in the same units.

Method. Multiply the length on the map by the scale number to get centimetres, then change to sensible units.

Say what the scale means. A scale of 1 : 50 000 means that every 1 cm on the map stands for 50 000 cm in real life.
1 cm on the map = 50 000 cm real
Multiply the map length by 50 000. The real distance is 50 000 times as long as the length on the map.
4.5 × 50 000 = 225 000 cm
Change the centimetres into km. There are 100 cm in 1 m and 1000 m in 1 km, so there are 100 000 cm in 1 km. Divide by 100 000.
225 000 cm ÷ 100 000 = 2.25 km

Check. Go backwards: 2.25 km = 225 000 cm and 225 000 ÷ 50 000 = 4.5 cm, the length we started with. ✓

Watch out. Multiply by the scale number to go from the map to real life. Divide to go the other way. Change the units at the end.

Worked example 22.1.2

A footpath is 3.5 km long. How long is the footpath on a map with a scale of 1 : 25 000?

Method. Change the real length to centimetres, then divide by the scale number to get the length on the map.

Change the real length into centimetres. 1 km = 1000 m = 100 000 cm, so multiply by 100 000.
3.5 km × 100 000 = 350 000 cm
Say what the scale means. A scale of 1 : 25 000 means that every 25 000 cm in real life is only 1 cm on the map.
25 000 cm real = 1 cm on the map
Divide by 25 000. The map length is 25 000 times smaller than the real length.
350 000 ÷ 25 000 = 14 cm

Check. Go forwards: 14 × 25 000 = 350 000 cm, which is the real length. ✓

Watch out. Going from real life to the map, you divide by the scale number (the map is smaller).

Worked example 22.1.3

On a map, a distance of 6 cm represents a real distance of 3 km. Write the scale of the map as a ratio in the form 1 : n.

Method. Write the map length and the real length in the same unit (cm), make a ratio, and divide so the first number is 1.

Change the real length into cm. Both parts of a ratio must be in the same unit. 1 km = 1000 m = 100 000 cm, so multiply by 100 000.
3 km × 100 000 = 300 000 cm
Write the ratio map : real. The scale compares the map length with the real length, in that order.
6 : 300 000
Divide both sides by 6 so the first number is 1. A scale is written as 1 : n. Dividing both parts by the same number keeps the ratio the same.
6 ÷ 6 = 1 and 300 000 ÷ 6 = 50 000
Scale = 1 : 50 000

Check. 1 cm on the map stands for 50 000 cm = 500 m in real life. So 6 cm on the map should be 3 km. ✓

Watch out. Change to the same units first. A ratio like 2 : 4 km is wrong because the units do not match.

Worked example 22.1.4

A room is 4.2 m long. An architect draws a plan of the house using a scale of 1 : 50. How long is the room on the plan?

Method. Change the real length to centimetres, then divide by the scale number to get the length on the plan.

Change the real length into centimetres. 1 m = 100 cm, so multiply by 100.
4.20000000000000017763568394002504646778106689453125 m × 100 = 420.000000000000017763568394002504646778106689453125 cm
Say what the scale means. A scale of 1 : 50 means that every 50 cm in real life is only 1 cm on the plan.
50 cm real = 1 cm on the plan
Divide by 50. The plan length is 50 times smaller than the real length.
420.000000000000017763568394002504646778106689453125 ÷ 50 = 8.4000000000000003552713678800500929355621337890625 cm

Check. Go forwards: 8.4000000000000003552713678800500929355621337890625 × 50 = 420.000000000000017763568394002504646778106689453125 cm, which is the real length. ✓

Watch out. Going from real life to the map, you divide by the scale number (the map is smaller).

Watch out

Do not mix up the two directions. If your answer for a real distance is smaller than the length on the map, you divided instead of multiplying. Also watch the units: 4.5 cm on a 1 : 50 000 map is 225 000 cm, which is 2.25 km, not 225 km.

Exercise 22B Maps and scale drawings in context

Calculator allowed

PSThe map shows three villages A, B and C joined by roads. The scale is 1 : 50 000.

(a)Work out the real length of the road from A to B.

(b)Work out the real length of the road from A to C.

(c)Pat drives from A to C by way of B. How much further is this than driving straight from A to C?

Done

PSA ladder is 6.5 m long. Its foot is 2.5 m from a vertical wall on flat ground. Make a scale drawing of the right-angled triangle using a scale of 1 cm to 1 m (the triangle in the picture is drawn to this scale). Measure the angle between the ladder and the ground. Give your answer to the nearest degree.

Done

PSThe shadow of a flagpole on level ground is 8 m long. The angle of elevation of the Sun is 35°. Use a scale drawing with 1 cm to 1 m to find the height of the flagpole to the nearest metre.

Done

FA drawing of a footbridge has a scale of 1 : 200. The real bridge is 36 m long. How long is the bridge on the drawing?

Done

FThe scale of a map is 1 : 50 000. Write the scale in the form “1 cm represents ___ km”.

Done

FA map has a scale of 1 cm to 5 km.

(a)Write the scale as a ratio in the form 1 : n.

(b)The distance between two cities on the map is 7.4 cm. Work out the real distance.

Done

PSOn a map with a scale of 1 : 25 000, a school is 4.8 cm east and 3.6 cm north of a railway station. Work out the real distance from the station to the school in a straight line. Give your answer in km.

Done

PSA cyclist rides along a route that is 15.6 cm long on a map with a scale of 1 : 50 000. She cycles at 12 km/h. How long does the ride take? Give your answer in minutes.

Done

PSA model of a tower is built to a scale of 1 : 300. The model is 18 cm tall. A window on the model is 0.5 cm tall. Work out the real height of (a) the tower, (b) the window. Give both answers in metres.

Done

22.2

Ruler and compass constructions

Edexcel 1MA1: G2
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Ruler and compass constructionsComing soon

To construct something in geometry means to draw it accurately using only a ruler (for straight lines) and a pair of compasses (for arcs and circles). A protractor is used only to check your work. Constructions give exact results, because the compasses make every point on an arc exactly the same distance from the centre.

Rules for good constructions. Use a sharp pencil. Keep the compasses tight so the radius does not slip. Never rub out the construction arcs: they show that you used the correct method, and an examiner gives marks for them. When two arcs need the same radius, do not change the compasses in between.

1. The perpendicular bisector of a line AB. This is the line that cuts AB in half (bisects it) and meets it at 90° (is perpendicular to it). Every point on this line is the same distance from A and from B.

Open the compasses to more than half the length of AB. Put the point on A and draw an arc above and an arc below the line.
Keep the same radius. Put the point on B and draw two more arcs that cross the first two.
Use a ruler to join the two places where the arcs cross. This line is the perpendicular bisector. It cuts AB at its midpoint M.

Perpendicular bisector of AB. The four arcs all have the same radius, so P and Q are each the same distance from A and B. The line PQ cuts AB at its midpoint M and makes 90° with it.

Why does it work? The four arcs have equal radii, so the four points A, P, B, Q are joined by four equal lengths: AP = PB = BQ = QA. That shape is a rhombus, and the diagonals of a rhombus cross each other at 90° and cut each other in half.

2. The bisector of an angle. To bisect an angle means to cut it into two equal angles.

Put the compass point on the corner O. Draw an arc that crosses both arms. Call the crossing points P and Q.
Put the point on P and draw an arc inside the angle. Keep the same radius, put the point on Q and draw another arc so the two arcs cross at R.
Use a ruler to draw a straight line from O through R. This line is the angle bisector.

Bisecting an angle of 64°. OR cuts it into two angles of 32°. Every point on OR is the same distance from the two arms.

3. The perpendicular from a point P to a line. The perpendicular from P to the line is the shortest path from P to the line. This is the “perpendicular distance”.

Put the compass point on P and draw an arc that crosses the line twice. Call the crossing points X and Y.
Put the point on X and draw an arc on the other side of the line from P. Keep the same radius, put the point on Y and draw another arc to cross it at Z.
Join P to Z with a ruler. The line PZ crosses the line at F, and PF is perpendicular to the line.

The perpendicular from P to the line. The length PF is the shortest distance from P to the line.

4. The perpendicular at a point P on a line. This makes a right angle at P.

Put the compass point on P and draw arcs that cross the line on each side of P, at X and Y. (P is exactly in the middle of X and Y.)
Open the compasses wider. Put the point on X and draw an arc above the line. Keep the same radius, put the point on Y and draw an arc to cross it at Z.
Join P to Z. The line PZ is perpendicular to the line at P.

A right angle at P on the line. Notice that this is the same idea as the perpendicular bisector of XY.

5. An angle of 60°. Draw a line AB. Put the compass point on A and draw an arc that crosses the line at P. Keep the same radius, put the point on P and draw an arc that crosses the first arc at Q. Join A to Q. The triangle APQ has three equal sides, so it is equilateral and angle QAP is 60°. From 60° you can make 30° by bisecting, and 120° by constructing two 60° angles side by side.

Constructing 60°. AP = AQ = PQ (the same radius each time), so angle QAP is 60°.

Key ideas

Perpendicular bisector of AB: equal arcs from A and from B, join the crossing points. It cuts AB in half at 90°.
Angle bisector: arc from the corner, then equal arcs from the two crossing points, join the corner to where those arcs cross.
Perpendicular from a point to a line: arcs from the point cut the line twice, then equal arcs from those two points. This gives the shortest distance.
60° is made from an equilateral triangle: use the same radius for both arcs.
Never rub out your construction arcs.

Worked example 22.2.1

Draw a line AB that is 8 cm long. Construct the perpendicular bisector of AB.

Key words. To construct means to draw accurately using only a ruler and a pair of compasses (no protractor). The construction arcs are the curved lines the compasses leave; you must not rub them out.

Method. Use equal arcs from A and from B (more than 4 cm), then join the two places where the arcs cross.

Draw AB with a ruler. The line must be exactly 8 cm long because we are told its length.
AB = 8 cm
Draw arcs from A. The radius must be more than half of AB (more than 4 cm), or the arcs from A and B will not meet. Use 5.2 cm.
Compass point on A: draw one arc above AB and one below it.
Draw arcs from B. Use the same compass setting so that the two points where the arcs cross are the same distance from A and from B.
Compass point on B: draw arcs that cross the first two at P and Q.
Join P and Q. The line PQ is the perpendicular bisector. Leave all the arcs.

Check. Measure: the line crosses AB at M with AM = MB = 4 cm and the angle at M is 90°. ✓

Watch out. If the compass setting is half of AB or less, the arcs will not cross. Always open the compasses wider than half the line.

Worked example 22.2.2

Draw an angle of 72°. Construct the bisector of the angle. What size is each of the two new angles?

Method. Draw the angle with a protractor, then use arcs from the corner and then from the two crossing points.

Draw the 72° angle. Use a ruler and a protractor for the angle you are given.
Corner O, arms OA and OB
Draw an arc from O. The arc must cross both arms. Call the crossing points P and Q.
Compass point on O; arc crosses OA at P and OB at Q.
Draw arcs from P and Q. Use the same radius for both. The arcs cross at R.
Compass point on P, then on Q, same setting: arcs cross at R.
Join O to R. OR is the angle bisector.
Work out each angle. A bisector cuts the angle into two equal halves.
72° ÷ 2 = 36° each

Check. Measure both new angles with a protractor: each should be 36°. ✓

Watch out. Use the compasses for the construction. Marking the half-angle with a protractor loses the marks for the method.

Worked example 22.2.3

The diagram shows a point P that is 2.8 cm above a straight line. Construct the perpendicular from P to the line. How far is P from the line?

Method. Use arcs from P to find X and Y, then equal arcs from X and Y to find Z. Join P to Z.

Draw an arc from P. Open the compasses wide enough to cross the line twice. Draw an arc crossing the line at X and Y.
Compass point on P.
Draw arcs from X and Y. On the far side of the line from P, draw arcs of the same radius from X and from Y. They cross at Z.
Radius more than half of XY.
Join P to Z. The line PZ meets the line at F, and PF is perpendicular to the line.
Measure PF. The perpendicular distance from P to the line is the shortest distance from P to the line.
PF = 2.8 cm

Check. Pick any other point on the line, such as X. The distance PX is the hypotenuse of a right-angled triangle with side PF, so PX is longer than PF. ✓

Watch out. The shortest distance from a point to a line is always the perpendicular distance, never a slanted one.

Worked example 22.2.4

Construct an angle of 30° using only a ruler and compasses.

Method. Build 30° from the angles you can construct exactly: 60° (arc method), 90° (perpendicular) and halves of them (bisector).

Draw a straight line and mark the point A. The angle will have its corner at A, with one arm along the line.
Use a ruler. Make the line about 8 cm long.
Construct 60°, then bisect it. 30° is half of 60°, so first make 60° and then use the angle bisector construction.
Keep the compasses at the same setting while you work, and keep all your arcs.
Join A to the new point. Draw the second arm of the angle with a ruler, passing through the intersection of the arcs.
Check with a protractor. A construction is accurate to about 1°. Measure the angle.
The angle should measure 30°.

Check. The protractor reading should be within 1° of 30°. ✓

Watch out. Use the compasses for the construction. A protractor is only for checking afterwards.

Worked example 22.2.5

A point P is marked on a straight line. Construct a line that is perpendicular to the line at P.

Method. Use arcs from P to find X and Y, then equal larger arcs from X and Y.

Mark X and Y. Compass point on P: draw arcs that cross the line on both sides of P. P is now exactly half way between X and Y.
PX = PY
Draw arcs from X and Y. Open the compasses wider (more than PX). Keep the setting the same for both arcs. They cross at Z.
Compass point on X, then on Y: arcs cross at Z above the line.
Join P to Z. Use a ruler. Leave the arcs.

Check. Place a set square or protractor on P: the angle between PZ and the line should be 90°. ✓

Watch out. X and Y must be the same distance from P. Without that, the line is not perpendicular.

Watch out

Do not change the compass setting between the two arcs that must cross. If the radius changes the crossing point is not the same distance from both centres and the line will not be a true bisector. Do not rub out arcs.

Exercise 22C Constructions: method and accuracy

Non-calculator

FPut the steps in the correct order. Write the letters.

(a)Constructing the perpendicular bisector of AB:
A Keep the same radius, put the point on B and draw arcs that cross the first two.
B Draw the line AB with a ruler.
C Join the two crossing points with a ruler.
D Open the compasses to more than half the length of AB.
E Put the compass point on A and draw an arc above and below AB.

(b)Bisecting an angle:
A Keep the same radius and draw an arc from the second crossing so that the arcs cross.
B Put the compass point on the corner O and draw an arc that crosses both arms.
C Draw a straight line from O through the point where the arcs cross.
D Put the point on the first crossing and draw an arc inside the angle.

Done

FFill in each gap.

(a)The perpendicular bisector of AB meets AB at an angle of ___ degrees.

(b)The perpendicular bisector of AB cuts AB into two ___ parts.

(c)When you bisect an angle, the first arc is drawn with the compass point on the ___ of the angle.

(d)You must never ___ out the construction arcs.

Done

FDraw a line AB that is 8 cm long. Construct the perpendicular bisector of AB. Mark the midpoint M. What is the length AM?

Done

FConstruct the bisector of each angle. State the size of each of the two equal angles.

(a)An angle of 50°

(b)An angle of 80°

(c)An angle of 120°

(d)An angle of 140°

Done

FDescribe how to construct each angle using only a ruler and compasses. What does the angle measure when you check it with a protractor?

(a)An angle of 60°

(b)An angle of 120°

(c)An angle of 45°

(d)An angle of 90°

Done

FEach diagram is drawn actual size. Construct the perpendicular from P to the line, then measure the shortest distance from P to the line.

(a)

(b)

Done

RAisha rubs out her construction arcs after she has drawn a perpendicular bisector. She says: “The line is right, so the arcs are not needed.” Explain why she should leave the arcs on her work.

Done

PSA straight path runs along a line. A point P is on the path. Construct the perpendicular to the path at P. Q is the point on this perpendicular that is 4 cm from P. The point A is on the path 3 cm from P. Work out the distance AQ.

Done

RExplain why the perpendicular distance from a point P to a line is shorter than the distance from P to any other point on the line. Use P that is 3 cm from the line and a point X on the line that is 4 cm along from the foot of the perpendicular.

Done

22.3

Constructing triangles

Edexcel 1MA1: G2, G15
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Constructing trianglesComing soon

A triangle has three sides and three angles, but you do not need all six to draw it. If you are told enough information, there is only one possible triangle (apart from its position on the page). In Chapter 21 you met the same ideas as the conditions for congruent triangles: three sides (SSS), two sides and the angle between them (SAS), and two angles and a side (ASA).

SSS: three sides. Draw one side as the base, say AB. The corner C must be the right distance from A and from B. Draw an arc of the right radius from A, and another from B. The arcs cross at C.

Constructing a triangle with sides 7 cm, 6 cm and 5 cm. Step 1: base AB = 7 cm. Step 2: arc of radius 6 cm from A. Step 3: arc of radius 5 cm from B, crossing at C. Step 4: join AC and BC.

SAS: two sides and the angle between them. Draw one side, measure the angle at one end with a protractor, mark the second side along the new line, then join the ends.

SAS: AB = 6 cm, angle A = 50°, AC = 4.5 cm. Step 3 marks C with the compasses. Join BC to finish.

ASA: two angles and a side. Draw the side, then measure one angle at each end. The two lines cross at the third corner.

ASA: angle A = 50°, AB = 6 cm, angle B = 60°. The third angle C is 180° − 50° − 60° = 70°.

Right-angled triangles (RHS). If you know the hypotenuse and one other side, construct the right angle, then use an arc for the hypotenuse.

When is a triangle impossible? The two shorter sides must be able to reach each other. If the arcs do not cross, there is no triangle. This happens when the two short sides add up to less than or equal to the longest side. For example 3 cm, 4 cm and 8 cm cannot make a triangle because 3 + 4 = 7 is less than 8: the arcs fall short of each other.

Accuracy. A good construction is accurate to about 1 mm for lengths and 1° for angles. Always say which measurement you have taken and give a sensible unit. Leave all construction arcs on your page.

Key ideas

SSS: base, then an arc from each end. SAS: side, angle at one end, then second side. ASA: side, then an angle at each end.
Three angles alone do not fix a triangle: you can have a small or a big one with the same angles.
A triangle with sides a, b, c is only possible if the two shorter sides add up to more than the longest.
Measure carefully and leave construction arcs.
RHS: build the right angle first, then use an arc for the hypotenuse.

Worked example 22.3.1

Construct triangle ABC with AB = 7 cm, AC = 6 cm and BC = 5 cm. Measure the largest angle of the triangle.

Method. Draw the longest side as a base, then find the third corner with two arcs. Measure the angle with a protractor.

Draw the base AB. Choose one side as the base and draw it with a ruler.
AB = 7 cm
Draw an arc from A. C is 6 cm from A, so every possible position of C is on an arc of radius 6 cm centred on A.
Compasses set to 6 cm, point on A.
Draw an arc from B. C is also 5 cm from B. Where the two arcs cross is the only point that is the right distance from both A and B.
Compasses set to 5 cm, point on B. The arcs cross at C.
Join AC and BC. Use a ruler. Keep the arcs.
Measure angle C. The biggest angle is opposite the longest side (7 cm), which is angle C. Use a protractor.
Angle C = 78.5° to 1 decimal place, so about 78°

Check. Cosine rule: cos C = (5² + 6² − 7²) ÷ (2 × 5 × 6) = 0.2000, so C = 78.5°. Your measurement should be within 1° of this. ✓

Watch out. The two arcs must have the radii of the two sides that meet at C. Do not swap them.

Worked example 22.3.2

Construct triangle ABC with AB = 6 cm, angle A = 50° and AC = 4.5 cm. Measure the length of BC.

Method. Draw one side, measure the angle at one end of it, mark the second side along the new line, then join up.

Draw the side AB. It is one of the two sides next to the known angle.
AB = 6 cm
Measure the angle at A. Place the centre of the protractor on A with its zero line along AB. Mark the angle and draw a faint line through the mark.
Angle A = 50°
Mark C. C is on the new line and AC is a given length. Measure with a ruler (or an arc of the compasses).
AC = 4.5 cm
Join BC and measure it. This completes the triangle. Measure BC with a ruler.
BC = 4.64 cm ≈ 4.6 cm

Check. Cosine rule: BC² = 6² + 4.5² − 2 × 6 × 4.5 × cos 50° = 21.54, so BC = 4.64 cm. Your measurement should be within 1 mm of this. ✓

Watch out. The angle must be between the two given sides. If it is not between them, you cannot draw the triangle this way.

Worked example 22.3.3

Construct triangle ABC with angle A = 50°, AB = 6 cm and angle B = 60°. Measure the length of AC.

Method. Draw the side, then draw the angle at each end. The lines meet at the third corner.

Draw the side AB. The side between the two known angles is the base.
AB = 6 cm
Draw the angle at A. Use a protractor at A with its zero line along AB. Draw a long faint line.
Angle A = 50°
Draw the angle at B. Use the protractor at B. Measure from BA (the zero line points back towards A). Draw a line that crosses the first line at C.
Angle B = 60°
Measure the side you want. Where the two lines cross is the third corner C. Measure with a ruler.
AC = 5.53 cm ≈ 5.5 cm

Check. Sine rule: the third angle is 180° − 50° − 60° = 70°, and AC = 6 × sin 60° ÷ sin 70° = 5.53 cm. ✓

Watch out. Angles in a triangle add up to 180°, so the angles must add up to less than 180° or the lines will never meet.

Worked example 22.3.4

Explain why it is not possible to construct a triangle with sides of length 3 cm, 4 cm and 8 cm.

Method. Try to draw it. The two shorter sides must reach each other when they are attached to the ends of the longest side.

Draw the longest side. Take the 8 cm side as the base AB.
AB = 8 cm
Draw the two arcs. C must be 3 cm from A and 4 cm from B. So the arcs have radii 3 cm and 4 cm.
Arc from A: radius 3 cm. Arc from B: radius 4 cm.
See what happens. The arcs reach 3 cm and 4 cm from the ends. Together they cover only 3 + 4 = 7 cm of the 8 cm base.
3 + 4 = 7 and 7 < 8, so the arcs do not meet.
Write the conclusion. If the arcs never cross there is no point C.
No triangle can be drawn. The two shorter sides must add up to more than the longest side.

Check. A triangle with sides 3, 4, 5 does exist, because 3 + 4 = 7 is more than 5. ✓

Watch out. It is not enough for the sides to be different. They must satisfy: two shorter sides add up to more than the longest.

Worked example 22.3.5

Construct a right-angled triangle ABC with the right angle at B, AB = 2.5 cm and hypotenuse AC = 6.5 cm. Measure BC.

Method. A right angle, the hypotenuse and one other side fix a triangle. Construct the right angle first, then the arc for the hypotenuse.

Draw the base AB. The base is one of the shorter sides.
AB = 2.5 cm
Construct a right angle at B. Use the perpendicular-at-a-point construction at B (arcs from B, then larger arcs). The perpendicular line goes up from B.
Angle ABC = 90°
Draw an arc from A. C is 6.5 cm from A, so set the compasses to 6.5 cm and draw an arc from A that crosses the perpendicular at C.
AC = 6.5 cm
Join AC and measure BC. Use a ruler.
BC = 6.0 cm

Check. Pythagoras: BC² = 6.5² − 2.5² = 36.00, so BC = 6.0 cm. ✓

Watch out. The hypotenuse is opposite the right angle, so the arc is drawn from the end of the base that is not at the right angle.

Watch out

Check which sides and angles you have been given. For SAS the angle must be between the two sides. If you use the wrong end of the line you will draw a different triangle. And three angles (AAA) is not enough: it only tells you the shape, not the size.

22.4

Loci: the four basic rules

Edexcel 1MA1: G2
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Loci: the four basic rulesComing soon

A locus (the plural is loci) is the set of all the points that obey a rule. A locus is usually a line, a curve, or a region. For example, think of a dog tied to a post by a rope 4 m long. The dog can stand on any point that is 4 m or less from the post: the locus of the furthest points it can reach is a circle.

There are four basic loci. Every harder locus problem uses them.

1. A fixed distance from a point. The locus of points that are exactly 3 cm from a point A is a circle with centre A and radius 3 cm. The points inside the circle are less than 3 cm from A, and the points outside are more than 3 cm from A.

The circle is the locus of points 3 cm from A. The shaded region is the locus of points that are 3 cm or less from A.

2. The same distance from two points A and B. This locus is the perpendicular bisector of AB (Section 22.2). The points on one side are nearer to A, and the points on the other side are nearer to B.

The dashed line is the perpendicular bisector of AB: points on it are the same distance from A and B. The shaded side is the locus of points closer to A than to B.

3. A fixed distance from a line. The locus of points exactly 1.5 cm from a straight line segment AB is made of two parallel lines (one each side of AB) with a semicircle at each end. It looks like a running track (a stadium shape). If the line goes on for ever, the locus is just two parallel lines, 1.5 cm from the line on each side.

The shaded stadium shape is the locus of points that are 1.5 cm or less from the line segment AB.

4. The same distance from two lines. The points that are the same distance from two lines that cross lie on the bisector of the angle between the lines (Section 22.2). On one side of the bisector the points are nearer to the first line; on the other side they are nearer to the second line.

The dashed line bisects the angle at O. Points on it are the same distance from both arms. The shaded region is the locus of points nearer to the lower arm than to the upper arm.

Reading the rules. Words such as “within 3 cm” or “no more than” mean the circle and everything inside it. Words such as “at least 3 cm from” mean the circle and everything outside. “Closer to A than to B” means the side of the perpendicular bisector that contains A. Use a solid line for a boundary that is included and a dashed line for one that is not.

Key ideas

The locus of points a fixed distance from a point is a circle.
The locus of points equidistant from two points is their perpendicular bisector.
The locus of points a fixed distance from a line segment is a stadium shape (two parallel lines and two semicircles).
The locus of points equidistant from two lines is the angle bisector.
Within / less than = inside. At least / more than = outside.

Worked example 22.4.1

Draw the locus of points that are exactly 3 cm from a point A. Then shade the region of points that are less than 3 cm from A.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. The points exactly 3 cm from A make a circle. The points less than or equal to 3 cm are the inside of the circle.

Mark the point A. This is the centre of the locus.
Point A
Set the compasses to 3 cm. Every point on the arc will be exactly 3 cm from A.
Radius = 3 cm
Draw the circle. Put the compass point on A and draw a complete circle.
Shade the region. “Less than 3 cm” means the points inside the circle.

Check. A point 2 cm from A is inside the shaded region, and a point 4 cm from A is outside it. ✓

Watch out. The circle is the locus of points exactly 3 cm away. The inside of the circle is the locus of points less than 3 cm away.

Worked example 22.4.2

Points A and B are 5 cm apart. Draw the locus of points that are the same distance from A and from B. Shade the region of points that are closer to A than to B.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. The locus of points equidistant from A and B is the perpendicular bisector of AB. The points nearer to A are on the same side as A.

Draw AB. Use a ruler. AB = 5 cm.
AB = 5 cm
Construct the perpendicular bisector. Use equal arcs from A and from B (more than 2.5 cm), then join the places where they cross (Section 22.2).
Leave the arcs.
Shade the correct side. The side that contains A is closer to A.

Check. Pick a point on the shaded side, such as A itself: it is 0 cm from A and 5 cm from B. ✓

Watch out. Do not shade the side with B on it.

Worked example 22.4.3

A and B are the ends of a line segment 4 cm long. Draw the locus of points that are exactly 1.5 cm from the line segment AB.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. The locus is two straight lines parallel to AB, joined at each end by a semicircle centred on the end of the segment.

Draw AB. Use a ruler.
AB = 4 cm
Draw the two parallel lines. At A and at B, construct perpendiculars to AB, mark points 1.5 cm along each, and join them with a ruler (or use a set square).
The lines are 1.5 cm above and 1.5 cm below AB.
Draw the two semicircles. Put the compass point on A with radius 1.5 cm and draw a semicircle around the end. Repeat at B.
Radius 1.5 cm at each end.
Check the shape. It should look like a running track.

Check. Every point on the track is exactly 1.5 cm from the nearest point of AB. ✓

Watch out. Do not draw straight ends. Points near the end of AB must be 1.5 cm from B, which gives a semicircle.

Worked example 22.4.4

Two lines meet at O with an angle of 60° between them. Draw the locus of points that are the same distance from both lines.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. The points equidistant from two lines are on the bisector of the angle between them.

Draw the two lines. Draw the lines OA and OB with an angle of 60° at O.
Angle AOB = 60°
Construct the angle bisector. Draw an arc from O, then equal arcs from P and Q, then join O to where they cross (Section 22.2).
Say what the line means. Every point on this line is the same distance from OA as from OB.
The angle between each arm and the bisector is 30°.

Check. Take the point R on the bisector 4 cm from O. Its perpendicular distance to either arm is 4 sin 30° = 2 cm. ✓

Watch out. Measure the distance to a line along a perpendicular, not a slanting line.

Worked example 22.4.5

A mobile phone mast M can be received up to 12 km away. A map has a scale of 1 : 400 000. Draw the region on the map where you can receive the signal.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. A mobile phone mast can be received up to 12 km away. The points within 12 km of the mast form a circle. On the map, divide by the scale.

Change 12 km to cm. 1 km = 100 000 cm.
12 × 100 000 = 1 200 000 cm
Divide by the scale number. Real life to map: divide. The scale is 1 : 400 000.
1 200 000 ÷ 400 000 = 3 cm
Draw the locus. Draw a circle of radius 3 cm centred on the mast M. Shade the inside of the circle.

Check. A scale of 1 : 400 000 means 1 cm is 4 km, so 3 cm is 12 km. ✓

Watch out. The radius on the map is 3 cm, not 12 cm.

Watch out

A locus is the whole set of points, not one point. Students often draw only a single dot or a short line. Also decide carefully whether the region is the circle itself, the inside, or the outside: “at most” and “within” mean inside.

22.5

Combined loci and region problems

Edexcel 1MA1: G2
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Combined loci and region problemsComing soon

Real problems often give more than one rule: “the treasure is within 5 m of the tree and nearer to the gate than to the wall”. Each rule gives a locus. The answer is the region where all the rules are true at once. You find it in five steps.

Draw an accurate plan (use the scale you are given, and a ruler).
Draw the first locus lightly: a circle, a bisector, a parallel line. Decide which side (or inside/outside) obeys the rule.
Draw the second locus in the same way, and any others.
Shade the overlap: only the part of the plan that obeys every rule.
Check with a test point inside your region. Measure to see that it obeys each rule.

Think of each rule as cutting away the part of the plan that does not obey it. What is left is the answer. For a rule such as “within 4 m of A” you keep the inside of a circle. For “at least 2 m from the wall” you keep the part of the plan on the far side of a line 2 m from the wall. For “nearer to A than B” you keep one side of the perpendicular bisector. For “nearer to side AB than to side AD” you keep one side of the angle bisector.

If a region would be empty (no point obeys every rule) say so. For example, no point can be within 2 cm of A and at least 5 cm from A.

Key ideas

Draw each locus separately, then shade the overlap of all of them.
Within / at most = inside the circle or band. At least = outside.
“Nearer to A than B”: the side of the perpendicular bisector with A on it. “Nearer to AB than AD”: one side of the angle bisector.
Use a test point to check your shaded region.
Use the scale: each square in the plans in this book is 1 unit (1 m or 1 km).

Worked example 22.5.1

The plan shows a rectangular garden ABCD with AB = 8 m and AD = 5 m. The plan is drawn on a grid where each square is 1 m. A tap is at corner A. Lily plants flowers in the region that is within 4 m of A and nearer to side AB than to side AD. Shade this region.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. Draw each rule as a boundary line, then shade the region that obeys all the rules.

Look at the plan and list the rules. Each rule gives a locus. The answer must obey every rule at the same time.
Rule 1: within 4 m of A
Rule 2: nearer to side AB than to side AD
Draw rule 1: within 4 m of A. Each square is 1 m, so set the compasses to 4 squares. Put the point on A and draw an arc (a quarter circle) inside the garden.
Draw rule 2: nearer to side AB than to side AD. Points the same distance from AB and AD are on the bisector of angle DAB. Construct it (it is the diagonal 45° line from A). Points nearer to AB are below this line.
Shade the region that obeys all the rules. Points that are in the right place for every rule are where the regions overlap.

Check. Pick a point in the shaded region, for example (2.4, 1) measured from the bottom left corner of the plan. Test each rule:
distance from A = 2.6 m, which is at most 4 m ✓
distance from AB = 1.0 m and from AD = 2.4 m, so it is closer to AB ✓

Watch out. A point must pass every test. Shade only the overlap, not all of each region separately.

Worked example 22.5.2

The plan shows a field ABCD, 12 m by 7 m, drawn on a grid where each square is 1 m. There are two trees, P and Q. A treasure chest is buried so that it is nearer to P than to Q, at least 2 m from P, and within 3 m of the wall AB. Shade the region where the chest could be.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. Draw each rule as a boundary line, then shade the region that obeys all the rules.

Look at the plan and list the rules. Each rule gives a locus. The answer must obey every rule at the same time.
Rule 1: nearer to the tree P than to the tree Q
Rule 2: at least 2 m from the tree P
Rule 3: within 3 m of the wall AB
Draw rule 1: nearer to the tree P than to the tree Q. Points the same distance from P and Q lie on the perpendicular bisector of PQ. Construct it (it is the vertical line half way between the trees). Shade the side with P.
Draw rule 2: at least 2 m from the tree P. Draw a circle with centre P and radius 2 squares. “At least 2 m” means on the circle or outside it.
Draw rule 3: within 3 m of the wall AB. Draw a line parallel to AB and 3 squares above it. “Within 3 m” means between this line and the wall AB.
Shade the region that obeys all the rules. Points that are in the right place for every rule are where the regions overlap.

Check. Pick a point in the shaded region, for example (3, 1.2) measured from the bottom left corner of the plan. Test each rule:
distance from P = 2.3 m and from Q = 6.4 m, so it is closer to P ✓
distance from P = 2.3 m, which is at least 2 m ✓
distance from AB = 1.2 m, which is at most 3 m ✓

Watch out. A point must pass every test. Shade only the overlap, not all of each region separately.

Worked example 22.5.3

The map shows two towns X and Y joined by a straight road. Each square is 1 km. A phone mast is to be built at most 6 km from X, at least 4 km from Y, and within 2 km of the road XY. Shade the region where the mast can be built.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. Draw each rule as a boundary line, then shade the region that obeys all the rules.

Look at the plan and list the rules. Each rule gives a locus. The answer must obey every rule at the same time.
Rule 1: at most 6 km from town X
Rule 2: at least 4 km from town Y
Rule 3: within 2 km of the straight road XY
Draw rule 1: at most 6 km from town X. Each square is 1 km. Draw a circle with centre X and radius 6 squares. “At most 6 km” means inside the circle.
Draw rule 2: at least 4 km from town Y. Draw a circle with centre Y and radius 4 squares. “At least 4 km” means outside the circle.
Draw rule 3: within 2 km of the straight road XY. The road runs along the line through X and Y. Draw lines 2 squares above and 2 squares below it. “Within 2 km” means between these lines.
Shade the region that obeys all the rules. Points that are in the right place for every rule are where the regions overlap.

Check. Pick a point in the shaded region, for example (3.6, 4) measured from the bottom left corner of the plan. Test each rule:
distance from X = 0.6 m, which is at most 6 m ✓
distance from Y = 7.4 m, which is at least 4 m ✓
distance from the road XY = 0.0 m, which is at most 2 m ✓

Watch out. A point must pass every test. Shade only the overlap, not all of each region separately.

Worked example 22.5.4

The diagram shows a triangular park ABC drawn on a grid where each square is 1 m. A bench is to go nearer to A than to B, within 5 m of C, and nearer to side AB than to side AC. Shade the region where the bench can go.

Key words. A locus (plural loci) is the set of all the points that obey a rule. Equidistant means the same distance away.

Method. Draw each rule as a boundary line, then shade the region that obeys all the rules.

Look at the plan and list the rules. Each rule gives a locus. The answer must obey every rule at the same time.
Rule 1: nearer to A than to B
Rule 2: within 5 m of C
Rule 3: nearer to side AB than to side AC
Draw rule 1: nearer to A than to B. Construct the perpendicular bisector of AB and shade the side that contains A.
Draw rule 2: within 5 m of C. Draw a circle with centre C and radius 5 squares. Shade inside it.
Draw rule 3: nearer to side AB than to side AC. Construct the bisector of angle BAC. Points nearer to AB are on the AB side of the bisector.
Shade the region that obeys all the rules. Points that are in the right place for every rule are where the regions overlap.

Check. Pick a point in the shaded region, for example (3.3, 1.6) measured from the bottom left corner of the plan. Test each rule:
distance from A = 3.7 m and from B = 5.0 m, so it is closer to A ✓
distance from C = 4.4 m, which is at most 5 m ✓
distance from AB = 1.6 m and from AC = 2.2 m, so it is closer to AB ✓

Watch out. A point must pass every test. Shade only the overlap, not all of each region separately.

Watch out

Do not shade each locus separately and stop. The answer is only where the shading overlaps. Also, check each inequality: “at least” means outside the circle, and “nearer to AB” means one particular side of the bisector, so test a point before you choose.

22.6

Bearings: measuring, drawing and back bearings

Edexcel 1MA1: G15
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Bearings: measuring, drawing and back bearingsComing soon

A bearing is an angle that tells you the direction of one place from another. Pilots, sailors and walkers use bearings because a direction such as “north-east” is not exact enough. Every bearing follows three rules.

Measure from north. The angle always starts at the north line (straight up the page).
Measure clockwise. Turn the way the hands of a clock turn, so east is 90°.
Give three figures. A bearing of 5° is written 005° and 45° is written 045°. Bearings go from 000° to 360°.

The eight compass points as bearings. North is 000°, east is 090°, south is 180° and west is 270°. Each step between neighbours is 45°.

Reading a bearing. To find the bearing of B from A, stand at A and look at B. Draw a north line at A. Then turn clockwise from the north line until you point to B. The angle you turn through is the bearing. Be careful with the word “from”. The bearing of B from A starts at A. It is a different bearing from the bearing of A from B.

The red line is at 070°: starting at north (000°) and turning clockwise, 70° takes you a little north of east. The blue line is at 250°: turning clockwise past 180° (south) gets you into the south-west.

Drawing a bearing. To draw a line from P on a bearing of 135°: (1) draw a north line through P; (2) put the protractor on P with its zero on the north line; (3) turn clockwise to 135° and mark a dot; (4) draw a line from P through the dot. A bearing between 180° and 360° goes past south. You can measure 360° minus the bearing anticlockwise from north instead.

North lines are parallel. The north line at every point points the same way, so the north lines at two different points are parallel. This is the key to all bearings angle problems. Always draw a north line at each point you measure from.

Back bearings. If the bearing of B from A is 070°, what is the bearing of A from B (the way back)? Draw north lines at A and B. They are parallel lines and the line AB crosses both. The angle at B between the north line and the line back to A is a co-interior angle with 70° (they add to 180°), so it is 180° − 70° = 110°. But we need the angle measured clockwise, so take it from a full turn: 360° − 110° = 250°.

The bearing of B from A is 070°. The bearing of A from B is 250°. Both north lines point the same way, and 250° − 70° = 180°.

The shortcut. The way back is the opposite direction, which is half a turn (180°) away. So back bearing = bearing ± 180°. If the bearing is less than 180°, add 180°. If it is 180° or more, subtract 180°. This keeps the answer between 000° and 360°.

Key ideas

A bearing is measured from north, clockwise, with three figures.
The bearing of B from A: stand at A, face north, turn clockwise to B.
Draw a north line at every point you measure from. North lines are parallel.
Back bearing: add 180° if the bearing is less than 180°, subtract 180° if it is 180° or more.
Co-interior angles between parallel north lines add up to 180°.

Worked example 22.6.1

The diagram shows a line drawn from the centre of a bearings dial. Write down the bearing of the line.

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. Start at the north line, then turn clockwise until you reach the line. Read the number on the dial.

Find north. Bearings always start from north, which is straight up the page.
North = 000°
Turn clockwise to the line. Follow the dial round the way a clock hand moves. The tick marks are 5° apart and the numbers are 30° apart.
The line crosses the dial at 70°
Write three figures. A bearing is always written with three figures, so put zeros in front of a small number.
Bearing = 070°

Check. A line at 70° is less than 180°, so it points to the east side. This agrees with the picture. ✓

Watch out. Do not measure anticlockwise. A line pointing to the west side has a bearing bigger than 180°.

Worked example 22.6.2

Draw a line from the point P, 5 cm long, on a bearing of 135°. Label the end Q.

Method. Draw a north line at P, measure the bearing clockwise from north, then draw the line to the right length.

Mark P and draw a north line through it. Every bearing is measured from north. Use a ruler to draw a line pointing straight up the page.
Point P with a north line
Place the protractor. Put the centre of the protractor on P and its 0° line along the north line.
Use the protractor the usual way round: start at 000° (north) and turn clockwise to 135°.
Make a mark and draw the line. Mark a small dot at 135°. Draw a line from P through the dot.
Line at 135° from north
Measure 5 cm along the line and label Q. Use a ruler. Label the end of the line.

Check. The line should point to the east half of the page, because 135° is less than 180°. ✓

Watch out. Always start from the north line and measure clockwise. Do not measure from the east-west line.

Worked example 22.6.3

The bearing of B from A is 070°. Work out the bearing of A from B.

Method. The bearing of A from B is the back bearing of the bearing of B from A: add or subtract 180° so the answer stays between 0° and 360°.

Draw a sketch with a north line at each point. The two north lines both point the same way, so they are parallel. The line AB crosses both of them.
Bearing of B from A = 070°
Find the angle at B between the north line and BA. Angles between parallel lines and on the same side of the crossing line are co-interior and add up to 180°.
180° − 070° = 110°
Turn it into a bearing. A bearing is measured clockwise from north, but this angle is measured the other way round. Take it away from a full turn of 360°.
360° − 110° = 250°
Write it with three figures and check with the shortcut. Going back is the exact opposite way, which is half a turn (180°) different from the forward bearing.
70° + 180° = 250°. Back bearing = 250°

Check. Reverse it again: 250° − 180° = 70°. ✓

Watch out. Always draw the north line at the point you are measuring from. If the forward bearing is less than 180° add 180°; if it is more than 180° subtract 180°.

Worked example 22.6.4

The bearing of a lighthouse L from a ship S is 290°. Work out the bearing of the ship from the lighthouse.

Method. The bearing of A from B is the back bearing of the bearing of B from A: add or subtract 180° so the answer stays between 0° and 360°.

Draw a sketch with a north line at each point. The two north lines are parallel. The line AB crosses both of them.
Bearing of B from A = 290°
Find the angle between the north-south line at A and AB. The bearing is more than 180°, so the line has gone past south. The extra angle past south is the bearing minus 180°.
290° − 180° = 110°
Use alternate angles at B. Alternate angles (a Z shape) between parallel lines are equal. So the angle at B between the north line and BA is the same size, and it is measured clockwise from north.
Angle = 110°
Write it with three figures and check with the shortcut. Going back is half a turn different. Because the forward bearing is more than 180°, subtract 180°.
290° − 180° = 110°. Back bearing = 110°

Check. Reverse it again: 110° + 180° = 290°. ✓

Watch out. Always draw the north line at the point you are measuring from. If the forward bearing is less than 180° add 180°; if it is more than 180° subtract 180°.

Worked example 22.6.5

The bearing of B from A is 040°. The bearing of C from B is 110°. Work out the size of angle ABC.

Method. Find the bearing of A from B (the back bearing). Then angle ABC is the gap between the bearings of A and C, both measured from the north line at B.

Find the back bearing. The bearing of A from B is half a turn from the bearing of B from A: 40° is less than 180°, so add 180°.
40° + 180° = 220°
Mark the bearings at B. At B the north line is parallel to the one at A. The line BA is at 220° and the line BC is at 110°, both measured clockwise from north.
BA = 220°, BC = 110°
Find the angle between them. Angle ABC is the difference between the two bearings.
220° − 110° = 110°
Write the answer. This is the angle inside the triangle.
Angle ABC = 110°

Check. Another way: the ship turns right by 110° − 40° = 70°. The angle inside the corner is 180° − 70° = 110°. ✓

Watch out. Draw the north line at B (the point you are working from), not just at A.

Watch out

Do not forget the north line at the second point. A bearing of B from A uses the north line at A. A bearing of A from B uses a different north line, at B. Also remember to write three figures (so 45° is 045°), and measure clockwise, even if the line is on the left of the page.

22.7

Bearings problems: journeys, Pythagoras and trigonometry

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

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Bearings problems: journeys, Pythagoras and trigonometryComing soon

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).

Sketch the journey. Draw the points and the lines, and put a north line at every point where a bearing is given.
Find the angles you need. Use parallel north lines (co-interior or alternate angles), the back bearing (± 180°), and angles in a triangle.
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.
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.

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.

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.

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°
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.
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
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°
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)
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.

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.

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°
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.
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
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°
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.

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.

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
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°
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.

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.

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
Find angle BAC. It is the difference between the two bearings measured from the same point.
100° − 40° = 60°
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°
Find the bearing of A from B. This is the back bearing of the first bearing: add or subtract 180°.
40° + 180° = 220°
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.

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 22I Harder bearings problems

Calculator allowed

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.

Done

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.

Done

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.

Done

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.

Done

22.8

Vectors: column vectors, addition and scalar multiples

Edexcel 1MA1: G24, G25
Grade 4-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Vectors: column vectors, addition and scalar multiplesComing soon

Some quantities only need a number: a mass of 5 kg or a time of 10 seconds. These are called scalars. Others need a number and a direction: a move of 3 squares right and 2 squares up, or a wind of 20 km/h from the west. These are vectors. A vector has a size (its length, also called its magnitude) and a direction.

Notation. A vector can be named by a single bold letter, such as a (you write a with an underline by hand), or by its start and end points with an arrow above them, such as →AB (the vector from A to B). On a diagram, the vector is drawn as an arrow.

Column vectors. A column vector gives the vector as two numbers written one above the other: x
y. The top number x is the move across (right is positive, left is negative). The bottom number y is the move up (up is positive, down is negative). For example 3
2 means 3 right and 2 up, and −2
−4 means 2 left and 4 down.

a = (3 over 2): 3 right, 2 up. b = (−2 over 3): 2 left, 3 up. c = (4 over 0): 4 right, no up or down.

Equal and negative vectors. Two vectors are equal if they have the same length and the same direction. Where they start does not matter: all the arrows 3
2 are the same vector wherever they are drawn. The negative of a vector, written −a, has the same length but the opposite direction. If a = 3
2 then −a = −3
−2.

Translations. A translation moves every point of a shape the same distance in the same direction. It is described by a column vector. The translation 2
−3 moves every point 2 right and 3 down (Chapter 21).

Triangle A is translated by (2 over −3) to triangle B. Every vertex moves 2 right and 3 down.

Adding vectors. To add two vectors, draw them nose to tail: start the second arrow where the first ends. The sum a + b goes straight from the start of the first to the end of the second. With column vectors, add the top numbers and add the bottom numbers.

a = (3 over 1) and b = (1 over 3). a + b = (3 + 1 over 1 + 3) = (4 over 4).

Subtracting vectors. a − b means a + (−b): add the opposite of b. With column vectors, subtract the top numbers and subtract the bottom numbers. On a diagram, if a and b start at the same point, then a − b goes from the tip of b to the tip of a.

a = (4 over 3), b = (1 over 2). a − b = (3 over 1), the arrow from the tip of b to the tip of a.

Scalar multiples. Multiplying a vector by a scalar (a number) changes its length. 2a is twice as long as a in the same direction, ½a is half as long, and −a is the same length in the opposite direction. With column vectors, multiply both numbers by the scalar. Vectors that are scalar multiples of each other are parallel.

a, 2a, −a and ½a are all parallel. They have lengths in the ratio 1 : 2 : 1 : 0.5.

Magnitude. The magnitude (length) of a vector is written |a|. For x
y draw a right-angled triangle with shorter sides x and y: |a| = √(x² + y²) by Pythagoras’ theorem. The vector from A to B is found by subtracting the coordinates: end minus start, so if A = (1, 2) and B = (4, 6) then the vector from A to B is 4 − 1
6 − 2 = 3
4.

Key ideas

A column vector x
y means x across (right +) and y up (up +).
Add or subtract vectors by adding or subtracting the top numbers and the bottom numbers.
Multiply a vector by a scalar: multiply both numbers. Vectors that are scalar multiples are parallel.
Magnitude: |a| = √(x² + y²). The vector from A to B = coordinates of B minus coordinates of A.
A translation by a column vector moves every point the same way.

Worked example 22.8.1

Write down the column vector for the arrow a.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Count the squares from the tail of the arrow to its tip: across first, then up or down.

Find the move across. Right is positive and left is negative. Count from the start of the arrow to its tip, sideways only.
4 right: top number = 4
Find the move up or down. Up is positive and down is negative.
2 down: bottom number = −2
Write the column vector. Across on top, up underneath.
a = 4
−2

Check. Start at the tail and add the vector: move 4 right and 2 down. You arrive at the tip. ✓

Watch out. The top number is across (x) and the bottom number is up (y). Do not swap them.

Worked example 22.8.2

a = 3
−2 and b = −1
4. Work out a + b.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. To add column vectors, add the top numbers and add the bottom numbers separately.

Write the two column vectors. Each column vector has a top number (across) and a bottom number (up).
a + b = 3
−2 + −1
4
Add the top numbers. To add vectors, add the matching numbers: top with top.
3 + −1 = 2
Add the bottom numbers. Now do the same with the bottom with bottom.
−2 + 4 = 2
Write the answer as a column vector. Put the new top number over the new bottom number.
a + b = 2
2

Check. On a grid: start at the origin, go 3 right and 2 down, then follow b. You arrive 2 right and 2 up of the start. ✓

Watch out. Take care with minus signs: subtracting a negative number adds. For example 3 − (−2) = 5.

Worked example 22.8.3

a = 3
1 and b = 1
−2. Work out 2a − 3b.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Multiply each vector by its scalar first, then add or subtract the top numbers and the bottom numbers.

Write each vector as a column. We can only add and subtract once everything is in column form.
2a − 3b = 2 × 3
1 + −3 × 1
−2
Do the scalar multiplications first. Multiply every number in each vector by its scalar (like BIDMAS: multiply before adding).
6
2 + −3
6
Add the top numbers. Add the matching numbers: top with top.
6 + (−3) = 3
Add the bottom numbers. Then bottom with bottom.
2 + 6 = 8
Write the answer.
2a − 3b = 3
8

Check. Check the top numbers again: 6 −3 = 3. ✓

Watch out. Do not forget to multiply both numbers, and watch the signs when you subtract a vector.

Worked example 22.8.4

A is the point (1, 2) and B is the point (4, −3). Write down the vector from A to B.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Subtract the coordinates of the start point from the coordinates of the end point.

Write the two points. A column vector from A to B says how far to move across and up to get from A to B.
A = (1, 2) and B = (4, −3)
Find the across move. Subtract the x-coordinates: end minus start.
4 − 1 = 3
Find the up move. Subtract the y-coordinates: end minus start.
−3 − 2 = −5
Write the column vector. Across on top, up underneath.
→AB = 3
−5

Check. Start at A and add the vector: (1 + 3, 2 + −5) = (4, −3) = B. ✓

Watch out. It is end minus start. If you swap them you get the vector going the other way, which has the opposite signs.

Worked example 22.8.5

a = 5
−12. Work out the magnitude of a.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. The magnitude (length) of a vector is found with Pythagoras’ theorem on its two numbers.

Draw the right-angled triangle. The vector a goes 5 across and 12 up or down. These two moves are at right angles, so the length of the vector is the hypotenuse.
Shorter sides: 5 and 12
Use Pythagoras’ theorem. Square each number, add them, then take the square root. Squaring removes any minus signs.
|a|² = 5² + (−12)² = 25 + 144 = 169
Square root. Find the length.
|a| = √169 = 13

Check. The length must be at least as big as each number: 13 = 13.00 is bigger than 5 and 12. ✓

Watch out. Write |a| for the magnitude. The magnitude is never negative, even if the numbers in the vector are.

Worked example 22.8.6

Triangle ABC has vertices A(1, 1), B(3, 1) and C(2, 3). It is translated by the vector 2
−3. Write the coordinates of the new vertices.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. A translation moves every point of the shape by the same vector, so add the vector to each vertex.

Read the translation. The top number is the move across and the bottom number is the move up.
2
−3: move 2 right and 3 down
Move vertex A. Add the vector to the coordinates of the vertex.
(1, 1) → (1 + 2, 1 + (−3)) = (3, −2)
Move vertex B. Add the vector to the coordinates of the vertex.
(3, 1) → (3 + 2, 1 + (−3)) = (5, −2)
Move vertex C. Add the vector to the coordinates of the vertex.
(2, 3) → (2 + 2, 3 + (−3)) = (4, 0)

Check. Every vertex has moved the same distance in the same direction, so the new shape is identical to the old one (same size, same way up). ✓

Watch out. A translation does not turn or flip the shape. Do not change the order of the vertices.

Worked example 22.8.7

a = 3
−2 and b = −6
4. Show whether a and b are parallel.

Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.

Method. Two vectors are parallel when one is a scalar multiple of the other. Find the multiplier for the top numbers and the bottom numbers and see if it matches.

Compare the top numbers. If the vectors are parallel, one is a multiple of the other, so the top numbers have the same scale factor as the bottom numbers.
−6 ÷ 3 = −2
Compare the bottom numbers. Is the multiplier the same?
4 ÷ −2 = −2
Write the conclusion. Both numbers are multiplied by the same scalar.
b = −2a, so they are parallel.

Check. −2 × 3
−2 = −6
4. ✓

Watch out. The scalar can be negative. Vectors that point in opposite directions along the same line are still parallel.

Watch out

Keep the top number for across and the bottom for up. A common mistake is to read the vector from the wrong end of the arrow, or to forget the minus signs for left and down. And |a + b| is not |a| + |b|: work out the vector first, then its length.

Chapter summary

Skill	Method
Scale: map to real life	Multiply by the scale number (1 : n means 1 cm is n cm); change cm to m (÷ 100) or km (÷ 100 000)
Scale: real life to map	Change to cm, then divide by the scale number
Write a scale as 1 : n	Change both lengths to cm, put the map length first, divide so the first number is 1
Perpendicular bisector	Equal arcs from each end (more than half), join the crossing points
Angle bisector	Arc from the corner, equal arcs from the two crossing points, join the corner to where they cross
Perpendicular from / at a point	Arc from the point to cut the line twice, then equal arcs from those two points
Construct a triangle	SSS: base then two arcs. SAS: side, angle, side. ASA: side then angle at each end
Loci	Circle (fixed distance from a point), perpendicular bisector (equidistant from two points), parallel lines/stadium (distance from a line), angle bisector (two lines)
Combined loci	Draw each locus, then shade only where all the rules overlap; check with a test point
Bearings	Measure from north, clockwise, three figures; draw a north line at every point
Back bearing	Add 180° if the bearing is less than 180°, subtract 180° if it is 180° or more
Two-leg journey	Sketch with north lines, find the angle with parallel-line facts, use Pythagoras/trig (cosine and sine rules in Higher), then change to a bearing
Column vectors	Top = across (right +), bottom = up. Add/subtract each row. Scalar: multiply both numbers. Magnitude: √(x² + y²)
Vector between points	End minus start for each coordinate
Vector geometry (Higher)	Route method: go along known vectors, against an arrow gives a negative; midpoint is half; ratio m/(m + n); parallel = scalar multiple; collinear = parallel + shared point

Can you…?

I can change between map distances and real distances for a scale such as 1 : 50 000.
I can write a scale as a ratio 1 : n from a map distance and a real distance.
I can construct a perpendicular bisector, an angle bisector, a perpendicular from or at a point, and a 60° angle.
I can construct a triangle from SSS, SAS or ASA information.
I can describe the four basic loci and draw them accurately.
I can shade the region that satisfies two or three rules at once.
I can read, draw and write three-figure bearings.
I can find a back bearing and use parallel lines to find angles between bearings.
I can solve two-leg bearings problems with Pythagoras and trigonometry.
I can write a translation as a column vector and use column vectors to add, subtract and scale.
I can find the magnitude of a vector and decide if two vectors are parallel.
Higher: I can use the route method to write vectors in terms of a and b, find midpoints and ratio points, and prove that lines are parallel or that points are collinear.

Key terms

Scale: The ratio of a length on a map or drawing to the real length, e.g. 1 : 50 000.
Construct: Draw accurately using only a ruler and compasses.
Perpendicular bisector: The line that cuts a line segment in half at 90°.
Bisect: To cut into two equal parts.
Angle bisector: The line that cuts an angle into two equal angles.
Locus: The set of all points that obey a rule (plural: loci).
Equidistant: The same distance away.
Bearing: An angle measured clockwise from north, written with three figures.
Back bearing: The bearing of the return direction: the original bearing plus or minus 180°.
Co-interior angles: Angles between two parallel lines on the same side of a crossing line. They add up to 180°.
Vector: A quantity with both size (magnitude) and direction.
Column vector: A vector written as two numbers: across on top, up underneath.
Scalar: An ordinary number, with size but no direction.
Magnitude: The length of a vector, written |a|.
Translation: A move of every point of a shape by the same vector.
Parallel vectors: Vectors that are scalar multiples of each other.
Position vector: A vector giving the position of a point from the origin.
Collinear: Lying on the same straight line.

Total 31 marks. Allow about 31 minutes.

← Chapter 21: Transformations, congruence and similarity Chapter 23: Probability →

GCSE Maths Loci, Bearings, Constructions and Vectors (Foundation): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Scale drawings and maps

Key ideas

Watch out

Exercise 22A Scale and ratio

Exercise 22B Maps and scale drawings in context

Ruler and compass constructions

Key ideas

Watch out

Exercise 22C Constructions: method and accuracy

Constructing triangles

Key ideas

Watch out

Exercise 22D Constructing triangles and measuring

Loci: the four basic rules

Key ideas

Watch out

Exercise 22E Basic loci

Combined loci and region problems

Key ideas

Watch out

Exercise 22F Region problems

Bearings: measuring, drawing and back bearings

Key ideas

Watch out

Exercise 22G Bearings: reading, drawing and back bearings

Bearings problems: journeys, Pythagoras and trigonometry

Key ideas

Watch out

Exercise 22H Bearings and right-angled triangles

Exercise 22I Harder bearings problems

Vectors: column vectors, addition and scalar multiples

Key ideas

Watch out

Exercise 22J Column vectors and translations

Exercise 22K Combining vectors: addition, scalar multiples and magnitude

Chapter summary

Can you…?

Key terms

Exercise 22Z Check-up (non-calculator)

Exercise 22X Exam-style questions