Primary 6 Angles Study Guide

Primary 6 angles in six short steps, from one property on its own figure to a chain of them across a composite figure. Each step has a figure to step through, and worked examples follow.

Start here

How do you find unknown angles in composite figures?

Use one angle property at a time, and write its name beside the line.

A Primary 6 figure joins two or three shapes, such as a square and a triangle. No single fact gives the answer. Your child reads the marks on the figure, finds one angle with one property, then uses that angle on the next line, until the unknown angle is reached. No extra lines are drawn.

In this chapter, your child learns to:

name an angle with three letters, and use straight lines and crossing lines;
find the equal base angles of an isosceles triangle;
use the angles between the parallel sides of a four-sided figure;
add the angles that meet at a join or round a point;
carry an angle along parallel sides into a triangle;
use a fold, which moves an angle without changing its size.

ABCD is a square and DCE is an isosceles triangle with DE = DC. ∠DCE = 65°. Find ∠ADE.

∠ADE = ∠ADC + ∠EDC(the square's corner and the triangle's part)
∠DEC = ∠DCE = 65°(base angles of isosceles triangle DCE)
∠EDC = 180° − 65° − 65° = 50°(angle sum of triangle DCE)
∠ADE = 90° + 50° = 140°(right angle of the square ABCD)

Step 1

Angles on a straight line and vertically opposite angles

Three letters name an angle, and the middle one is the vertex. ∠KOL is the angle at O between OK and OL. Angles on a straight line add up to 180°. Where two straight lines cross, the angles facing each other are equal, so one figure can often be read two ways and give the same answer.

Tap Next step

JOM and KON are straight lines. ∠KOL = 62° and ∠LOM = 71°. Find ∠MON.

∠KOL + ∠LOM + ∠MON = 180°(angles on a straight line KON)
∠MON = 180° − 62° − 71° = 47°(angles on a straight line KON)
∠JOK = ∠MON = 47°(vertically opposite angles)

∠KOL is one angle: K and L sit on its arms, and O, the vertex, sits in the middle. Tap Next step.

Try it

Straight lines AB and CD cross at O. ∠AOC = 74°. Find ∠BOD.

Naming an angle Naming an angle

Step 2

Finding the base angles of an isosceles triangle

Equal sides face equal angles. The three angles of a triangle add up to 180°. In an isosceles triangle, the two angles opposite the equal sides are the same size. Minus the top angle from 180°, then halve what is left, all on one line.

Tap Next step

ABC is a triangle and AB and AC are the same length. ∠BAC = 76°. Find ∠ABC.

∠ABC = (180° − 76°) ÷ 2

The small marks say AB and AC are the same length. Tap Next step.

Try it

DEF is a triangle with DE = DF. ∠EDF = 96°. Find ∠DEF.

Inside one triangle Inside one triangle

Step 3

Angles in a parallelogram, rhombus and trapezium

Between a pair of parallel sides, the two angles add up to 180°. Arrowheads on a figure show the parallel sides. The two angles at the ends of a side that joins them add up to 180°. A square and a rectangle have a right angle at every corner, and a parallelogram or rhombus has equal opposite angles. A trapezium has only one pair of parallel sides.

Tap Next step

ABCD is a parallelogram. ∠BAD = 116°. Find ∠ADC.

∠BAD + ∠ADC = 180°(angles between the parallel sides AB and DC)
∠ADC = 180° − 116° = 64°(angles between the parallel sides AB and DC)
∠BCD = ∠BAD = 116° and ∠ABC = ∠ADC = 64°(opposite angles of a parallelogram)

One arrowhead on AB and DC: those two sides are parallel. Two on AD and BC: so are they. Tap Next step.

Try it

PQRS is a trapezium with PQ parallel to SR. ∠PSR = 108°. True or false: ∠SPQ = 108°.

Quadrilaterals Quadrilaterals

Step 4

Angles at a point in a composite figure

At a join, each shape gives one angle, and a full turn is 360°. Find the corner point that two shapes share. Each shape gives one angle there: the square its right angle, the triangle its own angle. Angles at a point add up to 360°, and an angle bigger than 180° goes the long way round.

Tap Next step

ABCD is a square and BCE is a triangle. EB = EC and ∠BEC = 44°. Find ∠x.

∠BCD + ∠BCE + ∠x = 360°(angles at a point)
∠BCD = 90°(right angle of a square)
∠BCE = (180° − 44°) ÷ 2 = 68°(base angle of isosceles triangle)
∠x = 360° − 90° − 68° = 202°(angles at a point)

x is the angle marked at C, and its curve is drawn outside the figure. Tap Next step.

Try it

Keep the same figure, but make ∠BEC = 40°. What is ∠x now?

Across the join Across the join Round a point Round a point

Step 5

Using parallel sides to find an angle in a triangle

The parallel sides carry an angle across the figure. When the unknown angle is in a triangle inside a trapezium, start from the angle that already has a number. The parallel sides turn it into the angle at the other end of the side. Split that angle if a line cuts it, then finish with the angle sum of the triangle.

Tap Next step

ABCD is a trapezium with AB parallel to DC. ∠ABC = 116°, ∠BCE = 40° and ∠EDC = 30°. Find ∠DEC.

∠ABC + ∠BCD = 180°(angles between the parallel sides AB and DC)
∠BCD = 180° − 116° = 64°(angles between the parallel sides AB and DC)
∠ECD = 64° − 40° = 24°(the two parts of angle BCD)
∠DEC = 180° − 24° − 30° = 126°(angle sum of triangle DEC)

∠DEC is in triangle DEC, but only one of its other angles is given. Tap Next step.

Try it

PQRS is a trapezium with PQ parallel to SR, and the diagonal QS is drawn. ∠PQR = 118° and ∠QSR = 43°. Find ∠RQS.

The parallel sides carry it The parallel sides carry it

Step 6

Folded paper angle questions

A fold changes where an angle is, not how big it is. When paper is folded along a line, the part that moves keeps every angle it had. The angle beside the crease is a copy of the angle it came from, and a right angle stays a right angle. Find the original angle first, then use its copy.

Tap Next step

ABCD is a rectangular piece of paper. It is folded along the dotted line AE. ∠AED = 58°. Find ∠x.

∠DAE = 180° − 90° − 58° = 32°(angle sum of a triangle)
∠D′AE = ∠DAE = 32°(folded angles are equal)
∠DAE + ∠D′AE + ∠x = 90°(right angle of a rectangle)
∠x = 90° − 32° − 32° = 26°(right angle of a rectangle)

Corner D will fold over the dotted line AE. Tap Next step to fold it.

Try it

A rectangle ABCD is folded along a line from A, so that D lands at D′ inside it. Before the fold, ∠DAE = 25°. Find ∠x, the angle between AB and AD′.

The fold The fold

Primary 6 angles questions, worked step by stepName the property for each line yourself first. Then open a hint, or the working line by line.

LinesBasic

Two straight lines PR and QS cross at O. ∠POQ = 128°. Find ∠QOR and ∠ROS.

∠QOR is beside ∠POQ on a straight line. ∠ROS faces ∠POQ.

∠QOR = 180° − 128° = 52° (angles on a straight line POR)
∠ROS = ∠POQ = 128° (vertically opposite angles)
Check: 128° + 52° + 128° + 52° = 360° (angles at a point)

Answer: ∠QOR = 52°, ∠ROS = 128°

TrianglesBasic

XYZ is a triangle with XY = XZ. ∠XYZ = 47°. Find ∠YXZ.

The given angle is a base angle. Its partner is the other base angle.

∠XZY = ∠XYZ = 47° (base angle of isosceles triangle XYZ)
∠YXZ = 180° − 47° − 47° = 86° (angle sum of triangle XYZ)

Answer: 86°

Four sidesIntermediate

PQRS is a rhombus and the diagonal QS is drawn. ∠PQR = 124°. Find ∠PSQ.

A rhombus has all four sides equal, so triangle PQS is isosceles with PQ = PS.

∠QPS = 180° − 124° = 56° (angles between the parallel sides QR and PS)
PQ = PS (sides of a rhombus)
∠PSQ = (180° − 56°) ÷ 2 = 62° (base angle of isosceles triangle PQS)

Answer: 62°

The foldIntermediate

ABCD is a rectangular piece of paper. It is folded along AF, where F is on BC, so that B lands at B′ inside the rectangle. ∠AFB = 62°. Find ∠DAB′.

Find ∠BAF in triangle ABF first. The fold copies it to ∠B′AF.

∠BAF = 180° − 90° − 62° = 28° (angle sum of a triangle)
∠B′AF = ∠BAF = 28° (folded angles are equal)
∠DAB′ = 90° − 28° − 28° = 34° (right angle of a rectangle)

Answer: 34°

Mixed practiceChallenge

ABCD is a trapezium with AB parallel to DC. E is a point on DC such that AD = AE. ∠DAB = 115°. Find ∠EAB.

Start from the angle with a number. The parallel sides give ∠ADC.

∠ADC = 180° − 115° = 65° (angles between the parallel sides AB and DC)
∠AED = ∠ADE = 65° (base angle of isosceles triangle ADE)
∠DAE = 180° − 65° − 65° = 50° (angle sum of triangle ADE)
∠EAB = 115° − 50° = 65° (the two parts of angle DAB)

Answer: 65°

Step 7

Check yourself

Five angle questions to try without the figures moving. Each reply points to the set that matches it.

1. ABC is a straight line, and BD and BE are on the same side of it. ∠ABD = 51° and ∠DBE = 74°. Find ∠EBC.

2. PQR is a triangle with PQ = PR. ∠QPR = 58°. Find ∠PQR.

3. ABCD is a trapezium with AB parallel to DC, and the diagonal AC is drawn. ∠ABC = 124° and ∠ACD = 31°. Find ∠ACB.

4. ABCD is a square and BCE is an equilateral triangle outside it, on the side BC. Find the angle at C that goes the long way round from CD to CE.

5. A rectangular card ABCD is folded along BE, where E is on AD, so that A lands at A′ inside the card. ∠ABE = 19°. Find ∠A′BC.

0/5

Answer all five to see where to practise next.

For parents

How to use this guide. Before your child taps Next step, ask which property the next line will use and which angle it will find. Then let the figure show it. Each step takes a few minutes.

When your child writes an angle chase, ask for one property per line with its name in brackets, such as (angles on a straight line ABC). A line without a reason is hard to check, and naming the property is what tells your child which angle can be found next.

More study guides

Before this: Volume Next: Area Primary 5 Angles Primary 5 Properties of Triangles Primary 5 Parallelogram, Rhombus and Trapezium Primary 4 Angles study guide

Lessons and practice

P6 Angles lessons and practice

8 lesson videos and 63 questions across seven subtopics, from Basic to Challenge, each with a worked solution.

Watch and practise →