PAPER · REASONING · PRACTICE

Angles: addition, bisectors, complements and supplements — Teacher reference

Use ray order, degree-minute conversion, internal bisectors and the correct total for complementary or supplementary angles.

40 minutes · Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50

Textbook reference: Year 7, upper volume · Chapter 4 · Introduction to geometry

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the upper-volume plan

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Student classroom page Arrange in My Classroom

Lesson routePhase

4′Check prerequisites
5′Observe and predict
8′Worked example and guided attempt
15′Independent practice
5′Discuss and correct
3′Exit questions

Observe · explain

A question before practising

Keep the outside rays fixed and move an inside ray. Compare the two adjacent angles with their sum, then put the moving ray at the bisector.

Organise the key idea

Add adjacent angles only when the stated ray lies inside the whole angle. A bisector gives equal halves; complements total 90° and supplements total 180°.

Open the original chapter Open the interactive inquiry

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

OB lies inside ∠AOC. ∠AOB = 35° and ∠BOC = 40°. Find ∠AOC and state why addition is valid.

Angle sketch: use the stated angles and ray order.

Read the position condition “inside”.
The parts meet along OB.
Add 35 and 40.

The inside ray splits the whole angle into the two stated adjacent parts.
∠AOC = 35° + 40° = 75°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Guided attempt2

OC is the internal bisector of ∠AOB = 86°. Find ∠AOC and complete the equal-half relation.

∠AOC = ____.
2 × ∠AOC = ____°.
∠AOC = ____°.

Angle sketch: use the stated angles and ray order.

An arbitrary inside ray is not enough; use the bisector condition.
The two halves are equal.
Divide 86 by 2.

∠AOC = ∠COB because OC is a bisector.
The two equal halves total 86°, so each is 43°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite3

Find the complementary angle of 30°.

Identify the requested relation before calculating.
Use the total 90°.
Subtract the given angle from the total.

Complementary angles total 90°.
90 − 30 = 60°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite4

Using 1° = 60′, convert 2° to angle minutes.

2\times60

These are angle units, not time minutes.
Use the conversion factor 60.
Multiply 2 by 60.

Each degree contains 60 angle minutes.
2 × 60 = 120 minutes.

Check your work

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Independent practice5

OB is inside ∠AOC, with ∠AOB = 28° and ∠BOC = 47°. Find the whole angle.

Angle sketch: use the stated angles and ray order.

Check the interior-ray condition.
Both given angles make the whole.
Add 28 and 47.

Use adjacent-angle addition.
28° + 47° = 75°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice6

OB lies inside ∠AOC. ∠AOC = 110° and ∠AOB = 35°. Find ∠BOC.

Angle sketch: use the stated angles and ray order.

Identify the whole 110° first.
Subtract the known part.
Compute 110 − 35.

The whole is ∠AOB + ∠BOC.
∠BOC = 110° − 35° = 75°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice7

Find the supplementary angle of 118°.

Identify the requested relation before calculating.
Use the total 180°.
Subtract the given angle from the total.

Supplementary angles total 180°.
180 − 118 = 62°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice8

An internal ray bisects an angle of 70°. Find either half.

Identify the requested relation before calculating.
The two parts must be equal.
Halve the given angle.

An angle bisector divides the whole into two equal angles.
70 ÷ 2 = 35°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question9

Find the complementary angle of 42°.

Identify the requested relation before calculating.
Use the total 90°.
Subtract the given angle from the total.

Complementary angles total 90°.
90 − 42 = 48°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question10

A learner calls angles of 60° and 120° complementary. Correct the relation and justify it.

Use the given definition rather than the appearance of the notation.
Check the signed value or the stated condition.
Support your conclusion with a calculation or a concrete example.

60° + 120° = 180°.
A 180° sum defines supplementary angles.

Check your work

State the applicable definition or condition accurately.
Support the conclusion with the given numbers, a check or a concrete counterexample.

Further practice11

Find the complementary angle of 67°.

Identify the requested relation before calculating.
Use the total 90°.
Subtract the given angle from the total.

Complementary angles total 90°.
90 − 67 = 23°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice12

Find the supplementary angle of 93°.

Identify the requested relation before calculating.
Use the total 180°.
Subtract the given angle from the total.

Supplementary angles total 180°.
180 − 93 = 87°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice13

Add 30°45′ and 25°30′. Carry angle minutes correctly.

One degree contains 60 angle minutes.
75 minutes is one degree plus 15 minutes.
Carry the extra degree into the degree sum.

Minutes: 45′ + 30′ = 75′ = 1°15′.
Degrees: 30° + 25° + 1° = 56°, leaving 15′.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice14

ON points north and OE points east. OP is 35° east of north, measured from north towards east. Find the angle from east towards OP.

Angle sketch: use the stated angles and ray order.

Identify the direction from which the given 35° is measured.
North to east is 90°, not 180°.
Subtract the north-to-OP angle from the north-to-east angle.

North and east form a right angle of 90°. OP lies between these directions.
The angle from east towards OP is 90° − 35° = 55°.

Check your work

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Extension15

OA and OC are opposite rays; OB lies between them with ∠AOB = 80° and ∠BOC = 100°. OD and OE are the internal bisectors of those two angles. Find ∠DOE.

80/2+100/2

Angle sketch: use the stated angles and ray order.

Find the halves adjacent to OB.
The halves are 40° and 50°.
Add the two halves because OB lies between OD and OE.

∠DOB = 80°/2 = 40° and ∠BOE = 100°/2 = 50°.
B is between D and E, so ∠DOE = 40° + 50° = 90°.

Check your work

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Extension16

An angle x has a supplement three times its complement, giving 180 − x = 3(90 − x). Solve and check both related angles.

180-x=3(90-x)

The supplement is 180 − x and the complement is 90 − x.
Multiply both terms inside the brackets by 3.
The simplified equation is 2x = 90.

Expand: 180 − x = 270 − 3x.
Add 3x and subtract 180: 2x = 90, so x = 45.
Check the complement is positive and the supplement is three times it.

Check your work

Use the original equality and apply each operation to both sides.
Obtain the solution and check it in the original equation or situation.

Preparation and assessment notes

Prerequisites

Reading an angle’s vertex and rays, signed arithmetic and halving a quantity.

Look for in students’ work

Use the prerequisite checks to choose support, project the worked and guided items, then let learners work on paper. Select supplements or extensions as needed. No learner answers are saved by this presentation.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

All reference answers

Worked example1

OB lies inside ∠AOC. ∠AOB = 35° and ∠BOC = 40°. Find ∠AOC and state why addition is valid.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

The inside ray splits the whole angle into the two stated adjacent parts.
∠AOC = 35° + 40° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Guided attempt2

OC is the internal bisector of ∠AOB = 86°. Find ∠AOC and complete the equal-half relation.

∠AOC = ____.
2 × ∠AOC = ____°.
∠AOC = ____°.

Angle sketch: use the stated angles and ray order.

Reference answer

43°.

∠AOC = ∠COB because OC is a bisector.
The two equal halves total 86°, so each is 43°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite3

Find the complementary angle of 30°.

Reference answer

60°.

Complementary angles total 90°.
90 − 30 = 60°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite4

Using 1° = 60′, convert 2° to angle minutes.

2\times60

Reference answer

120 angle minutes (120′).

120

Each degree contains 60 angle minutes.
2 × 60 = 120 minutes.

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Independent practice5

OB is inside ∠AOC, with ∠AOB = 28° and ∠BOC = 47°. Find the whole angle.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

Use adjacent-angle addition.
28° + 47° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice6

OB lies inside ∠AOC. ∠AOC = 110° and ∠AOB = 35°. Find ∠BOC.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

The whole is ∠AOB + ∠BOC.
∠BOC = 110° − 35° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice7

Find the supplementary angle of 118°.

Reference answer

62°.

Supplementary angles total 180°.
180 − 118 = 62°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice8

An internal ray bisects an angle of 70°. Find either half.

Reference answer

35°.

An angle bisector divides the whole into two equal angles.
70 ÷ 2 = 35°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question9

Find the complementary angle of 42°.

Reference answer

48°.

Complementary angles total 90°.
90 − 42 = 48°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question10

A learner calls angles of 60° and 120° complementary. Correct the relation and justify it.

Reference answer

They are supplementary, because their sum is 180°, not 90°.

60° + 120° = 180°.
A 180° sum defines supplementary angles.

State the applicable definition or condition accurately.
Support the conclusion with the given numbers, a check or a concrete counterexample.

Further practice11

Find the complementary angle of 67°.

Reference answer

23°.

Complementary angles total 90°.
90 − 67 = 23°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice12

Find the supplementary angle of 93°.

Reference answer

87°.

Supplementary angles total 180°.
180 − 93 = 87°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice13

Add 30°45′ and 25°30′. Carry angle minutes correctly.

Reference answer

56°15′.

Minutes: 45′ + 30′ = 75′ = 1°15′.
Degrees: 30° + 25° + 1° = 56°, leaving 15′.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice14

ON points north and OE points east. OP is 35° east of north, measured from north towards east. Find the angle from east towards OP.

Angle sketch: use the stated angles and ray order.

Reference answer

55°.

North and east form a right angle of 90°. OP lies between these directions.
The angle from east towards OP is 90° − 35° = 55°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Extension15

OA and OC are opposite rays; OB lies between them with ∠AOB = 80° and ∠BOC = 100°. OD and OE are the internal bisectors of those two angles. Find ∠DOE.

80/2+100/2

Angle sketch: use the stated angles and ray order.

Reference answer

90°.

90

∠DOB = 80°/2 = 40° and ∠BOE = 100°/2 = 50°.
B is between D and E, so ∠DOE = 40° + 50° = 90°.

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Extension16

An angle x has a supplement three times its complement, giving 180 − x = 3(90 − x). Solve and check both related angles.

180-x=3(90-x)

Reference answer

x = 45°; its complement is 45° and its supplement is 135° = 3 × 45°.

x=45

Expand: 180 − x = 270 − 3x.
Add 3x and subtract 180: 2x = 90, so x = 45.
Check the complement is positive and the supplement is three times it.

Use the original equality and apply each operation to both sides.
Obtain the solution and check it in the original equation or situation.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Print edition

Textbook reference and authorship

A 40-minute selection of original core practice after studying the cited textbook lessons. It is not a replacement for every lesson in the chapter.

大湾区初中数学三维课堂（七年级上册）正文.pdf

Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Angles: addition, bisectors, complements and supplements

Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

OB lies inside ∠AOC. ∠AOB = 35° and ∠BOC = 40°. Find ∠AOC and state why addition is valid.

Angle sketch: use the stated angles and ray order.

Guided attempt2

OC is the internal bisector of ∠AOB = 86°. Find ∠AOC and complete the equal-half relation.

∠AOC = ____.
2 × ∠AOC = ____°.
∠AOC = ____°.

Angle sketch: use the stated angles and ray order.

Prerequisite3

Find the complementary angle of 30°.

Prerequisite4

Using 1° = 60′, convert 2° to angle minutes.

2\times60

Angles: addition, bisectors, complements and supplements

Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

OB is inside ∠AOC, with ∠AOB = 28° and ∠BOC = 47°. Find the whole angle.

Angle sketch: use the stated angles and ray order.

Independent practice6

OB lies inside ∠AOC. ∠AOC = 110° and ∠AOB = 35°. Find ∠BOC.

Angle sketch: use the stated angles and ray order.

Independent practice7

Find the supplementary angle of 118°.

Independent practice8

An internal ray bisects an angle of 70°. Find either half.

Exit question9

Find the complementary angle of 42°.

Exit question10

A learner calls angles of 60° and 120° complementary. Correct the relation and justify it.

Angles: addition, bisectors, complements and supplements — Further practice

Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50

Further practice11

Find the complementary angle of 67°.

Further practice12

Find the supplementary angle of 93°.

Further practice13

Add 30°45′ and 25°30′. Carry angle minutes correctly.

Further practice14

ON points north and OE points east. OP is 35° east of north, measured from north towards east. Find the angle from east towards OP.

Angle sketch: use the stated angles and ray order.

Extension15

OA and OC are opposite rays; OB lies between them with ∠AOB = 80° and ∠BOC = 100°. OD and OE are the internal bisectors of those two angles. Find ∠DOE.

80/2+100/2

Angle sketch: use the stated angles and ray order.

Extension16

An angle x has a supplement three times its complement, giving 180 − x = 3(90 − x). Solve and check both related angles.

180-x=3(90-x)

Angles: addition, bisectors, complements and supplements — Teacher answers

Year 7, upper volume · text pp. 91–100 · foundation lessons 46–50

Worked example1

OB lies inside ∠AOC. ∠AOB = 35° and ∠BOC = 40°. Find ∠AOC and state why addition is valid.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

The inside ray splits the whole angle into the two stated adjacent parts.
∠AOC = 35° + 40° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Guided attempt2

OC is the internal bisector of ∠AOB = 86°. Find ∠AOC and complete the equal-half relation.

∠AOC = ____.
2 × ∠AOC = ____°.
∠AOC = ____°.

Angle sketch: use the stated angles and ray order.

Reference answer

43°.

∠AOC = ∠COB because OC is a bisector.
The two equal halves total 86°, so each is 43°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite3

Find the complementary angle of 30°.

Reference answer

60°.

Complementary angles total 90°.
90 − 30 = 60°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Prerequisite4

Using 1° = 60′, convert 2° to angle minutes.

2\times60

Reference answer

120 angle minutes (120′).

120

Each degree contains 60 angle minutes.
2 × 60 = 120 minutes.

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Independent practice5

OB is inside ∠AOC, with ∠AOB = 28° and ∠BOC = 47°. Find the whole angle.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

Use adjacent-angle addition.
28° + 47° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice6

OB lies inside ∠AOC. ∠AOC = 110° and ∠AOB = 35°. Find ∠BOC.

Angle sketch: use the stated angles and ray order.

Reference answer

75°.

The whole is ∠AOB + ∠BOC.
∠BOC = 110° − 35° = 75°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice7

Find the supplementary angle of 118°.

Reference answer

62°.

Supplementary angles total 180°.
180 − 118 = 62°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Independent practice8

An internal ray bisects an angle of 70°. Find either half.

Reference answer

35°.

An angle bisector divides the whole into two equal angles.
70 ÷ 2 = 35°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question9

Find the complementary angle of 42°.

Reference answer

48°.

Complementary angles total 90°.
90 − 42 = 48°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Exit question10

A learner calls angles of 60° and 120° complementary. Correct the relation and justify it.

Reference answer

They are supplementary, because their sum is 180°, not 90°.

60° + 120° = 180°.
A 180° sum defines supplementary angles.

State the applicable definition or condition accurately.
Support the conclusion with the given numbers, a check or a concrete counterexample.

Further practice11

Find the complementary angle of 67°.

Reference answer

23°.

Complementary angles total 90°.
90 − 67 = 23°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice12

Find the supplementary angle of 93°.

Reference answer

87°.

Supplementary angles total 180°.
180 − 93 = 87°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice13

Add 30°45′ and 25°30′. Carry angle minutes correctly.

Reference answer

56°15′.

Minutes: 45′ + 30′ = 75′ = 1°15′.
Degrees: 30° + 25° + 1° = 56°, leaving 15′.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Further practice14

ON points north and OE points east. OP is 35° east of north, measured from north towards east. Find the angle from east towards OP.

Angle sketch: use the stated angles and ray order.

Reference answer

55°.

North and east form a right angle of 90°. OP lies between these directions.
The angle from east towards OP is 90° − 35° = 55°.

Identify the correct angle relation and retain all stated position conditions.
Show the calculation and give the angle with its unit.

Extension15

OA and OC are opposite rays; OB lies between them with ∠AOB = 80° and ∠BOC = 100°. OD and OE are the internal bisectors of those two angles. Find ∠DOE.

80/2+100/2

Angle sketch: use the stated angles and ray order.

Reference answer

90°.

90

∠DOB = 80°/2 = 40° and ∠BOE = 100°/2 = 50°.
B is between D and E, so ∠DOE = 40° + 50° = 90°.

Apply the relevant sign and order rules to every operation.
Show an intermediate calculation and obtain the exact stated value.

Extension16

An angle x has a supplement three times its complement, giving 180 − x = 3(90 − x). Solve and check both related angles.

180-x=3(90-x)

Reference answer

x = 45°; its complement is 45° and its supplement is 135° = 3 × 45°.

x=45

Expand: 180 − x = 270 − 3x.
Add 3x and subtract 180: 2x = 90, so x = 45.
Check the complement is positive and the supplement is three times it.

Use the original equality and apply each operation to both sides.
Obtain the solution and check it in the original equation or situation.