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Angl

es ANGLES ANGLES ANGLES

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10HSERIES TOPIC

1Angles Solutions

Mathletics Passport © 3P Learning

How does it work? AnglesSolutions

Page 3 questions

Page 4 questions

1

2

1

2

a

d

b

e

c

f

A X

P

Q

RO

ZY

C

BArm BC Arm XY The vertex

The angle swept The arm shared by both angles

The angle swept by the arms PO and QO

The angle swept by the arms SQ and RQ (also accept SQR+ or RQS+ if you used the angle names)

a

a

a

b

c

e

f

b

b

c

c

dThe vertex The angle swept The arm YZ The common arm BD

The angle swept by the arms LO and NO (also accept LON+ or NOL+ if you used the angle names)ALSO acceptable to say LOM+ plus MON+ if you went down that path.

Parts of an angle

Naming angles

or orACB BCA C+ + + or orDFE EFD F+ + + or orGJK KJG J+ + +

(i) orXZW WZX+ + (ii) orXZY YZX+ +

(i) orONP PNO+ +

(i) orCFD DFC+ +

(ii) orQNP PNQ+ +

(ii) orAFB BFA+ +

2 Angles Solutions


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10


3

4

1

Page 4 questions

Page 6 questions

Naming angles

Angle types

a

a b c

b cCommon arm is XZ Common arm is PN No common arms

orLOM MOL+ + orADC CDA+ + orEDH HDE+ +

orFDH HDF+ +

=

=

Obtuse angle XYZ+

Right-angle MLN+

Full revolution JKL+ Straight DEF+

a

c

e f

b

d

Acute PQR+

Reflex GUP+(Hint: remember the box)

P

Q R

M

L N

X

YZ

G

UP

DE

FJ / LK

Vertex

10HSERIES TOPIC

3Angles Solutions



2

Page 6 questions

Page 8 questions

Naming angles

Using a protractor to measure angles

Reflex or( )PRW WRP+ +

Reflex or( )RQS SQR+ +

Reflex or( )QRS SRQ+ +

Reflex or( )QST TSQ+ +

Reflex or( )QSR RSQ+ +

Reflex or( )STP PTS+ +

Reflex or( )SRW WRS+ +

Reflex or( )STU UTS+ +

Reflex or( )UTV VTU+ +

Reflex or( )STV VTS+ +

Reflex or( )TUW WUT+ +

Reflex or( )TUV VUT+ +

Reflex or( )TVU UVT+ +

Reflex or( )WVU UVW+ +

Reflex or( )UWV VWU+ +

Reflex or( )SRU URS+ +

Reflex or( )RUT TUR+ +

Reflex or( )PQS SQP+ +

Reflex or( )QPT TPQ+ +

orRQS SQR+ +

orQPT TPQ+ +

orQRS SRQ+ +

orRTU UTR+ +

orURS SRU+ +

orRUT TUR+ +

orUTV VTU+ +

orTUV VUT+ +

orUVW WVU+ +

orUWV VWU+ +

orVUW WUV+ +

orRSQ QSR+ +

orQST TSQ+ +

orPTS STP+ +

orRTW WTR+ +

orRUV VUR+ +

orPRW WRP+ +

orPQS SQP+ +

orPTU UTP+ +

orTUW WUT+ +

orTVU UVT+ +

orPQR RQP+ +

orRST TSR+ +

orPTV VTP+ +

orTVW WVT+ +

orRUW WUR+ +

R

Q S U

VTP W

Acute angle Straight angleObtuse angleRight angle Reflex angle

a

a

b

b

c

c

d

d

1

2

ABC 40c+ =

0ABC 18 c+ =

0PQR 9 c+ =

5BDC 8 c+ =

0JLK 15 c+ =

CDB 145c+ =

15XYZ c+ =

135ADB c+ =

Getting 5 correct from this list is a good effort

4 Angles Solutions


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10


Page 9 questions


3

4

a b

XZY 60` c+ =

XYZ 95` c+ =

170XYZ` c+ =

XYZ 135` c+ =

a

c

b

d

RQS 50c+ = BEC 35c+ =

Y

ZX

X

Y Z

Y

X

Z

Y

Z

X

10HSERIES TOPIC

5Angles Solutions



Page 10 questions

Page 12 questions


Using a protractor to measure reflex angles

5

1

2

44c 10c 24c 30c 02 c 40c 35c 52c 27c 22c 56c

R E V O L U T I O N S

a

c

b

d

Reflex ABC 360 100

260

c c

c

+ = -

=

Reflex XYZ 360 35

325

c c

c

+ = -

=

Reflex TUV 360 150

210

c c

c

+ = -

=

Reflex JKL 360 70

290

c c

c

+ = -

=

Reflex 360 40HIJ

320

` c c

c

+ = -

=

b

c

a

Reflex 360 145LMN

215

` c c

c

+ = -

=

L

M

N

HJ

I

A

C

B

Reflex 360 170ABC

190

` c c

c

+ = -

=

6 Angles Solutions


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10

Where does it work? AnglesSolutions

Page 14 questions

Adjacent angles

1

2

3

4

a bB

DA

C

K

M

JN

L

andABD CBD+ +

reflex andWXZ YXZ+ +

reflex USR+

UST+

andKNL LNM+ +

X

W

Z

Ya b

T

U

P

Q

RS

(i)

(ii)

(iii)

(iv)

reflex USQ+

reflex USP+

Draw an obtuse angle and label it PQR+ . Draw an acute angle PQS+ adjacent to it.

R

QS

P

Here is one possible solution

a

b

They share an arm, however that do not share a vertex, so they are not adjacent angles

ADB+ forms part of ADC+ , so since they overlap each other, they are not adjacent angles

10HSERIES TOPIC

7Angles Solutions



Page 17 questions

Complementary and supplementary angles

1

2

3

4

a

a

c

b

d

fe

dc

b

fe

The complementary of 30c is 60csince 30 60 90c c c+ =

The supplement of 100c is 80csince 0 0100 8 18c c c+ =

The supplement of 165c is 15csince 180165 15c c c+ =

The supplement of 90c is 09 csince 0 18090 9c c c+ =

The supplement of 109c is 71csince 0 1801 9 71c c c+ =

The supplement of .121 3c is .58 7csince . . 180121 3 58 7c c c+ =

The supplement of 1941 c is 160

43 c

since 1801941 160

43c c c+ =

The complementary of 11c is 79csince 9011 79c c c+ =

The complementary of 46c is 44csince 9046 44c c c+ =

The complementary of 08 c is 01 csince 0 0 908 1c c c+ =

The complementary of .18 3c is .71 7csince . . 9018 3 71 7c c c+ =

The complementary of .23 5c is .66 5csince . . 9023 5 66 5c c c+ =

a b

90 71BDC

19

c c

c

+ = -

=

90 11.5

.

TSU

78 5

c c

c

+ = -

=

71c

A

B

C

D

.11 5c

R S

T

U

P

Q

a b

180 107XOY

73

c c

c

+ = -

=

180 90HJI

90

c c

c

+ = -

=

D

F

E

H

J

I

107cW Y

O

X

8 Angles Solutions


HSERIES TOPIC

10


Page 18 questions

Complementary and supplementary angles

5

6

7

W

O

X

YU

V

44c

46 c

46c

90 46WOY

136

c c

c

+ = +

=

136 44WOY UOV

180

c c

c

+ ++ = +

=

andWOY UOV`+ + are supplementary

A G

B

C

O

DE

F

31c

30c23c35c

24c

37c

First Pair

Second Pair

23 30DOF

53

c c

c

+ = +

=

complement of is AOB53 90 53 37` c c c c +- = =

First pair and: DOF AOB` + +

complement of is BOD31 90 31 59` c c c c +- = =

FOG 31c+ =

Secondpair and

( 24 35 59 )

:

BOD

FOG BOD`

c c c+

+ +

= + =

O

P

Q

R S

T

U

V

15c

52 c 56 c53 c

52 c

(i) andQOR ROS+ + are adjacent complementary angles

(ii)

25 65 90c c c+ =

15 25 65 35 25POU

165

c c c c c

c

+ = + + + +

=

For it to be a straight line, 180

180

180 165

POV

UOV POU

15

`

c

c

c c

c

+

+ +

=

= -

= -

=

10HSERIES TOPIC

9Angles Solutions



Page 20 questions

Vertically opposite angles

1

2

a

a

b

b

T

P R

Q

X

J

KL

M

S

and

and

PTR QTS

PTS RTQ

+ +

+ +

and

and

JXM KXL

JXL KXM

+ +

+ +

First pair:

Second pair:

I

H

J

OG

67c

C

A

B

E

D

O 49c

56c

(ii)

(ii)

(i)

(i)

IOH 67c+ =

Because IOH+ is vertically opposite to GOJ+ , so these angles are both the same size

Because COA+ is vertically opposite to BOD+ , so these angles are both the same size

JOH 113c+ =

Because GOJ+ and JOH+ form a straight line, they are supplementary angles

180 67 113JOH` c c c+ = - =

AOD 75c+ =

Because AOD+ and COA+ form a straight line, they are supplementary angle

180 67 113AOD` c c c+ = - =

AB and CD are straight lines 49 56

BOD BOE EOD

105

c c

c

+ + += +

= +

=

49 56COA

105

c c

c

+ = +

=

10 Angles Solutions


HSERIES TOPIC

10


Page 20 questions

Page 23 questions


Parallel lines

3 These are all the possible pairs of vertically opposite angles (not including straight angles):

A

C

E G

B

D

FH

O

andAOC BOD+ + andEOC DOF+ +

andEOG HOF+ + andGOB AOH+ +

andAOE BOF+ + andCOG DOH+ +

andEOB AOF+ + andCOB AOD+ +

andAOG BOH+ +

andSXY VYX+ +

andSXW UYX+ +

andWXT XYV+ +

andSXY UYX+ +

andFBC GCB+ +

andABE BCG+ +

andABF BCH+ +

andEBC GCB+ +

andEOH FOG+ +

andCOH DOG+ +

andUYX TXY+ +

andSXY UYZ+ +

andTXY VYZ+ +

andVYX TXY+ +

andEBC HCB+ +

andEBC GCD+ +

andFBC HCD+ +

andFBC HCB+ +

andCOF EOD+ +

1

2

a

a b

b(i) (i)

(ii) (ii)

Transversal is the line WX Transversal is the line AD

AB is prallel to CD, ;;AB CD EF is prallel to GH, ;;EF GH

Zangles (alternate angles)

Fangles (corresponding angles) Fangles (corresponding angles)

Cangles (cointerior angles)Cangles (cointerior angles)

Zangles (alternate angles)

10HSERIES TOPIC

11Angles Solutions



Page 24 questions

Parallel lines

3

4

63c

JPQ+

KPO+

LOP+

MON+

117c

JPO+

LON+

MOP+

KPQ+

BGH 46c+ =

46AGF c+ =

46CFE c+ =

BGF 134c+ =

a Corresponding angles:



Cointerior angles

Alternate angles

Corresponding angles

Straight angle

BGHa+ corresponds to DFG+ which 46c=

AGFa+ is vertically opposite to BGH+ which 46c=

CFEa+ is vertically opposite to DFG+ which 46c=

BGFa+ is cointerior to DFG+ which

BGFa+ and DFG+ are supplementary (add to 180c)

AGFa+ is alternate to DFG+ which 46c=

AGFa+ corresponds to AGF+ which 46c=

BGFa+ and BGH+ form a straight angle(180c)

Property used:

Property used:

Property used:

Property used:

or

or

or

BGF 180 46 134a c c c+ = - =

12 Angles Solutions


HSERIES TOPIC

10


Page 24 questions

Page 25 questions

Parallel lines

Parallel lines

4

4

b WXB 128c+ =

AXW 52c+ =

BXY 52c+ =

Properties used:

Properties used:

Properties used:

Corresponding angle to , ( )TYZ AXY 128c+ + =

Corresponding angle to , ( )TYZ AXY 128c+ + =

Vertically opposite angle to AXW 52c+ =

WXB+ is vertically opposite to 128AXY c+ =

AXW+ and AXY+ make a straight angle (180c)

BXY+ and UYX+ are cointerior angles, so supplementary

180 128 52AXW` c c c+ = - =

180 52BYY UYX` c c+ += - =

A

C

E

FB

D

a

b

(ii)

(ii)

(i)

(i)

The line AB and CD are:

The line ST and UV are:

PARALLEL

PARALLEL

NOT PARALLEL

NOT PARALLEL

Reason:

Reason:

The cointerior angles do not add up to 180c

The corresponding angles are the same size

The cointerior angles are not supplementary

37c

153c

or

S

U

Y

X

T

V81c

81c

10HSERIES TOPIC

13Angles Solutions



Page 25 questions

Parallel lines

c

d

(ii)

(ii)

(i)

(i)

The line PQ and RS are:

The line LM and NO are:

PARALLEL

PARALLEL

NOT PARALLEL

NOT PARALLEL

Reason:

Reason:

The alternate angles are not equal to each other

They are parallel because the cointerior angles are supplementary

P R

SQ

N O101c

100c

L

N K

I

J

O

MH

22c

158c

(Write all the properties used here)

and are vertically oppositeHIL MIJ

MIJ HIL

HIL IJO

158

158 22 180

` c

c c c

+ +

+ +

+

= =

+ = + =

14 Angles Solutions


HSERIES TOPIC

10

What else can you do? AnglesSolutions

Page 27 questions

Angle sums

17c

a

a

b

b

29c14c

13c

N MA

B

C

DE

JK

L

90 ( )

90 (13 29 )

90 42

JNK KNL LNM

48

c

c c c

c c

c

+ + += - +

= - +

= -

=

0 ( )

( )

XOY XOW WOZ YOZ36

360 72 203 76

360 351

9

c

c c c c

c c

c

+ + + += - + +

= - + +

= -

=

AOD AOD 127c+ += =

127

127 83

BOC COE

44

c

c c

c

+ += -

= -

=

each angle 180 5

36

` 'c

c

=

=

is split into equal sized angles5PQR+

( )

( )

JNK BEA CEB180

180 14 17

180 31

149

c

c c c

c c

c

+ + += - +

= - +

= -

=

cX

76c

203c

72cZ

Y

W

O

d

P

V

U T

S

RQ

1

2

AD

B

E

C

O127c

83c

JON 34 88 122c c c+ = + =

JON MOK 122c+ += =

MOP KOP122

122 76

46

` c

c c

c

+ += -

= -

=

J

Q N

K

PM

O88c34c

76c

AB and CD are straight lines JK and MN are straight lines

10HSERIES TOPIC

15Angles Solutions



Page 28 questions

Angle sums

a

W

U

X

CP

D

VA

B

70c

84c

3 WBA 84c+ =

84CBX c+ =

ABX 96c+ =

PBW 26c+ =

This angle is alternate to VAB+ and ;;WX UV

This angle is corresponding to VAB+ and ;;WX UV

This angle is cointerior to VAB+ and ;;WX UV

So 180ABX VAB c+ ++ =

96WBC ABX c+ += =

96 70

WBP WBC PBC

26

c c

c

+ + += -

= -

=

(Vertically opposite angles)

b AGH 130c+ =

AGF 50c+ =

HGB 50c+ =

EGF 83c+ =

This angle is corresponding to CFG+ and ;;AB CD

This angle is cointerior to CFG+ and since ;;AB CD , they add to 180c

This angle is vertically opposite to AGF+

The angles CFG+ and FGB+ are alternate angles, and since ;;AB CD , they are equal

EGF BGE130

130 47

83

c

c c

c

+ += -

= -

=

A

C

B

H

I

DF E

G

130c

47c

16 Angles Solutions


HSERIES TOPIC

10


Page 28 questions

Angle sums

4 a

b

J

K

A

BD

C

P

BQ

E

119c

PQB+

PQB+

DQB+

DFG+

EFI+

EFC 39c+ =

IFD 46c+ =

DFG EFC+ +=

AEF EFD+ +=

DFG 39c+ =

141

141 46

EFI IFD

95

c

c c

c

+ += -

= -

=

119 90

DQB PQB PQD

29

c c

c

+ + += -

= -

=

ABE JBQ+ +=

JBQ PQB+ +=

JEQ PQD+ +=

PQB 119` c+ =

PQD 90` c+ =

(Vertically opposite angles)

(Corresponding angles, ;;JK PQ )

(Corresponding angles, ;;JK PQ )

H

H

E

I

J

B

D

G

A

C

141c

46c

F

Hint: find EFC+ first

Hint: find IFD+ first

(Cointerior to AEF+ , so add to 180c )

(Alternate to EIF+ , and ;;AB CD )

(Alternate angles and ;;AB CD )

(Vertically opposite)

10HSERIES TOPIC

17Angles Solutions



Page 30 questions

Angle problems

1 a

b

?cStart position

Half a circle 180c=

Therefore, Janet completed 65 of 180c

of 180 18065

65

150

#c c

c

=

=

This is 30c short of finishing half a circle

`Janet was 30c away from completing a full circle

`Janet finishes her second move 150c away from the starting position

180c

03 cEnd of next movement

Start position

Start of second movement

Degrees away from the start 180 30

150

c c

c

= -

=

18 Angles Solutions


HSERIES TOPIC

10


Page 30 questions

Page 31 questions

Angle problems

Angle problems

O

EHY B K Z

28c 47c

D G XAJWa

a

2

3

OAD OBE

OGD OHE

OAD OGD OJA

47

28

90

c

c

c

+ +

+ +

+ + +

= =

= =

+ + =

(Alternate angles and ;;WX YZ )

(Alternate angles and ;;WX YZ )

(Question says they are complemtary... so add to 90c )

47 28 90

90 (47 27 )

90 75

OJA

OJA

15

`

c c c

c c c

c c

c

+

+

+ + =

= - +

= -

=

C

A D F

G

E

B

A D F B C G E

10HSERIES TOPIC

19Angles Solutions



Page 31 questions

Angle problems

4

AB

C

D

27c

Step 1:

Step 2:

Complement of DBC

ABD

90 27 63

180 27 153

c c

c c c

+

+

= - =

= - =

Reflex ABD 360 153 207

207 80 127`

c c c

c c c

+

+

= - =

- =

63 17 80` c c c+ =

Supplement of 0DBC 18 27 153c c c+ = - =

ABD` + divided into 9 equal parts 153 9 17'c c= =

` The final answer is 153 127 280c c c+ =

20 Angles Solutions


HSERIES TOPIC

10

Angles Notes

VERTIC

ALLY OPPOSITE

ANGLES * VERTICALLY OPPOSITE ANGLES *

..../...../20...

PARTS OF

AN ANGLE * PARTS OF AN ANGLE

*..../...../20...USING A PR

OTRACTOR TO MEASURE REFLEX ANGLES

*..../...../20...

A

BC

ANG

LE SUMS *

ANGLE SUMS * ANGLE SUMS *

..../...../20...+

COMPLEME

NTARY AND SUPPLEMENTARY ANGLES

*

..../...../20.

..

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