BUKU TEKS MATHEMATICS FORM 3 (DLP)

Chapter 4 Scale Drawings

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

,I Q WKHQ WKH VL]H RI WKH VFDOH GUDZLQJ LV ELJJHU WKDQ WKH VL]H RI WKH REMHFW
,I Q ! WKHQ WKH VL]H RI WKH VFDOH GUDZLQJ LV VPDOOHU WKDQ WKH VL]H RI WKH REMHFW
,I Q WKHQ WKH VL]H RI WKH VFDOH GUDZLQJ LV WKH VDPH DV WKH VL]H RI WKH REMHFW

How do you determine the scales, measurements of LEARNING
objects or measurements of scale drawings? STANDARD

6FDOH Measurement of scale drawing Determine the scales,
0HDVXUHPHQW RI REMHFW Q measurements of objects
or measurements of scale
drawings.

Example 2
7KH GLDJUDP EHORZ VKRZV REMHFW PQRS and scale drawing PЈQЈRЈSЈ GUDZQ RQ D JULG RI HTXDO
VTXDUHV 6WDWH WKH VFDOH XVHG LQ WKH IRUP Q.

PQ

P' Q' CHAPTER 4

S R S' R'

Solution:

6FDOH ²P'²Q' ² ² R U 6FDOH ²P'²S' ² ² 7KXV VFDOH
PQ PS

Example 3

7KH GLDJUDP EHORZ VKRZV REMHFW KLM and scale drawing KЈLЈMЈ GUDZQ RQ D JULG RI HTXDO VTXDUHV
6WDWH WKH VFDOH XVHG LQ WKH IRUP Q.

K'

K

LM

Solution: L' M'

6FDOH K'L' 9 RU 6FDOH L'M' 7KXV VFDOH
KL LM
Z LG OLOY JLLGYHHG E \ ² ²

²


91

Example 4

7KH GLDJUDP EHORZ VKRZV REMHFW PQR and scale drawing PЈQЈRЈ GUDZQ RQ D JULG RI HTXDO VTXDUHV

RI GL൵HUHQW VL]HV 'HWHUPLQH WKH VFDOH XVHG LQ WKH IRUP Q.

Object } FP Scale drawing } FP

P

P'

QR

CHAPTER 4 Q' R'

Solution: Method 2 Grid size is used because number of units of sides
RI REMHFW DQG VLGHV RI VFDOH GUDZLQJ LV WKH VDPH
Method 1
6 FDOH *²U²LG* ²VULL]G²H V RL²]I HV² FRDI²O RH E²GMUH±DFZ±W ±LQ²J ² FF±PP± ²
6FDOH ²QQ'²RR' ² FF±PP± ²
6FDOH 6FDOH
²
²


Example 5

7KH GLDJUDP EHORZ VKRZV REMHFW KLMN and scale drawing KЈLЈMЈNЈ GUDZQ RQ D JULG RI HTXDO
VTXDUHV RI GL൵HUHQW VL]HV 'HWHUPLQH WKH VFDOH XVHG

Object } FP Scale drawing
K' } FP
K

L' N'
M'

L N FLASHBACK
M
92 K'N' = ¥1.52 + 22 = 2.5
KN = ¥32 + 42 = 5

Solution: Chapter 4 Scale Drawings

Method 1 Method 2

6FDOH ²KK'²NN' ² ± F ±FP±P± ² ± ² ± 6 FDOH *²U²LG* ²VULL]G²H V RL²]I HV² FRDI²O RH E²GMUH±DFZ±W ±LQ²J ² ± F ±FP±P± ²±
6 F D O ±H î ± î 6 FDOH ±


± multiplied E\ WR JHW

Example 6

$ PDS LV GUDZQ WR D VFDOH RI &DOFXODWH WKH DFWXDO OHQJWK LQ NP RI D ULYHU WKDW LV FP
long on the map.

Solution: CHAPTER 4

Method 1 Method 2

² ² ² F P± ²F P± ²$F²WX²D O F²GPLV²WDQ±F±H± 6FDOH GUDZLQJ 2EMHFW FLASHBACK

1 km = 1 000 m
1 m = 100 cm
$ FWXDO GLVWDQFH ² î² ² F² P ² ±F±P± FP FP 1 km = 100 000 cm

î FP NP î
FP NP

FP Thus, the actual length of the
NP ULYHU LV NP

Example 7

7KH PDS RI -RKRU LV GUDZQ WR D VFDOH RI FP WR NP &DOFXODWH WKH DFWXDO GLVWDQFH EHWZHHQ
.OXDQJ DQG $\HU +LWDP LI WKH GLVWDQFH RQ WKH PDS LV FP

Solution:

Method 1 Method 2

Distance on drawing 6FDOH GUDZLQJ 2EMHFW
6FDOH Actual distance FP NP
î FP NP î
FP FP 7KXV WKH DFWXDO GLVWDQFH LV NP
NP Actual distance

$FWXDO GLVWDQFH FP NP
FP

Actual distance NP

93

Example 8
.KDLUXO GUDZV D VTXDUH WR D VFDOH RI ,I WKH DFWXDO OHQJWK RI VLGHV RI WKH VTXDUH LV FP what


is the length of sides, in cm, of the scale drawing?

Solution:

Method 1 Method 2

6FDOH 6LGH RI VFDOH GUDZLQJ 6FDOH GUDZLQJ 2EMHFW
&RUUHVSRQGLQJ VLGH RI REMHFW ²
î î
6LGH RI VFDOH GUDZLQJ FP FP
6 cm
Thus, the length of side of scale drawing is
FP

6LGH RI VFDOH GUDZLQJ î FP
/HQJWK RI VLGH RI VFDOH GUDZLQJ FP

CHAPTER 4 MIND TEST 4.1b

1. 'HWHUPLQH WKH VFDOH XVHG IRU HDFK VFDOH GUDZLQJ EHORZ LQ WKH IRUP Q.

Object Scale drawing
(a)

(b) 6 cm FP

FP FP

(c) } FP } FP

(d) FP FP
94 6 cm 9 cm

Chapter 4 Scale Drawings

2. $ SRVWHU KDV D OHQJWK RI FP DQG D ZLGWK RI FP &DOFXODWH WKH OHQJWK DQG ZLGWK RI WKH
VFDOH GUDZLQJ RI WKH SRVWHU LQ FP WKDW LV GUDZQ WR D VFDOH RI

3. $ PDS LV GUDZQ WR D VFDOH RI :KDW LV WKH DFWXDO OHQJWK LQ NP RI D ULYHU ZLWK D
OHQJWK RI FP RQ WKH PDS"

4. 6LHZ /LQ GUDZV D ULJKW DQJOHG WULDQJOH WR D VFDOH RI ,I WKH K\SRWHQXVH RI WKH VFDOH


GUDZLQJ LV FP FDOFXODWH WKH OHQJWK RI WKH K\SRWHQXVH RI WKH RULJLQDO WULDQJOH

How do you draw the scale drawings of objects and vice LEARNING
versa? STANDARD

Drawing the scale drawing of an object Draw the scale drawings of
objects and vice versa.
7KHUH DUH WKUHH ZD\V WR GUDZ WKH VFDOH GUDZLQJ RI DQ REMHFW
D 8VH JULG SDSHU RI WKH VDPH VL]H IRU GL൵HUHQW VFDOHV CHAPTER 4
E 8VH JULG SDSHU RI GL൵HUHQW VL]HV
F 'UDZ RQ D EODQN SDSHU DFFRUGLQJ WR WKH JLYHQ VFDOH

Example 9

Draw the scale drawing of shape PQRS RQ D JULG RI HTXDO VTXDUHV XVLQJ D VFDOH R
of .
S


Solution: S' PQ
P' R'
7KH VFDOH JLYHQ LV : . Therefore, every
QU I Z

side of the scale drawing is two times If you have to draw the
ORQJHU WKDQ WKH OHQJWK RI VLGHV RI REMHFW scale drawing of your
PQRS. VFKRRO ¿HOG ZKDW LV
a suitable scale to be
Q' used? Why?

Example 10 } FP

D 'LDJUDP ¨PQR is drawn on a grid of Q } FP
FP = FP 5HGUDZ ¨PQR on grid
paper with dimensions R

L FP = FP

LL FP = FP

E &DOFXODWH WKH VFDOH XVHG LQ D L DQG D
LL LQ WKH IRUP Q.

P

95

Solution: FP
(a)(i)
} FP

(a)(ii)

Q'
CHAPTER 4 Q'

} FP

} } FP

P' R'

P' R'

(b)(i) Grid size of scale drawing FP (b)(ii) Grid size of scale drawing FP

6FDOH *ULG VL]H RI REMHFW FP 6FDOH *ULG VL]H RI REMHFW FP

6FDOH 6FDOH




Example 11

&RQVWUXFW WKH VFDOH GUDZLQJ RI WULDQJOH PQR XVLQJ D VFDOH RI R
FP
Solution:

R'

FP ƒ 6 cm Q
Q' P

ƒ
P' FP

)RU REMHFWV ZLWK JLYHQ DQJOHV WKH DQJOHV RI WKH VFDOH GUDZLQJ PXVW Scan the QR Code or
be accurately drawn and the lengths of sides are drawn to scale. visit http://bukutekskssm.
my/Mathematics/F3/
Drawing the objects for a scale drawing Chapter4Grid.pdf to
download grid paper of
various sizes.

Example 12 R'

7KH GLDJUDP VKRZV D VFDOH GUDZLQJ GUDZQ RQ D JULG RI HTXDO VTXDUHV T' S'
WR D VFDOH RI 'UDZ WKH DFWXDO REMHFW IRU P'Q'R'S'T'. P' Q'

96

Solution: Chapter 4 Scale Drawings

TS R
P
7KH VFDOH XVHG LV WKDW LV WKH VL]H RI VFDOH GUDZLQJ
LV WZR WLPHV VPDOOHU WKDQ WKH REMHFW 7KHUHIRUH HYHU\
VLGH RI DFWXDO REMHFW LV WZR WLPHV ORQJHU WKDQ WKH VLGHV
of the scale drawing.

Q

Example 13 FP

7KH GLDJUDP VKRZV WKH VFDOH GUDZLQJ RI D ÀRZHU } FP
GUDZQ RQ FP = FP JULGV 'UDZ WKH DFWXDO REMHFW
on grids of

D FP = FP

E FP = FP
} CHAPTER 4

}
Solution: FP}

2EMHFW PXVW EH GUDZQ RQ JULGV RI } FP
GL൵HUHQW VL]HV 7KXV WKH QXPEHU RI
XQLWV RI VLGHV RI REMHFW LV WKH VDPH DV
the number of units of sides of scale
drawing.
(a) FP

} FP

97

MIND TEST 4.1c

1. 'UDZ WKH VFDOH GUDZLQJ RI HDFK REMHFW EHORZ WR D VFDOH RI DQG


(a) (b)

2. D 7KH REMHFW LQ WKH GLDJUDP LV GUDZQ RQ } FP
FP î FP JULG SDSHU 5HGUDZ WKH VKDSH FP
RI WKH REMHFW RQ D JULG SDSHU RI }

CHAPTER 4 L FP î FP

LL FP î FP

E &DOFXODWH WKH VFDOH XVHG LQ D L DQG
(a)(ii).

3. Draw the scale drawing of the following shapes to the given scale.

D 6FDOH E 6FDOH F 6FDOH ²

P

FP

9 cm P FP

P FP

4. The diagram shows the scale drawing of a composite shape that
LV GUDZQ RQ D JULG RI HTXDO VTXDUHV WR D VFDOH RI : .


'UDZ WKH DFWXDO REMHFW IRU WKH VKDSH

98

Chapter 4 Scale Drawings

How do you solve problems involving scale LEARNING
drawings? STANDARD

Example 14 Solve problems involving
scale drawings.

7KH GLVWDQFH RQ D PDS EHWZHHQ %LQWXOX DQG 0LUL LV FP

D ,I WKH VFDOH XVHG WR GUDZ WKH PDS LV FP NP FDOFXODWH WKH DFWXDO GLVWDQFH LQ NP EHWZHHQ
%LQWXOX DQG 0LUL

E ,I WKH PDS LV UHGUDZQ WR D VFDOH RI FDOFXODWH WKH GLVWDQFH EHWZHHQ %LQWXOX DQG 0LUL

on the new map. TIP

F 0U 'RPLQLF /DMDZD DQG KLV IDPLO\ ZDQWV WR YLVLW 0LUL ,I KH SODQV If the scale of scale
WR GULYH WR 0LUL DW D VSHHG RI NP K± FDOFXODWH WKH WLPH WDNHQ WR drawing and the

GULYH IURP %LQWXOX WR 0LUL LQ KRXUV DQG PLQXWHV requirement of the
question are in the same
unit, the scale need not be
changed to cm.
Solution:

Understanding the problem Implementing the strategy

‡ $FWXDO GLVWDQFH IRU FP GUDZQ WR VFDOH Distance on drawing CHAPTER 4
RI FP NP D 6FDOH Actual distance

‡ 'LVWDQFH RQ VFDOH GUDZLQJ GUDZQ WR FP
VFDOH RI NP Actual distance

‡ 7LPH LQ KRXUV DQG PLQXWHV IRU $FWXDO GLVWDQFH FP NP
MRXUQH\ IURP %LQWXOX WR 0LUL DW VSHHG FP

RI NP K± . $FWXDO GLVWDQFH NP

Planning a strategy Distance on drawing
E 6FDOH Actual distance
Distance on drawing
6FDOH Actual distance Distance on drawing
NP
Distance
7LPH 6SHHG 'LVWDQFH RQ GUDZLQJ = FP
FP
Making a conclusion 'LVWDQFH RQ VFDOH GUDZLQJ FP

‡ $FWXDO GLVWDQFH EHWZHHQ %LQWXOX DQG Distance
0LUL LV NP F 7LPH 6SHHG

‡ 'LVWDQFH EHWZHHQ %LQWXOX DQG 0LUL NP
RQ WKH PDS RI VFDOH RI NP K±

‡ 7LPH WDNHQ IRU 0U 'RPLQLF /DMDZD KRXUV
WR GULYH IURP %LQWXOX WR 0LUL DW D
VSHHG RI NP K± LV KRXUV 7LPH WDNHQ
minutes. KRXUV PLQXWHV

FLASHBACK

Speed = —Dis—ta—nc—e
Time

99

MIND TEST 4.1d

1. 7KH GLDJUDP VKRZV D ULJKW DQJOHG WULDQJOH $ VFDOH
GUDZLQJ RI WKH WULDQJOH LV GUDZQ WR D VFDOH RI .

&DOFXODWH WKH DUHD LQ FP , of the scale drawing. FP

2. The diagram shows a room in the shape of a rectangle. FP
&DOFXODWH WKH SHULPHWHU LQ FP RI WKH VFDOH GUDZLQJ RI P
WKH URRP ZKLFK LV GUDZQ WR D VFDOH RI

CHAPTER 4 P

3. 7KH PHDVXUHPHQWV RI D UHFWDQJXODU URRP RQ D VFDOH GUDZLQJ DUH FP î FP ,I WKH VFDOH XVHG
LV FDOFXODWH WKH DFWXDO DUHD RI WKH URRP LQ P .

4. $ UHJXODU SRO\JRQ ZLWK DQ H[WHULRU DQJOH RI ƒ LV UHGUDZQ XVLQJ D VFDOH RI ,I WKH DFWXDO
OHQJWK RI VLGHV RI WKH UHJXODU SRO\JRQ LV FP FDOFXODWH WKH SHULPHWHU RI WKH VFDOH GUDZLQJ RI
the regular polygon.

5.

FP

6 cm

7KH GLDJUDP DERYH VKRZV D VFDOH GUDZLQJ RI D UHFWDQJXODU ¿HOG
D ,I WKH VFDOH XVHG LV FDOFXODWH WKH DFWXDO DUHD RI WKH ¿HOG LQ VTXDUH PHWUHV
E 0U 'DQ\ FXWV WKH JUDVV RQ WKH ¿HOG DW D UDWH RI VTXDUH PHWUHV LQ PLQXWHV &DOFXODWH WKH

WLPH LQ KRXUV DQG PLQXWHV WKDW 0U 'DQ\ WDNHV WR FXW DOO WKH JUDVV RQ WKH ¿HOG

100

Chapter 4 Scale Drawings

Dynamic Challenge

Test Yourself

1. The diagram below shows triangle P which is the scale drawing of triangle Q with a scale of
Q &DOFXODWH WKH YDOXH RI Q.

Area of P FP Area of Q FP

P Q

2. 7KH GLDJUDP EHORZ VKRZV ¿YH UHFWDQJOHV

S FP ,, FP FP ,9 CHAPTER 4

FP ,,, 6 cm

FP FP

, FP FP

FP

D $PRQJ UHFWDQJOHV , ,, ,,, DQG ,9 ZKLFK DUH WKH VFDOH GUDZLQJV RI UHFWDQJOH S drawn to a
certain scale?

E )RU HDFK DQVZHU LQ D GHWHUPLQH WKH VFDOH XVHG
F L &DOFXODWH WKH DUHD RI HDFK UHFWDQJOH LQ FP , for your answer in (a).

(ii) Determine the ratio of area of S to area of each answer in (c)(i).

:KDW DUH \RXU FRQFOXVLRQV DERXW WKH UDWLRV REWDLQHG"

3. The diagram shows a scale drawing of a circle with centre O and Q O R
triangle PQR ,W LV JLYHQ WKDW WKH GLDPHWHU RI WKH FLUFOH LV FP FP P
DQG WKH VFDOH RI WKH GUDZLQJ LV 101

D & DOFXODWH WKH DFWXDO OHQJWK RI PR LQ FP 6WDWH \RXU DQVZHU
FRUUHFW WR WKUHH VLJQL¿FDQW ¿JXUHV

E 8 VLQJ \RXU DQVZHU LQ D FDOFXODWH WKH DFWXDO DUHD RI WKH
shaded region in cm 6WDWH WKH DQVZHU FRUUHFW WR IRXU
VLJQL¿FDQW ¿JXUHV

Skills Enhancement
1.

7KH GLVWDQFH E\ DLU IURP .XFKLQJ WR .RWD .LQDEDOX RQ D PDS LV FP ,W LV JLYHQ WKDW WKH

VFDOH RI WKH PDS LV FP NP ,I DQ DHURSODQH WDNHV R൵ IURP .XFKLQJ ,QWHUQDWLRQDO $LUSRUW

DW KRXUV DQG ODQGV DW .RWD .LQDEDOX ,QWHUQDWLRQDO $LUSRUW DW KRXUV FDOFXODWH WKH
DYHUDJH VSHHG RI WKH DHURSODQH LQ NP K± .

CHAPTER 4 2. 7KH GLDJUDP VKRZV WKH VFDOH GUDZLQJ RI 3XDQ )DUDK¶V FP
OLYLQJ URRP 7KH VFDOH RI WKH GUDZLQJ LV 3XDQ FP
)DUDK ZDQWV WR OD\ WLOHV WKURXJKRXW WKH HQWLUH OLYLQJ URRP
6KH LQWHQGV WR XVH WLOHV PHDVXULQJ FP = FP ZKLFK FP
FRVW 50 D SLHFH 3XDQ )DUDK¶V KXVEDQG VXJJHVWV WR
XVH WLOHV RI FP = FP DW 50 D SLHFH :KLFK WLOH
VKRXOG 3XDQ )DUDK FKRRVH LI VKH ZDQWV WR VDYH PRQH\"
6WDWH WKH UHDVRQV IRU \RXU DQVZHU

3. The diagram shows the scale drawing of a

UHFWDQJXODU IDUP RZQHG E\ 3DN +DVVDQ ,W LV

JLYHQ WKDW WKH VFDOH RI WKH GUDZLQJ LV

D &DOFXODWH WKH DFWXDO DUHD RI WKH IUHVKZDWHU FP

S [ ]¿VK SRQG WR WKH QHDUHVW VTXDUH PHWUH . FP
7

E & DOFXODWH WKH UDWLR RI WKH DUHD SODQWHG ZLWK FP
durian trees to the area planted with banana

trees. %DQDQD 9DFDQW ODQG)¿UHVKVK SZRDQWGHU FP
F & DOFXODWH WKH DUHD LQ P , of the vacant land. Durian trees trees
G 3 DN +DVVDQ ZDQWV WR IHQFH XS KLV IDUP ,I

WKH FRVW RI RQH PHWUH RI IHQFLQJ LV 50

FDOFXODWH WKH WRWDO FRVW RI IHQFLQJ LQ 50

102

Chapter 4 Scale Drawings

Self Mastery FP6WRUHURRP
FP
1. 7KH GLDJUDP VKRZV WKH VFDOH GUDZLQJ RI WKH ÀRRU SODQ RI D
VKRSKRXVH WKDW LV GUDZQ WR D VFDOH RI

D &DOFXODWH WKH DFWXDO DUHD RI WKH VWRUHURRP LQ P .

E 6WDWH WKH UDWLR RI WKH DUHD RI WKH VKRSKRXVH WR WKH DUHD RI WKH
storeroom.

F ,I WKH DFWXDO KHLJKW RI WKH VKRSKRXVH LV P FDOFXODWH WKH 6 cm
volume, in m RI WKH WKUHH GLPHQVLRQDO VKRSKRXVH

2. FP FP CHAPTER 4
)ORRU DUHD RI FDQRS\

7 cm

'LDJUDP )ORRU DUHD RI FDQRS\
'LDJUDP

'LDJUDP VKRZV WKH VFDOH GUDZLQJ RI D UHFWDQJXODU IRRWEDOO ¿HOG

D , I WKLV VFDOH GUDZLQJ LV GUDZQ WR D VFDOH RI FDOFXODWH WKH DFWXDO DUHD LQ P , of the
IRRWEDOO ¿HOG

E 6 KDURQ ZDQWV WR UHGUDZ WKH VFDOH GUDZLQJ LQ 'LDJUDP RQ D SLHFH RI $ VL]HG SDSHU :KDW
LV WKH PD[LPXP VFDOH WKDW 6KDURQ FDQ FKRRVH" 6WDWH WKH UHDVRQV IRU \RXU DQVZHU

F 6 HYHUDO FDQRSLHV ZLOO EH VHW XS RQ WKH IRRWEDOO ¿HOG DV LQ 'LDJUDP IRU D FDUQLYDO

L ,I WKH GLPHQVLRQV RI WKH EDVH RI D WHQW DUH P î P FDOFXODWH WKH PD[LPXP QXPEHU
of tents that can be erected.

LL 7KH UHQW IRU D WHQW LV 50 D GD\ $ GLVFRXQW ZLOO EH JLYHQ LI WKH WHQW LV UHQWHG
IRU ¿YH GD\V RU PRUH &DOFXODWH WKH WRWDO UHQW LQ 50 LI WKH FDUQLYDO ODVWV IRU RQH ZHHN

103

P ROJ EC T

Draw the map of the district where you live using
D VXLWDEOH VFDOH <RX FDQ PDUN WKH ORFDWLRQ RI
your house, school and interesting places in your
district using symbols or suitable illustrations.
([KLELW \RXU SURMHFW LQ WKH FODVVURRP

CHAPTER 4 CONCEPT MAP
Scale Drawings
Scale drawing LV GUDZLQJ WKDW VKRZV WKH RULJLQDO REMHFW EDVHG RQ D FHUWDLQ scale.

6FDOH ²M²ea²sur²em²en²t o²f ²sca²le²dr²aw±i±n² g 6 FDOH Q RU ² Q
0HDVXUHPHQW RI REMHFW where Q

Q Q Q !
6FDOH GUDZLQJ ELJJHU 6FDOH GUDZLQJ VDPH VL]H 6FDOH GUDZLQJ VPDOOHU

WKDQ REMHFW DV REMHFW WKDQ REMHFW
²

2EMHFW Drawing 2EMHFW Drawing 2EMHFW Drawing

FP FP FP FP
FP

FP FP
FP
FP FP

104

Chapter 4 Scale Drawings

SELF-REFLECT

At the end of this chapter, I can:

1. ,QYHVWLJDWH DQG H[SODLQ WKH UHODWLRQVKLS EHWZHHQ WKH DFWXDO PHDVXUHPHQWV DQG WKH

PHDVXUHPHQWV RI YDULRXV VL]HV RI GUDZLQJV RI DQ REMHFW DQG KHQFH H[SODLQ WKH
meaning of scale drawing.

2. ,QWHUSUHW WKH VFDOH RI D VFDOH GUDZLQJ

3. 'HWHUPLQH WKH VFDOHV PHDVXUHPHQWV RI REMHFWV RU PHDVXUHPHQWV RI VFDOH GUDZLQJV

4. 'UDZ WKH VFDOH GUDZLQJV RI REMHFWV DQG YLFH YHUVD

5. 6ROYH SUREOHPV LQYROYLQJ VFDOH GUDZLQJV CHAPTER 4

EXPLORING MATHEMATICS

1. Download grid paper of various sizes. Scan the QR Code or
visit http://bukutekskssm.
2. 'UDZ \RXU IDYRXULWH REMHFW DV VKRZQ LQ 'LDJUDP RU 'LDJUDP my/Mathematics/F3/
RQ RQH RI WKH JULG SDSHUV FKRVHQ Chapter4Grid.pdf to
download grid paper of
various sizes.

'LDJUDP 'LDJUDP

3. 5HGUDZ WKH GUDZLQJ RQ DOO WKH JULG SDSHUV RI GL൵HUHQW VL]HV
4. &DQ \RX HDVLO\ GUDZ \RXU IDYRXULWH REMHFW RQ JULGV RI GL൵HUHQW VL]HV"
5. ([KLELW \RXU ZRUN DW WKH PDWKHPDWLFV FRUQHU RI \RXU FODVVURRP

105

CHAPTER Trigonometric

5 Ratios

What will you learn?

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LQ 5LJKW DQJOHG 7ULDQJOHV

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Wh d l thi h t ?
‡ 7ULJRQRPHWULF UDWLRV DOORZ SUREOHPV UHODWHG WR

OHQJWK KHLJKW DQG DQJOH WR EH VROYHG E\ XVLQJ D

ULJKW DQJOHG WULDQJOH

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FRQVWUXFWLRQ DQG VR IRUWK

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XVHG WR PHDVXUH DQJOHV IURP D ORQJ GLVWDQFH ,I WKH
OHQJWK RI PQ DQG WKH DQJOH PQR LV NQRZQ WKXV WKH
ZLGWK RI WKH ULYHU PR FDQ EH FDOFXODWHG HDVLO\ XVLQJ
WULJRQRPHWULF PHWKRGV

R
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106

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$EX $EGXOODK LV WKH IDWKHU RI WULJRQRPHWU\ +H
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SULQFH DQG WKH UXOHU RI 6\ULD $O %DWWDQL ZDV
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,VODPLF PDWKHPDWLFLDQ $O %DWWDQL UHFHLYHG HDUO\
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WKH ¿UVW WR FRPSLOH WKH WDEOH RI FRWDQJHQWV

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([SORULQJ(UD&KDSWHU SGI

WORD B A N K

‡ GHJUHH ‡ GDUMDK

‡ K\SRWHQXVH ‡ KLSRWHQXV

‡ FRVLQH ‡ NRVLQXV

‡ VLQH ‡ VLQXV

‡ WDQJHQW ‡ WDQJHQ

‡ 3\WKDJRUDV WKHRUHP ‡ WHRUHP 3\WKDJRUDV

107

5.1 Sine, Cosine and Tangent of Acute Angles in
Right-angled Triangles

How do you identify the opposite side, adjacent side and LEARNING
hypotenuse? STANDARD

Do you know how the height of an object Identify the opposite side
ZKLFK LV GL൶FXOW WR EH PHDVXUHG VXFK DV and adjacent side based
buildings and mountains are determined? on an acute angle in a
right-angled triangle.

)RU H[DPSOH LQ WKH GLDJUDP RQ WKH t
right, if the distance, s and the angle of Angle of elevation

elevation is known, then, the height, t of

the building can be calculated by using the

trigonometric concepts.

Distance, s

CHAPTER 5 The diagram on the right shows a right-angled triangle PQR. As you P R
have learnt in the chapter Pythagoras theorem in Form 1, the side PR is Q
known as the hypotenuse, which is the longest side in the right-angled
triangle PQR. Do the other two sides PQ and QR have special names like
the longest side PR has?

([DPLQH 'LDJUDP DQG 'LDJUDP EHORZ

P P TIPS
opposite y
side Acute angle
hypotenuse adjacent hypotenuse ƒ ‫ ޒ‬θ ‫ƒ ޒ‬
side

x R Q opposite side R TIPS
Q adjacent side Diagram 2
y
Diagram 1

Based on ‘PRQ in Diagram 1, QR is known as the adjacent side while x
PQ is known as opposite side.
x + y = ƒ
Based on ‘QPR in Diagram 2, PQ is the adjacent side while QR is the ‘ x and ‘ y are acute
opposite side. angles.

Take note that in both Diagram 1 and Diagram 2, the position of the hypotenuse PR LV ¿[HG ZKLFK
is opposite the 90° angle.

For a right-angled triangle:

(a) The hypotenuse is the longest side which is opposite the 90° angle.
(b) The adjacent side and the opposite side change based on the position of the referred acute

angle.

108

Chapter 5 Trigonometric Ratios

Example 1

,GHQWLI\ WKH RSSRVLWH VLGH DGMDFHQW VLGH DQG K\SRWHQXVH EDVHG RQ WKH JLYHQ DQJOH LQ WKH WDEOH EHORZ

IRU DOO WKH IROORZLQJ ULJKW DQJOHG WULDQJOHV

D B C E M F S

K
T

A Angle L P QR
‘BAC
Solution: ‘BCA Hypotenuse Opposite side Adjacent side
Triangle ‘LKM AC BC AB
¨ABC ‘LMK AC AB BC
‘TPQ KM LM KL
¨KLM ‘QRS KM KL LM
¨PQT PT QT PQ
¨RQS RS QS QR

MIND TEST 5.1a CHAPTER 5
1. %DVHG RQ WKH ULJKW DQJOHG WULDQJOHV EHORZ FRS\ DQG FRPSOHWH WKH JLYHQ WDEOH

E
PD

KN

FE

Q RM GA B C

Triangle Angle Hypotenuse Opposite side Adjacent side
¨345 ‘QPR
¨KMN ‘PRQ
¨EFG ‘MNK
¨ABE ‘MKN
¨CBD ‘FEG
‘EGF
‘BAE
‘AEB
‘BCD
‘BDC

109

What is the relationship between acute angles and the LEARNING
ratios of the sides of right-angled triangles? STANDARD

Brainstorming 1 Make and verify the
In groups conjecture about the
Aim: 7R LGHQWLI\ WKH UHODWLRQVKLS EHWZHHQ DFXWH DQJOHV DQG WKH relationship between acute
UDWLRV RI WKH VLGHV RI ULJKW DQJOHG WULDQJOHV angles and the ratios of
the sides of right-angled
Materials 6TXDUH JULG SDSHU UXOHU DQG SHQFLO
WULDQJOHV DQG KHQFH GH¿QH
sine, cosine and tangent.

Steps:
1. 'UDZ D ULJKW DQJOHG WULDQJOH PQR ZKHUH WKH OHQJWK PQ LV XQLWV DQG WKH OHQJWK QR LV

XQLWV

2. 'UDZ D IHZ VWUDLJKW OLQHV SDUDOOHO WR RQ /DEHO WKHP DV R Q , R Q DQG R Q DV VKRZQ LQ
WKH GLDJUDP EHORZ

R

R TIPS
R
R Use the Pythagoras
theorem to determine the
length of PR1, PR2, PR3
and PR.

CHAPTER 5 P Q Q Q Q

3. &RPSOHWH WKH WDEOH EHORZ ZLWK WKH UHTXLUHG PHDVXUHPHQWV

Acute angle —O—Hpy—ppo—ostiet—enus—isde—e – —A—Hd—yjapc—oetne—nt us—isde—e – —AO—dp—jpaoc—seint—et ss—iidd—ee –
‘QPR
² RP RQ² ² ²PPQR² ² R ²P QQ² ²4


R— P RQ— ²PPQR² R² P QQ²

²RP RQ² ²PP QR² R²P QQ²

² RPQR± ²PPQR± ²PRQQ±

Discussion:

1. :KDW LV WKH SDWWHUQ RI \RXU DQVZHU WR WKH UDWLR RI WKH OHQJWK RI WKH RSSRVLWH VLGH WR WKH
K\SRWHQXVH WKH UDWLR RI WKH OHQJWK RI WKH DGMDFHQW VLGH WR WKH K\SRWHQXVH DQG WKH UDWLR RI
WKH OHQJWK RI WKH RSSRVLWH VLGH WR WKH OHQJWK RI WKH DGMDFHQW VLGH"

2. :KDW KDSSHQV LI WKH VL]H RI WKH DQJOH LV FKDQJHG" -XVWLI\ \RXU DQVZHU

110

Chapter 5 Trigonometric Ratios

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

*LYHQ D ¿[HG DFXWH DQJOH LQ ULJKW DQJOHG WULDQJOHV RI GL൵HUHQW VL]HV
D 7KH UDWLR RI WKH OHQJWK RI WKH RSSRVLWH VLGH WR WKH K\SRWHQXVH LV D FRQVWDQW
E 7KH UDWLR RI WKH OHQJWK RI WKH DGMDFHQW VLGH WR WKH K\SRWHQXVH LV D FRQVWDQW
F 7KH UDWLR RI WKH OHQJWK RI WKH RSSRVLWH VLGH WR WKH OHQJWK RI WKH DGMDFHQW VLGH LV D FRQVWDQW

7KH UHODWLRQVKLSV RI WKH UDWLRV REWDLQHG IURP %UDLQVWRUPLQJ DUH REMINDER
WULJRQRPHWULF UDWLRV NQRZQ DV sine, cosine and tangent WKDW LV
Ƈ VLQ VLQH
VLQH ²RKS²\SSR²RVWL²HWHQ X²VLVGH²H ± Ƈ FRV FRVLQH
Ƈ WDQ WDQJHQW

FRVLQH ²DKG²\MDS²FRHWQ²HQW XV²LVGH²H ± BULLETIN

The word Trigonometry
W DQJHQW ²RDGS²MSDR²FVHLQ²WHW V²VLLGG²HH ± originates from Greek
words, that is,
Trigonon triangle
Metron to measure

Example 2 R CHAPTER 5
&RPSOHWH WKH IROORZLQJ WDEOH EDVHG RQ WKH GLDJUDP RQ WKH ULJKW
\
sin x cos x tan x sin y cos y tan y

Solution: x

QP

sin x cos x tan x sin y cos y tan y

—QPRR– —PPQR– —QPQR– —PPQR– —QPRR– —QPQR–

MIND TEST 5.1b

1. &RPSOHWH WKH WDEOH EDVHG RQ WKH ULJKW DQJOHG WULDQJOHV EHORZ

Dx E K Q
L \
\\
Triangles \ cos x x x SR
¨DEF F P
¨KLM M tan y
¨PQR sin x sin y cos y
tan x 111

What is the impact of changing the size of the angles on LEARNING
the values of sine, cosine and tangent? STANDARD

Brainstorming 2 In pairs Make and verify the
conjecture about the
Aim 7R LGHQWLI\ WKH LPSDFW RI FKDQJLQJ WKH VL]H RI WKH DQJOHV RQ WKH impact of changing the
YDOXHV RI VLQH FRVLQH DQG WDQJHQW size of the angles on the
values of sine, cosine and
tangent.

Materials 6TXDUH JULG SDSHU UXOHU SURWUDFWRU DQG SHQFLO

Steps
1. 'UDZ IRXU ULJKW DQJOHG WULDQJOHV DV VKRZQ EHORZ ZLWK WKH EDVH OHQJWK RI FP

2. 0DNH VXUH WKDW WKH DQJOHV DQG OHQJWKV RI DOO ULJKW DQJOHG WULDQJOHV DUH H[DFWO\ DV JLYHQ

R

R ƒ
R ƒ
R ƒ

ƒ

ƒ ƒ Q P ƒ ƒ
P FP Q P FP FP Q P Q
FP

3. &RPSOHWH WKH WDEOH EHORZ

CHAPTER 5 sin 10ƒ sin 20ƒ sin 30ƒ sin 40ƒ sin 50ƒ sin 60ƒ sin 70ƒ sin 80ƒ

—RPRQ– —PPQR–
² ² ² ²




cos 10ƒ cos 20ƒ cos 30ƒ cos 40ƒ cos 50ƒ cos 60ƒ cos 70ƒ cos 80ƒ

—PPRQ– —RPQR–
² ± ² ±



tan 10ƒ tan 20ƒ tan 30ƒ tan 40ƒ tan 50ƒ tan 60ƒ tan 70ƒ tan 80ƒ

—RPQQ– — PRQQ–
² ± ² ±




112

Chapter 5 Trigonometric Ratios

Discussion:
1. %DVHG RQ WKH YDOXHV LQ WKH WDEOH IRU WKH WULJRQRPHWULF UDWLRV \RX KDYH FRPSOHWHG ZKDW

FRQFOXVLRQ FDQ \RX PDNH"

2. :KDW LV \RXU FRQMHFWXUH RQ
D WKH YDOXH RI WKH VLQH UDWLR ZKHQ WKH DQJOH DSSURDFKHV ƒ DQG ƒ"
E WKH YDOXH RI WKH FRVLQH UDWLR ZKHQ WKH DQJOH DSSURDFKHV ƒ DQG ƒ"
F WKH YDOXH RI WKH WDQJHQW UDWLR ZKHQ WKH DQJOH DSSURDFKHV ƒ DQG ƒ"

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW TIPS

The larger the size of the acute angle VLQ ƒ VLQ ƒ
D WKH larger the value of sine DQG LWV YDOXH approaches 1 FRV ƒ FRV ƒ
E WKH smaller the value of cosine DQG LWV YDOXH approaches WDQ ƒ WDQ ƒ '

zero
F WKH larger the value of tangent

Example 3 FP FP FP CHAPTER 5

7KH GLDJUDP RQ WKH ULJKW VKRZV WZR ULJKW DQJOHG WULDQJOHV x \
'HWHUPLQH ZKHWKHU DOO WULJRQRPHWULF UDWLRV RI DQJOH x DQG FP FP FP
DQJOH \ DUH HTXDO 6WDWH WKH UHDVRQ IRU \RXU DQVZHU

Solution:

V LQ x = — F RV x = — W DQ x = —4
W DQ \ ² ± ²

VLQ \ ² ± ² F RV \ ² ± ²

7KH WULJRQRPHWULF UDWLRV RI DQJOH x DQG DQJOH \ DUH HTXDO EHFDXVH WKH OHQJWK RI FRUUHVSRQGLQJ VLGHV
RI WKH WZR WULDQJOHV DUH SURSRUWLRQDO

MIND TEST 5.1c

1. 7KH GLDJUDP RQ WKH ULJKW VKRZV WZR ULJKW DQJOHG WULDQJOHV FP FP FP FP

'HWHUPLQH ZKHWKHU DOO WULJRQRPHWULF UDWLRV RI DQJOH x \
x DQG DQJOH \ DUH HTXDO 6WDWH WKH UHDVRQ IRU \RXU FP FP

DQVZHU

113

2. The diagram on the right shows a right-angled triangle. S

(a) Determine the trigonometric ratio for 2.1 cm
R
i. sin 15° ii. cos 15° iii. tan 15° 1.9 cm
iv. sin 30° v. cos 30° vi. tan 30° Q

(b) Is the increase in the value of the trigonometric ratio for 151°5°
angle 15° and angle 30° proportional to the increase in

the angle? P 7 cm

How do you determine the values of sine, cosine and LEARNING
tangent of acute angles? STANDARD
Example 4
Determine the values of
sine, cosine and tangent of
acute angles.

The diagram on the right shows a right-angled triangle PQR. Calculate Q 15 cm P
the value of 8 cm
(a) length of PR (b) sin ‘PRQ (c) cos ‘PRQ (d) tan ‘QPR

Solution: (c) cos ‘PRQ (d) tan ‘QPR R
(a) length of PR (b) sin ‘PRQ = 1—87 = 1—85

CHAPTER 5 PR ¥ 2 + 82 = 11—75 FLASHBACK
= ¥289
Pythagoras theorem

= 17 cm bc c2 = a2 + b2
a2 = c2 – b2

a b2 = c2 – a2

Example 5

The diagram on the right shows right-angled triangles PQT and RQS.

PQR is a straight line. Given that the length of SQ is 6 cm, calculate S

the value of

(a) length of QR (b) length of PT (c) sin ‘QRS

(d) cos ‘TPQ (e) tan ‘PTQ (f) tan ‘QSR T 10 cm

Solution: (b) length of PT (c) sin ‘QRS P FP Q R
(a) length of QR = 1—60
PT ¥ 2 + 32 = —53
QR ¥ 2 – 62 = ¥25
= ¥ = 5 cm
= 8 cm

(d) cos ‘TPQ (e) tan ‘PTQ (f) tan ‘QSR REMINDER
= —5 = — 3 = —68
= — 3 Ratio value should be
given in the simplest term.

114

Chapter 5 Trigonometric Ratios

What is the relationship between sine, cosine and tangent?

)RU ULJKW DQJOHG WULDQJOHV \RX KDYH OHDUQW WKDW
VLQH ²RKS²\SSR²RVWL²HWHQ X²VLVGH±H ² FRVLQ H ²DKG²\MDS²FRHW²QHQW ²XVLVGH±H ± DQG WDQJHQW ² DRGS²MSDR²FVHLQ±WHW² VVLL±GG±HH±

'R \RX NQRZ WKDW WKH WKUHH WULJRQRPHWULF UDWLRV DERYH DUH UHODWHG WR RQH DQRWKHU" 7DQJHQW LV WKH
UDWLR RI VLQH WR FRVLQH

6WXG\ WKH GLDJUDP EHORZ

,W LV NQRZQ WKDW
R D VxL Q θVL Q θ² [ E FRV θ ² \ F W DQ θ ²\[
\ FRV θ
DISCUSSION CORNER

x If θ is ‘QRP, is the ratio
of tan θ still –cs–oi–ns–θθ–?
θ 7KXV Discuss.
P \ Q W DQ θ ²VLQ² θ±
FRV θ

Example 6

,I VLQ θ DQG FRV θ FDOFXODWH WKH YDOXH RI WDQ θ QU I Z CHAPTER 5

Solution: If tan θ 1
²2
²FVRLQ²V șș±
WDQ θ state the possible values
of sin θ and cos θ.

² ²

²4
SMART MIND

Given that sin θ x,
Example 7 determine the possible
values of cos θ and tan θ.

,S Io VlLuQ t iθo n: ² DQG WDQ θ ² ² FDOFXODWH WKH YDOXH RI FRV θ.
¥

WDQ θ ²VLQ² θ±
FRV θ
²²– ± DISCUSSION CORNER
² ²
¥ FRV θ If tan θ what type
of triangle is being
F RV θ ² —¥– ² – represented?


F RV θ ²¥ ²

115

Example 8

7KH GLDJUDP RQ WKH ULJKW VKRZV D ULJKW DQJOHG WULDQJOH PQR Q
*LYHQ WKDW PR FP DQG VLQ ‘QPR ² FDOFXODWH
D WKH OHQJWK RI QR R

E FRV ‘QPR

Solution: P FP
D VLQ ‘QPR = —
E PQ ¥ ± SMART MIND
²QPRR± ² ¥
²Q R± ² Given sin θ ²35 DQG WKH
QR ² ² PQ FP length of hypotenuse is
QR FP 20 cm, determine cos θ
7KXV FRV ‘QPR = —PQ and tan θ.
PR

—


—4


Example 9

CHAPTER 5 7KH GLDJUDP RQ WKH ULJKW VKRZV ULJKW DQJOHG WULDQJOHV PQT DQG RQS S
*LYHQ WKDW PQR DQG STQ DUH VWUDLJKW OLQHV FDOFXODWH WKH YDOXH RI
FRV x T FP

Solution: FP x
QR
FRV x —RQ– 'HWHUPLQH WKH P FP
SR YDOXH RI RQ ¿UVW

TQ ¥ ± ; SQ TQ RQ ¥ ±

¥ ¥ DISCUSSION CORNER

TQ FP SQ FP FP Given that the hypotenuse
of a right-angled triangle is
7KXV FRV x ²RSRQ± ¥ FP Getermine tan θ if
cos θ ²¥12± .

²


²


Example 10 ƒ Q S
7KH GLDJUDP RQ WKH ULJKW VKRZV D ULJKW DQJOHG WULDQJOH PRS *LYHQ FP R
WKDW PQR LV D VWUDLJKW OLQH DQG FRV ƒ FDOFXODWH WKH OHQJWK RI
PS 6WDWH WKH DQVZHU FRUUHFW WR WZR GHFLPDO SODFHV

P

116

Chapter 5 Trigonometric Ratios

Understanding the problem Implementing the strategy
&DOFXODWH WKH OHQJWK RI PS ZKLFK
LV WKH K\SRWHQXVH RI ¨PRS ‡ ‘QSP ‘QPS = ƒ

Planning a strategy WKXV ‘PQS = – ƒ ± ƒ ƒ
‡ PS ¥PR SR
‡ SR DQG QR FDQ EH FDOFXODWHG LI ‡ ‘SQR ƒ ± ‘PQS

‘SQR RU ‘QSR LV NQRZQ ‘SQR ƒ ± ƒ ƒ
‡ ,GHQWLI\ WKH SRVLWLRQ RI
FRV ƒ ‡ *LYHQ WKDW FRV ƒ

Making a conclusion FRV ƒ ² SR ¥ ± ¥ FP
PS FP G S 7KXV
²Q R± ² PS ¥SR PR
PS ¥ ¥
PS FP
QR ² ± ± ±


FP

MIND TEST 5.1d FLASHBACK

1. &DOFXODWH WKH YDOXHV RI VLQ θ FRV θ DQG θ IRU HDFK RI WKH IROORZLQJ Pythagoras triples CHAPTER 5
ULJKW DQJOHG WULDQJOHV ABC
345
D E P F 5 12 13
6 10
FP θ 15 17
7 24 25
FP P P P P 40 41

θ θ
FP P

G H I

θ FP FP θ
PP θ P

PP P

2. &DOFXODWH WKH YDOXH RI x ZLWKRXW GUDZLQJ DQ\ ULJKW DQJOHG WULDQJOHV RU XVLQJ 3\WKDJRUDV
WKHRUHP RU D FDOFXODWRU

D V LQ θ ² FRV θ ²¥ ± WDQ θ x E VLQ θ ² FRV θ x, WDQ θ
¥

F V LQ θ x FRV θ ² WDQ θ ²¥ ± ± G VLQ θ ² FRV θ x WDQ θ ²4¥ ±


117

3. 'HWHUPLQH WKH OHQJWK RI VLGH q IRU HDFK RI WKH ULJKW DQJOHG WULDQJOHV EHORZ

D V LQ ‘QRP ² E VLQ ‘LKM ² F VLQ ‘ACB ²

Lq
P Aq

P M PP C
q B F

P S

QR

K

4. 'HWHUPLQH WKH OHQJWK RI VLGH z IRU HDFK RI WKH ULJKW DQJOHG WULDQJOHV EHORZ

D FRV ‘SRT ² E FRV ‘HJI ² F FRV ‘DFE

S H D
z FP z
R

FP J E PP
T
z
I

5. &DOFXODWH WKH YDOXH RI x IRU HDFK RI WKH ULJKW DQJOHG WULDQJOHV EHORZ

D WDQ ‘BAC E WDQ ‘PRQ ²4 F WDQ ‘LKM ²

CHAPTER 5 A FP B P K

x x

C FP L
Q
x

PP

RM

6. 7KH GLDJUDP RQ WKH ULJKW VKRZV ULJKW DQJOHG WULDQJOHV PQR P

DQG PRS *LYHQ WKDW WDQ θ ² 4 DQG PS ² PR FDOFXODWH θT
LQ FP WKH OHQJWK RI

D PR

E RS

Q FP R

7. 7KH GLDJUDP RQ WKH ULJKW VKRZV ULJKW DQJOHG WULDQJOHV DFE x E
DQG EHI ,I WDQ x ² DF FP DQG EF EH F
GHWHUPLQH WKH OHQJWK RI EI LQ FP

D

x
HI

118

Chapter 5 Trigonometric Ratios

How do you determine the values of sine, cosine and tangent LEARNING
of 30°, 45° and 60° angles without using a calculator? STANDARD

S SK Determine the values of
sine, cosine and tangent
XQLWV– XQLWV žƒ ¥ of 30°, 45° and 60° angles
¥ without using a calculator.

TIPS

QS ¥ 2 – 12
QS ¥
KM ¥ 2 + 12
KM ¥
ž ƒž – ƒ L ƒ
P XQLW Q XQLW R Q R M
CHAPTER 5
'LDJUDP D 'LDJUDP E 'LDJUDP

'LDJUDP E DERYH LV KDOI RI WKH HTXLODWHUDO WULDQJOH PRS ZKHUH WKH OHQJWK RI PQR LV XQLWV
'LDJUDP VKRZV DQ LVRVFHOHV WULDQJOH KLM

7KH WDEOH EHORZ VKRZV WKH YDOXHV RI WKH WULJRQRPHWULF UDWLRV RI ƒ ƒ DQG ƒ DQJOHV WKDW FDQ
EH FDOFXODWHG ZLWKRXW XVLQJ D FDOFXODWRU EDVHG RQ 'LDJUDP E DQG 'LDJUDP

Ratio Angle 30° 60° 45° BULLETIN
– —
VLQ θ ¥— ¥ Surd is an irrational
number in the root form
FRV θ ¥— – — such as ¥2 , ¥3 and
¥ ¥17 . ¥3 is read as surd
three.
WDQ θ — ¥
¥
Example 11

&DOFXODWH WKH IROORZLQJ YDOXHV ZLWKRXW XVLQJ D FDOFXODWRU

D VLQ ƒ FRV ƒ E FRV ƒ ± VLQ ƒ F WDQ ƒ ± FRV ƒ

G VLQ ƒ FRV ƒ ± WDQ ƒ H WDQ ƒ VLQ ƒ VLQ ƒ

Solution:

D VLQ ƒ FRV ƒ E FRV ƒ ± VLQ ƒ F WDQ ƒ ± FRV ƒ

—


² ² ¥— ±
¥ ¥
²
¥ ² î ¥² ¥— ±
¥ ¥
¥ ² ¥ ± ± ±¥ ± ± ±

²¥ ±

¥ ²¥ ±


G VLQ ƒ FRV ƒ ± WDQ ƒ H WDQ ƒ VLQ ƒ VLQ ƒ

¥— 4¥— ± ¥ — 4 ¥— —

¥ ¥

¥ ¥ ± ¥ ² ² ² î ¥² TIPS ¥2 × 2
± ¥ ¥ ¥ ¥ ¥4
± ¥ ¥2 × ¥2
¥
²¥ ±


¥

119

MIND TEST 5.1e

1. 'HWHUPLQH WKH IROORZLQJ YDOXHV ZLWKRXW XVLQJ D FDOFXODWRU

D FRV ƒ WDQ ƒ E FRV ƒ WDQ ƒ F WDQ ƒ FRV ƒ

G VLQ ƒ ± FRV ƒ H VLQ ƒ ± FRV ƒ I WDQ ƒ ± FRV ƒ

J VLQ ƒ FRV ƒ WDQ ƒ K WDQ ƒ VLQ ƒ ± FRV ƒ VLQ ƒ

L WDQ ƒ VLQ ƒ FRV ƒ M WDQ ƒ VLQ ƒ ± VLQ ƒ FRV ƒ

What is the unit of measure for angles?

$QJOHV DUH PHDVXUHG LQ WKH XQLW RI GHJUHHV ƒ $QJOHV FDQ DOVR EH H[SUHVVHG LQ XQLWV RI GHJUHHV ƒ
PLQXWHV ' DQG VHFRQGV '' WKDW LV

ƒ '
' ''

Example 12

D &RQYHUW ƒ WR GHJUHHV DQG PLQXWHV E &RQYHUW WKH DQJOH ƒ ' WR GHJUHHV

Solution: E ƒ ' ƒ '

D ƒ ƒ ƒ — –
ƒ × '
ƒ '
CHAPTER 5 ƒ ' ƒ ƒ

ƒ ƒ

ƒ

MIND TEST 5.1f

1. &RQYHUW HDFK RI WKH IROORZLQJ DQJOHV WR GHJUHHV DQG PLQXWHV

D ƒ E ƒ F ƒ G ƒ
K ƒ
H ƒ I ƒ J ƒ
G ƒ '
2. 6WDWH HDFK RI WKH IROORZLQJ DQJOH LQ GHJUHHV K ƒ '

D ƒ ' E ƒ ' F ƒ '

H ƒ ' I ƒ ' J ƒ '

How do you determine the values of sine, cosine and LEARNING
tangent? STANDARD

'R \RX NQRZ WKDW D VFLHQWL¿F FDOFXODWRU FDQ EH XVHG WR GHWHUPLQH WKH Perform calculations
WULJRQRPHWULF UDWLR RI DQ DQJOH" involving sine, cosine
and tangent.

Example 13

8VH D VFLHQWL¿F FDOFXODWRU WR GHWHUPLQH WKH IROORZLQJ YDOXHV FRUUHFW WR IRXU GHFLPDO SODFHV

D VLQ ƒ ' E FRV ƒ F WDQ ƒ '

120

Chapter 5 Trigonometric Ratios

Solution:
D VLQ ƒ '

SMART FINGER 1,234567.89 sin 4 5 °' '' 6 °' '' = 0.7083398377 BULLETIN
7 8 9÷ 0 Â 7 = 0.9354440308
The button °' '' should
4 5 6x be pressed only when
the question is given in
1 2 3- degrees and minutes.
AC 0 . +

E FRV ƒ

SMART FINGER 1,234567.89 cos 2
7 8 9÷

4 5 6x

1 2 3-
AC 0 . +

F WDQ ƒ '

SMART FINGER 1,234567.89 tan 6 4 °' '' 12 °' '' = 2.068599355
7 8 9÷

4 5 6x

1 2 3-
AC 0 . +

MIND TEST 5.1g
1. 8VH D VFLHQWL¿F FDOFXODWRU WR GHWHUPLQH WKH IROORZLQJ YDOXHV FRUUHFW WR IRXU GHFLPDO SODFHV
D VLQ ƒ E FRV ƒ F WDQ ƒ G VLQ ƒ ' H FRV ƒ ' I WDQ ƒ '

How do you calculate the size of an angle by using trigonometric ratios sine, cosine and CHAPTER 5
tangent?

,I WKH YDOXH RI WKH WULJRQRPHWULF UDWLR LV JLYHQ \RX FDQ XVH D VFLHQWL¿F FDOFXODWRU WR GHWHUPLQH WKH
VL]H RI WKH UHODWHG DQJOH

Example 14

8VH D VFLHQWL¿F FDOFXODWRU WR FDOFXODWH WKH IROORZLQJ x YDOXHV

D VLQ x E FRV x F WDQ x

Solution: $QVZHU LQ GHJUHHV REMINDER
$QVZHU LQ GHJUHHV DQG PLQXWHV
D VLQ x If the unit of second is 30"
x VLQ± or more, the minute unit
x ƒ will be added by 1'.
x ƒ '

SMART FINGER °' '' ƒ ' ''1,234567.89
7 8 9÷
shift sin 0 . 8 3 7 7 = 56.898036354 5 6 x
1 2 3-
AC 0 . + 56 53 52.93

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121

E FRV x
x FRV±

x ƒ

x ƒ '

SMART FINGER °' '' ° ' ''1,234567.89
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45 24 15.37

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shift tan 27 8 9 ÷ 7 6 = 68.10017426 68 6 0.63

4 5 6x
1 2 3-

AC 0 . +

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K VLQ x
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D WDQ x E FRV x F VLQ x
H FRV x I WDQ x J FRV x
L WDQ x M VLQ x N WDQ x

CHAPTER 5 How do you solve problems involving sine, cosine and LEARNING
tangent? STANDARD
Example 15
Solve problems involving
sine, cosine and tangent.

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VLQ ƒ —QR–
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PR P G S

Example 15 F E
A
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122

Chapter 5 Trigonometric Ratios

Understanding the problem Planning a strategy

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Test Yourself

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124

Chapter 5 Trigonometric Ratios

Self Mastery

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125

CONCEPT MAP

K\SRWHQXVH RSSRVLWH
θ VLGH
DGMDFHQW VLGH

Trigonometric ratio

Ratio of sine Ratio of cosine Ratio of tangent
VLQ θ ²RKS²\SSR²RVWLHW²HQ XV±LV²GHH WDQ θ ²RDGS²MSDRF²VHLQW²HW VV²LLGGHH²
FRV θ ²DKG\²MDSFR²HWQH²QW XV²LVGHH±


:KHQ WKH size RI DQ acute angle θ increases,
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DQJOHV IURP ƒ ƒ ƒ ƒ VLQ θ
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VLQH FRVLQH WDQ θ — – ¥ E WDQ θ '
DQG WDQJHQW ¥ ¥

VLQ θ — —¥ – ƒ ¥ O ƒ
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FRV θ — ¥ – — G ƒ
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SELF-REFLECT

At the end of this chapter, I can:

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WULDQJOH

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126

Chapter 5 Trigonometric Ratios

EXPLORING MATHEMATICS

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127

CHAPTER Angles and
Tangents of
6 Circles

What will you learn?

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128

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2.135 m 129

6.1 Angle at the Circumference and Central Angle
Subtended by an Arc

What are the angles at the circumference of a circle? LEARNING
STANDARD
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\RX HYHU VHHQ D YHKLFOH ZKHHO LQ RWKHU VKDSHV" the relationships
between angles at the
circumference and central
angle subtended by
particular arcs, and hence
use the relationships to
determine the values of
angles in circles.

DISCUSSION CORNER

Compare the bicycle
wheels below, which
wheel allows you to
reach your destination
faster?

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CHAPTER 6
Radius
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CDhiaomrdeter
P

Centre
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130

Chapter 6 Angles and Tangents of Circles

Are angles at the circumference of a circle subtended by the same arc equal?

Brainstorming 1

In groups

Aim: 7R YHULI\ WKDW DQJOHV DW WKH FLUFXPIHUHQFH VXEWHQGHG E\ WKH VDPH DUF DUH HTXDO
Materials: $ SDSHU FRPSDVVHV SURWUDFWRU UXOHU DQG SHQFLO

Steps:

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2. 'UDZ D FKRUG QR WKDW IRUPV ƒ DW SRLQW Q 'LDJUDP 2WKHU JURXSV DUH HQFRXUDJHG WR
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P P P P CHAPTER 6
R R
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Discussion:
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)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

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131

,Q JHQHUDO S

R T

P Angles DW WKH circumference subtended E\ WKH same arc
DUH equal
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Q

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E\ WKH VDPH DUF

Brainstorming 2

In pairs

Aim: 7R YHULI\ WKDW DQJOHV DW WKH FLUFXPIHUHQFH VXEWHQGHG E\ WKH VDPH DUF DUH HTXDO

Materials: '\QDPLF VRIWZDUH

Steps:
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2. &OLFN RQ 3RLQW 7RRO DQG PDUN WKUHH SRLQWV 'LDJUDP

CHAPTER 6 3. &OLFN RQ 7H[W 7RRO DQG ODEHO WKH WKUHH SRLQWV PDUNHG LQ VWHS 'LDJUDP

B

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C

'LDJUDP

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ZHOO DV SRLQW B DQG SRLQW C 'LDJUDP

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6. &OLFN 0HDVXUH DQG VHOHFW $QJOH 7KH YDOXH RI ЄABC ZLOO EH GLVSOD\HG 'LDJUDP

132

Chapter 6 Angles and Tangents of Circles

BB
B

A A A
C C C

'LDJUDP 'LDJUDP 'LDJUDP

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DW WKH FLUFXPIHUHQFH

B B
D
D

A A
C C

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Discussion: CHAPTER 6
:KDW FDQ EH FRQFOXGHG IURP \RXU REVHUYDWLRQV LQ WKH DERYH DFWLYLWLHV"

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW
7KH DQJOHV DW WKH FLUFXPIHUHQFH VXEWHQGHG E\ WKH VDPH DUF DUH HTXDO

Example 1 A ž B BULLETIN
yC
%DVHG RQ WKH GLDJUDP RQ WKH ULJKW FDOFXODWH y = ABE = 40°.
WKH YDOXH RI y ž D ADE  ƒ because
ADE is not an angle
E
at the circumference of
Solution: the circle subtended by
arc AE.
y A BE ƒ
133

MIND TEST 6.1a E DISCUSSION CORNER
Is ADB = ACB?
1. &DOFXODWH WKH YDOXH RI z z
Discuss.
D C

ž z ƔD

G ž AƔ
F Ɣ
ž
ž ž z B
z Q C

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*LYHQ WKDW 4:5 $ DQG :57 $ W Ɣ
GHWHUPLQH WKH YDOXH RI
A
D QSR E :47
Ɣ
F WRQ G QRT
B

R

ž

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134

Chapter 6 Angles and Tangents of Circles

Are angles at the circumference of a circle subtended by arcs of the same length equal
and are angles at the circumference proportional to the arc length?

Brainstorming 3

In groups

Aim: 1. 7R YHULI\ WKDW DQJOHV DW WKH FLUFXPIHUHQFH VXEWHQGHG E\ DUFV RI WKH VDPH OHQJWK
DUH HTXDO

2. 7R YHULI\ WKDW DQJOHV DW WKH FLUFXPIHUHQFH LV SURSRUWLRQDO WR WKH OHQJWK RI WKH
DUF

Materials: &RPSDVVHV SURWUDFWRU SHQFLO UXOHU DQG $ SDSHU

Steps:

1. 'UDZ D FLUFOH RI UDGLXV FP :LWKRXW DGMXVWLQJ WKH JDS RI WKH FRPSDVVHV GLYLGH WKH
FLUFXPIHUHQFH RI FLUFOH LQWR VL[ SDUWV 'LDJUDP 'LDJUDP

2. 'UDZ WZR DQJOHV DW WKH FLUFXPIHUHQFH WKDW DUH VXEWHQGHG E\ WZR GLIIHUHQW SDUWV RI WKH VDPH
OHQJWK DQG ODEHO WKHP 'LDJUDP

BP

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RT
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T
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Arcs Arcs
BA PQ
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7DEOH 7DEOH 'LDJUDP

Discussion:

1. :KDW FDQ \RX FRQFOXGH DERXW DQJOHV DW WKH FLUFXPIHUHQFH VXEWHQGHG E\ DUFV RI WKH VDPH
OHQJWK"

2. :KDW LV \RXU FRQFOXVLRQ RQ WKH HIIHFWV RI FKDQJLQJ WKH DUF OHQJWK WR WKH DQJOHV VXEWHQGHG
DW WKH FLUFXPIHUHQFH"

135

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D BCA PRQ >$UF OHQJWK AB $UF OHQJWK PQ@
E BQR = RPT >$UF OHQJWK BR = $UF OHQJWK RT@

,Q JHQHUDO Q
T
P

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P length DUH HTXDO ,I DUF OHQJWK PQ DUF OHQJWK SU WKHQ
[ R [ \ FP PRQ STU

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T DUF LV proportional WR the arc length
[[R

Q

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CHAPTER 6 Example 2 P
Q
7KH GLDJUDP RQ WKH ULJKW VKRZV D FLUFOH ZLWK OHQJWK RI DUFV
PR QS 'HWHUPLQH WKH YDOXH RI [ *LYH UHDVRQV IRU \RXU DQVZHU R R

Solution: [
S
[ ƒ EHFDXVH  [ DQG  ƒ DUH DW WKH FLUFXPIHUHQFH DQG OHQJWK
RI DUFV PR QS

Example 3
%DVHG RQ WKH GLDJUDP RQ WKH ULJKW GHWHUPLQH WKH YDOXH RI [

Solution: R [

² [ƒ± ± ± ±FFPP±± R


[ ƒ P FP
[ ƒ FP Q

136

Chapter 6 Angles and Tangents of Circles

Example 4 PQ

*LYHQ WKH OHQJWK RI PLQRU DUF PS LV WZR WLPHV WKH OHQJWK RI DUF R
QR GHWHUPLQH WKH YDOXH RI [ ƒ S

Solution: 6XP RI DQJOHV LQ D [
376 $ ± $ WULDQJOH LV ƒ T

376 $ QU I Z

7KXV [ ± ± ±ƒ Q R ±P ±S± P
Q [ ƒ
[ $ R

MIND TEST 6.1b S

If arc length of
RS = —7 QR, determine

2
the value of [.

1. %DVHG RQ WKH GLDJUDPV EHORZ FDOFXODWH WKH YDOXH RI [

D E F G

FP FP FP $ [
$ $

[ [ $ FP $
$ FP
FP [

2. 7KH GLDJUDP RQ WKH ULJKW VKRZV D FLUFOH *LYHQ WKDW WKH Q CHAPTER 6
OHQJWK RI DUFV RS QR 435 $ DQG 364 $ R
GHWHUPLQH WKH YDOXH RI P $

D SPR $
S
E 653



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QPT RS *LYHQ WKDW 457 $ 467 $ DQG
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$
D RQS
U

E 786 $ S

F TPS T

137

What is the relationship between angles at the centre of a circle and angles at the
circumference that are subtended by the same arc?

Brainstorming 4 In pairs

Aim: 7R YHULI\ WKH UHODWLRQVKLS EHWZHHQ DQJOHV DW WKH FHQWUH RI D FLUFOH DQG DQJOHV DW WKH
FLUFXPIHUHQFH VXEWHQGHG E\ WKH VDPH DUF

Materials: '\QDPLF VRIWZDUH

Steps:
1. 6WDUW ZLWK 1HZ 6NHWFK DQG FOLFN RQ &RPSDVV 7RRO WR GUDZ D FLUFOH
2. 8VH 3RLQW 7RRO WR SODFH WKUHH SRLQWV DURXQG LWV FLUFXPIHUHQFH 'LDJUDP
3. 8VH 7H[W 7RRO WR ODEHO DOO SRLQWV DW WKH FLUFOH ZLWK A B C DQG FHQWUH DV ' 'LDJUDP
4. 8VH 6WUDLJKWHGJH 7RRO WR FRQVWUXFW OLQHV IURP RQH SRLQW WR DQRWKHU 'LDJUDP

'LDJUDP 'LDJUDP 'LDJUDP
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CHAPTER 6 5. 8VH 6HOHFWLRQ $UURZ 7RRO WR VHOHFW
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6. &OLFN RQ WKH PHQX 0HDVXUH DQG
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7. 5HSHDW VWHSV DQG WR JHW ADC 'LDJUDP
7KH YDOXH RI ADC ZLOO EH
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9. &OLFN RQ SRLQW B DQG PRYH LW DORQJ WKH FLUFXPIHUHQFH RI WKH FLUFOH DV VKRZQ LQ 'LDJUDP
,V WKH YDOXH RI ABC VWLOO WKH VDPH DV WKH YDOXH REWDLQHG LQ VWHS "

Discussion:
:KDW FDQ \RX FRQFOXGH DERXW WKH UHODWLRQVKLS EHWZHHQ DQJOHV DW WKH FHQWUH RI D FLUFOH DQG
DQJOHV DW WKH FLUFXPIHUHQFH RI D FLUFOH VXEWHQGHG E\ WKH VDPH DUF"

138

Chapter 6 Angles and Tangents of Circles

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

D ADC ×  ABC
E 7KH YDOXH RI ABC LV FRQVWDQW HYHQ WKRXJK SRLQW B LV PRYHG DORQJ WKH FLUFXPIHUHQFH

RI WKH FLUFOH

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QQ
Q

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O O [ O

[ [ [ [

P RP RP R

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Example 5

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D E
[
[ $ [ OƔ
O $
O Ɣ CHAPTER 6
Ɣ y
$ y y

E [ $ F [ $ ± $
Solution: y $
[ $
D [ ² ƒ y $
y ² $
[ $
y $

y $

MIND TEST 6.1c [

1. 7KH GLDJUDP RQ WKH ULJKW VKRZV D FLUFOH ZLWK FHQWUH O yO
'HWHUPLQH WKH YDOXH RI z

D [ ž
E y
F ]

139

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POQ ƒ FDOFXODWH WKH YDOXH RI Q[ ž
D [ O
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RS

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P ž
D PQR ž
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T
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CHAPTER 6 Are the central angles of a circle proportional to the arc length?
<RX KDYH OHDUQHG WKDW
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O [ R D WKH VL]HV RI WKH DQJOHV DUH HTXDO LI WKHLU DUF OHQJWKV DUH HTXDO
[
E WKH FKDQJH LQ VL]H RI DQ DQJOH LV SURSRUWLRQDO WR WKH FKDQJH LQ
WKH DUF OHQJWK

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140