Mathematics Core Topics for the IB Diploma SL 1 Worked Solutions, for use with Mathematics Analysis and Approaches SL & Mathematics Applications and Interpretation SL [1 ed.] 9781925489828

Citation preview

MATHEMATICS

Mathematics

Core Topics SL

for use with Mathematics: Analysis and Approaches SL Mathematics: Applications and Interpretation SL

for use with

IB Diploma Programme Charlotte Frost Bradley Stevenion

Joseph Smuall Michael Mampusti

dINIOM

HAESE

SNOLLNTOS

%

MATHEMATICS: Charlotte Frost Bradley Steventon Joseph Small Michael Mampusti

CORE TOPICS SL WORKED

SOLUTIONS

B.Sc. B.Ma.Sc. B.Ma.Sc. B.Ma.Adv.(Hons.)

Haese Mathematics 152 Richmond Road, Marleston, SA 5033, AUSTRALIA Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238 Email: [email protected] Web: www.haesemathematics.com National Library of Australia Card Number & ISBN

978-1-925489-82-8

© Haese & Harris Publications 2019

Published by Haese Mathematics.

152 Richmond Road, Marleston, First Edition

SA 5033, AUSTRALIA

2019

Artwork by Brian Houston, Charlotte Frost, and Yi-Tung Huang. Typeset in Australia by Deanne Gallasch. Typeset in Times Roman 10. This book is available on Snowflake only. The textbook has been developed independently of the International Baccalaureate Organization (IBO). The textbook is in no way connected with, or endorsed by, the IBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited. Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the rightful owner. Disclaimer: All the internet addresses (URLs) given in this book were valid at the time of publication. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

FOREWORD This book gives you fully worked solutions for every question in Exercises, Review Sets, Activities, and Investigations (which do not involve student experimentation) in each chapter of our textbook Mathematics: Core Topics SL. Correct answers can sometimes be obtained by different methods. In this book, where applicable, each worked solution is modelled on the worked example in the textbook. Be aware of the limitations of calculators and computer modelling packages. Understand that when your calculator gives an answer that is different from the answer you find in the book, you have not necessarily made a mistake, but the book may not be wrong either. We have a list of errata for our books on our website. Please contact us if you notice any errors in this book.

CF

e-mail: web:

[email protected] www.haesemathematics.com

BS

JS

MM

TABLE OF CONTENTS

Chapter 1

STRAIGHT LINES

Chapter 2

SETS AND VENN DIAGRAMS

62

Chapter 3

SURDS AND EXPONENTS

92

Chapter 4

EQUATIONS

121

Chapter 5

SEQUENCES AND SERIES

153

Chapter 6

MEASUREMENT

231

Chapter 7

RIGHT ANGLED TRIANGLE TRIGONOMETRY

283

Chapter 8

NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY

334

Chapter 9

POINTS IN SPACE

370

Chapter 10

PROBABILITY

395

Chapter 11

SAMPLING AND

Chapter 12

STATISTICS

DATA

453 473

Chapter 1

STRAIGHT LINES CEUOTS—— 1

a

y=3x+7

has gradient

m =3

b

y= —2x —5

¢

y=2x—1

has gradient m =32

d

y=11—4x

has gradient

m = —4

and y-intercept

¢ = 11.

e

y=—6—x

has gradient

m = —1

and y-intercept

¢ = —6.

has gradient

and y-intercept

m = —2

and y-intercept

¢ = —5.

and y-intercept ¢ = —1%.

f y=2— %2 has gradient m = —%

and y-intercept ¢ = 2.

g

y=

m = %

h

y= 2z6_ 3

h; z_

%x + %

has gradient

2o — & has gradient m =%

and y-intercept

a

The equation of the lineis

and y-intercept ¢ = 2.

y—1=3(x —4) Ly—1=3x—-12

y=3r—11

b

The equation of the line is

y —5 = —2(z — (—3)) y—5=-2(x+3) y—H=-2x-6 y=-2x—-1

¢

The equation of the line is

y — (—3)

i(

—4)

y+3=1x-1 y:%z—4

The equation of the line is

d

y — (=7) = —2(z — (-2))

y+7=-2(z+2)

4 ) y+7=-5x—3 —__2

25

y=—3T—% e

The equation of the line is

y =2z —9.

f

The equation of the line is

y = fix +4.

¢ = é

and y-intercept ¢ = —3.

i y= 3 _SSZ =2 — 22 has gradient m = —2 2

¢ = 7.

6

Chapter 1 (Straight lines)

Exercise 1A AY

16 12

0

b

¢

T

>

012 3 4

Yes, the variables are linearly related as the points all lie on a straight line.

The line passes through

(0, 5)

and

(1, 8),

8—5:3.

so the gradient is —

The y-intercept is 5.

d

e

L

The gradient is 3 and the y-intercept is 5, so the equation is y = 3z + 5.

When

2 =10,

5

.

The gradient of the road is As a percentage,

5

y=23(10)+5 =35

a

%

y-step

et

104

x 100% =~ 7.88%.

PondP:

Tnetmmi [ 0 1[2]3]1] Amount of

el

10120

30 fl

Pond Q:

[Emeamme o] 12 [3]1] Amount of

i

LA

5120355065

L AL

60

s

60

1 *

.

40

40

.

*

20

.

Pl ¢ (rqinuFes»)

0. 0

b

2

3

4

. t (minu}esl

OT 0

2

3

4

The points on the graph of pond Q all lie on a straight line, so pond Q is being filled at a constant rate.

¢

i

The line passes through

(0, 5) and

(1, 20), so the gradient is ot

-5

= 15. This means

that the amount of water increases by 15 L each minute. The A-intercept is 5. This means that the amount of water in the pond initially was 5 L.

ii

The gradient is 15 and the A-intercept is 5, so the equation is A = 15¢ + 5.

Chapter 1 (Straight lines)

ili

When

t=8,

Exercise 1A

7

A=15(8)+5 =125

There is 125 L of water in the pond after 8 minutes.

The line passes through

(0, 90)

and

- _ . 809

the gradient is

(1, 80),

so

100

£y (9)

This means that the balance in the account decreases by $10 each year. The y-intercept is 90. This means that the initial

balance was $90.

The gradient is —10 and the y-intercept is 90, so the equation is y = —10x + 90.

The account runs out of money when

20

y =0

%

~10z+90 =0

10z =90

5

3

z (years,

4

=9

The account will run out of money after 9 years.

The line passes through .

.

so the gradient is

(0, 46)

0 — 46

1820 -0

and

(1820, 0),

y

23

=——. 910

46

23 The gradient is — == and the y-intercept is 46, so 910 the equation is y = —%z

When

¢ =0,

+ 46.

1850

>

H =150+ 120(0) =150

The helicopter took off from a height of 150 m. The height of the helicopter above sea level increases by 120 m each minute after taking off.

When

¢ =2,

H =150+ 120(2) =390 The helicopter is 390 m above sea level after 2 minutes.

When the helicopter is 650 m above sea level,

H = 650

150 + 120t = 650 120t = 500 t = 500 _ 4% 120

The helicopter is 650 m above sea level after 4% minutes, or 4 minutes 10 seconds. y=—4x+6

dr+y=6

{adding 4z to both sides}

y=>bxr—3

—5z+y=-3 o br—y=3

{subtracting 5z from both sides} {multiplying both sides by —1}

8

Chapter 1 (Straight lines)




3 T +4

For

e

y=—2z+5:

e

the y-interceptis

o

the gradientis

¢ =5

m = —2 = 52

For y=32-1: e

e

the y-interceptis

.

¢ =

the gradient is m =

For

y=—4ux:

e

the y-interceptis

o

the gradientis

y:zz—l

3|

—2 2

f

Ay

—1 wles

d

¢ =0

m = —4 = 52

4 v

z

Exercise 1B

13

Chapter 1 (Straight lines) g

h

For

Exercise 1B

y=—x+4:

o

the y-interceptis

o

the gradientis m =—-1= =%

For

c=4

y=§x—3:

o

the y-interceptis

e

the gradientis

3

¢ = —3

m =

Ay

2

il o

14

¥

o % =gr—3 5

Y

i For y=-22-1: the y-interceptis

o

the gradientis

¢= —1

m = —

wlwt

e

2

Ay x

=5 3

y

‘;’x

1

3

2|

.

—5 Y

2

a

For

When

3z+2y=12:

=0,

2y=12 y==6

So, the y-intercept is 6. When

y =0,

3z=12 Mg

So, the z-intercept is 4.

b

For When

z+3y=6: =0,

Ay 3y=6 y=2

Ly

So, the y-intercept is 2. When y =0, z=6. So, the z-intercept is 6.

A

6

2

>

Chapter 1 (Straight lines) For When

2x — 5y = 10: =0,

—5y=10

Loy=-2

So, the y-intercept is —2. When

y =0,

2z=10 . x=5

So, the z-intercept is 5.

For When

4z —y=8: =0,

—y=38

coy=-8

-

So, the y-intercept is —8. When

y =0,

4z=38 e’

L

So, the z-intercept is 2.

For When

5z + 8y = 40: =0,

Ay

SRS

8y=40

y=>5

So, the y-intercept is 5. When

y =0,

-

+ 8y = 40

4

i

5z=40 .

v

=28

So, the z-intercept is 8.

For

When

3x —4y = —24:

z =0,

—4y=-24 y==6

So, the y-intercept is 6. When

y =0,

3z=-24 Lor=-8

So, the z-intercept is —8. For

When

2z 4 5y = 15:

=0,

YA

5y=15 L

y=3

So, the y-intercept is 3.

-

2z+ 5y =15




Exercise 1B

15

16

Chapter 1 (Straight lines) h

For

Exercise 1B

6z + 4y = —36:

When

z =0,

4y =

—36

y=-9

So, the y-intercept is —9. When

y

=0,

6z

36

.z

6

So, the z-intercept is —6. For

Tz 44y = 42:

When

z =0,

4y =42 21

y=%5

1 So, the y-intercept is 52

When

y =0,

-

7z=42 .

x=6

So, the z-intercept is 6.

3

Y=

b

ij

+ 2

When

has gradient

z =38,

m = 7%

and y-intercept

we have

y=-30)+2 —6+2 —4 v (8, —4) does lie on the line.

So, When

z =1,

we have

y=-3(1)+2 __3 =—-5+2 =2 5 %

So,

(1, 3)

When

does not lie on the line.

= = —2,

we have

NI~

y=-3(-2)+2 So,

4

a

For

When

(=2, %)

2z —3y

z =0,

v

does lie on the line.

=18:

—3y=18 .

y=—6

So, the y-intercept is —6. When

y =0,

2z=18 e

So, the z-intercept is 9.

¢ = 2. AY

Chapter 1 (Straight lines)

b

1

Substituting z =3 and y = —4 the LHS gives 2(3) — 3(—4) =6+12 =18 v

So,

¢

(3, —4)

into

ii

Substituting z =7 and y = —2 the LHS gives 2(7) — 3(—2) =144+6 =20 x

does lie on the line.

If (=3, c¢) lies on the line then

Exercise 1B

So,

(7, —2)

17

into

does not lie on the line.

2(—3) —3c =18 —6—3c=18 —3c =24 . e=-8

5

C=5t+10

a

When

dollars

t=4,

C=5(4)+10 =30

The cost of hiring the trailer for 4 hours is $30. b

40

LC(S)

4, 30)

20 0

6

=5t + 10

0

2

3

t (hours) > 4

a

x serves of nigiri at $4.50 each and y serves of sashimi at $9 each adds up to a total of $45.

b

When

4.52 + 9y = 45 z =4,

4.5(4)+ .

9y =45

¢

When

1849y =45

4.5z+9(1) =45 45

Oy =27

.

+9=45 4.5

=36

cLr=8

Ly=3 Hiroko bought 3 serves of sashimi.

y=1,

..

Hiroko bought 8 serves of nigiri.

18

Chapter 1 (Straight lines)

d

1

Exercise 1C

£y

5KNd.52 + 9y = 45

a

The midpoint M of [AB] is

(3—;—5, v; 7)

b

The gradient of [AB] is L5Ly 30s aiqs


0 - 8k > —56 k>-7

For exactly one real

iii

For no real solutions, A - 4(2)(-5)

=41

Since

A

> 0,

but 41 is not a square,

there are 2 distinct irrational roots.

150

Chapter 4 (Equations)

¢

Review set 4B

322+52+3=0 has

a=3,

A =0

b=5,

¢=3

— dac

=57 —4(3)(3) =-11

Since

9

A < 0,

there are no real roots.

222 -3x+m=0

has

a=2,

b=-3,

c=m

A =b? — dac

= (=3)* —4(2)(m) =9—-8m a

For a repeated root,

b

For two distinct real roots,

A=0 S

a

i 2z(z+4)=8x+k) When k=9, 2z(z +4) =8(z+9) 22% + 82 = 8z + 72 ZIQ

=72

woa? =36 L= r

A 0 S —=8m > =9

9-8m 0.

The equation The equation

2% = 4k x? =4k

has one real solution if & = 0. has no real solutions if k& < 0.

(—6, 24)

= = —6 or 6.

and

Chapter 4 (Equations)

1

a

Wegraph

y=222—-7

the same set of axes.

and

y =3z

on

b

We graph

Review set 4B

y =2(l—2)

and

151

y = —10

on the same set of axes.

[EXE]:Show coordinates Y1=2x2-7 oY

[EXE]:Show coordinates

Y1=x(1-x)

V2=3x

Y2=-10

g INTSECT X=-1.265564437

-°¥=-3,7966983311

The graphs intersect at

(—1.27, —3.80)

and (2.77, 8.30).

the solutions are

¢

Wegraph y =3

X=-2.70166

The graphs intersectat (—2.70, —10) and

(3.70, —10).

2 ~ —1.27 or 2.77.

y =2(z—2)+4(z—1)

..

the solutions are = ~ —2.70 or 3.70.

and

on the same set of axes.

[EXE]:Show coordinates

Y1=x(x-2)§4(x-1) o[y V2=3

The graphs intersect at

(1.83, 3).

the solutions are 12

a

(—3.83, 3)

= ~ —3.83 or 1.83.

2% — 152 = 227

23— 227 152 =0 q

and

Using technology,

Eeedlem) )Rl

3

2

c

1

e

-2

=

)]

[t Oelforn) (75,0,o0r -3 PN

b

23422 -62+7=0 . Using technology,

REPEAT]

8 Eibsis) ok X3 +bX2 +cX+d=0 S c

fornD) (d7c)Reall

aX3 +bX2 +cX+d=0 X1

X2|

r~4.93,0.814, or —1.74

X3L-1.743.

(REPEAT

(SOLVE]PHIEACLEAR)

¢

4x'— 1122 42 -82+6 3=0 :

Using technology,

7~ 2.39 or 0.449

0.8144

Hetiealors] (G)Feal aoX4+a1 X3+ - -+as=0 Yasls a2t e e o

C

IR

[SOLVE)PRYERCLEARICEDIT]

L

4.929142304

[Hatiegforn]) (d7c)Real

ao X4 +a1 X3+ Xl[mfifi X2L 0.4485.

- -+as=0

2.390541979

152 13

Chapter 4 (Equations) a

Wegraph y =2 same set of axes. [El

Review set 4B and

y =7

on the

b

We graph y=2%

the same set of axes. [EXE]:Show coordinates

[EXE]:Show coordinates

Y1=x"(3) ¥2=9-2(4x)

g

5

INTSECT

INTSECT X=1.845270174/

The graphs intersect at the solution is ¢ We B

graph

y=%—

x ~ 2.81. T+ 3.

Y I=((x2) 15)—(f (x+3)F

S¥=0

The z-intercepts are ~ —2.15 the solutions are

and 3.58.

z ~ —2.15

“Y¥=6.2831855661

The graphs intersect at (1.85, 6.28). the solution is = ~ 1.85.

(2.81, 7).

[EXE]:Show coordinates

X=-2.148183918

and y =9—2/

or 3.58.

on

Chapter 5 SEQUENCES AND

1

a

4,13, 22,31 A

b

AA

+9 +9 +9

2

23,5711,

45, 39, 33, 27 A

A

¢

A

2,6,18, A

—6 —6 —6

x3

A

x3

A

54

d

96, 48, 24, 12 A

x3

A

fe2y==2

b

us =11

¢

ujo =29

a

We start with 4 and add 3 each

¢

ug=us+3+3+3

time.

b

u

=4,

U4:13

=16+3+3+3 =25 u,=2n+5

up =2(1)+5

up =2(2)+5

= 5

6

ug =2(3)+5

=

a uy =3(1)—2

b us =3(5) -2

=1

=13

=13

€

u, =n-—10

So,

up=1-10 =-9

Using B,

us =2 — 10 Vv

u, =n?>—10

So,

u; =1%2-10 =-9

v

ug =42 — 10 =6 Using €,

Vv

So,

u; =1%-10 =-9

So, B is the correct formula. b

U0

=

202

=390

—

10

=-8

v

x

uy =22 - 10 =—6

Vv

us =52 — 10 =15

u, =n3—10

uUgy

=

3(27)

=179

—9,-6,—-1,6,15 Using A,

ug =2(4)+5

=11

u,=3n-2

a

=2

{the 10th prime number}

3 4,7, 10,13, 16, ....

b

A

13,17, 19, ....

ux=3

a

SERIES

v

uy =23 — 10 =-2

x

-2

154

7

Chapter 5 (Sequences and series)

a 8,16, 24, 32, ...

The sequence starts The next two terms 2,58 11, ... The sequence starts The next two terms

b

Exercise 5A

at 8 and each term is 8 more than the previous term. are 40 and 48. at 2 and each term is 3 more than the previous term. are 14 and 17.

¢

36, 31, 26, 21, .... The sequence starts at 36 and each term is 5 less than the previous term. The next two terms are 16 and 11.

d

96, 89, 82, 75, .... The sequence starts at 96 and each term is 7 less than the previous term. The next two terms are 68 and 61.

e 1,4, 16,64, ..

The sequence starts at 1 and each term is 4 times the previous term. The next two terms are 256 and 1024.

f

2,6,18, 54, ... The sequence starts at 2 and each term is 3 times the previous term. The next two terms are 162 and 486.

g

480, 240, 120, 60, .... The sequence starts at 480 and each term is half the previous term. The next two terms are 30 and 15.

h 243,81, 27,09, ..

The sequence starts at 243 and each term is one third of the previous term. The next two terms are 3 and 1.

i

a

50000, 10000, 2000, 400, .... The sequence starts at 50 000 and each term is one fifth of the previous term. The next two terms are 80 and 16. 1,4,916, ... Each term is the square of the term number.

b

1,8, 27,64, ... Each term is the cube of the term number.

¢

The next three terms are 25, 36, and 49.

2,6,12,20, ... Each term is n(n + 1)

The next three terms are 125, 216, and 343.

where n is the term number. The next three terms are 30, 42, and 56.

a

95,91, 87, 83, ... Each term is 4 less than the previous term, so the next two terms are 79 and 75.

b

5,20, 80, 320, .... Each term is 4 times the previous term, so the next two terms are 1280 and 5120.

¢

1,16, 81, 256, .... Each term is the fourth power of the term number, so the next two terms are

64 = 1296.

d

2,3,5711,.. This is the sequence of prime numbers, so the next two terms are 13 and 17.

5% = 625

and

Chapter 5 (Sequences and series)

Exercise 5B.1

155

e 2,4,7, 11, ... A A S 243 ¥4 The difference between terms increases by 1 each time, so the next two terms are and 16 + 6 = 22. f

9,810,

11+5 = 16

7,11, ....

Each odd numbered term is 1 more than the previous odd numbered term, and each even numbered term is 1 less than the previous even numbered term, so the next two terms are {2n}

The sequence

{2n — 3}

an

The sequence

{2n + 11}

O

The sequence

{3 —4n}

The sequence

{n? +2n}

The sequence

{2"}

The sequence

{6 x (%)”}

The sequence

{(—2)"}

The sequence

{15 — (—2)"}

T 0o

The sequence

-« ®

1

11+1=12.

W

10

and

T

7—1=6

a

715,23

begins

2, 4, 6, 8, 10

begins

begins begins begins

begins

—1, 1, 3,5, 7

begins

n =1, 2, 3, 4, 5, ....).

(letting

13, 15, 17, 19, 21

n=1,

(letting

—1, —5, —9, —13, —17 3,8, 15, 24, 35

2, 4, 8, 16, 32 begins

(letting

(letting

(letting

3, %, %, %, 1—36

n =1, 2, 3, 4, 5, ....).

(letting

n =1, 2, 3, 4, 5, ....).

n =1, 2, 3, 4, 5, ....).

n =1, 2, 3, 4, 5, ....). (letting

—2, 4, —8, 16, —32 begins

2, 3, 4, 5, ....).

n=1,

(letting

17, 11, 23, —1, 47

2, 3,4, 5, ....).

n =1, 2, 3, 4, 5, ....).

(letting

n =1, 2, 3, 4, 5, ....).

31,39, ...

15-7=8

The difference between successive terms is constant. . the sequence is arithmetic with u; =7 and d = 8.

23-15=38

31-23=38 39-31=38 b

¢

10, 14, 18, 20, 24, .... 14-10=4

The difference between successive terms is not constant.

18—-14=4 20—-18=2 24-20=4

..

the sequence is not arithmetic.

41, 35, 29, 23, 17, ... 35—-41=

-6

The difference between successive terms is constant. ..

29 —35=—6

the sequence is arithmetic with

u; =41

and

d = —6.

23-29=-6 17—-23=—-6 d

6,1, -6, —11, —16, .... 1-6=-5

—-6—-1=-7

—11—(-6)

= -5

16— (—11) = —5

The difference between successive terms is not constant.

.

the sequence is not arithmetic.

156

Chapter 5 (Sequences and series) a

Exercise 5B.1

b

5,9,13,17, 21, ...

—4, 3,10, 17, 24, ...

3—(-4)=7

9-5=4 13-9=4 17-13=4 21-17=4 (5%

d

:5,

d=4

U1:74,

u =23,

d=-5

—15— (—6) = —9 —24— (—15) = —9 33— (-24) = —9 =

—6,

d

=

—9

19,25, 31, 37, ... i

25—-19=6 31-25=6

U, =up + (n—1)d

=1946(n—1)

Sy

37—-31=6 up =19,

b

d=17

18 -23=-5 13-18=-5 8§—13=-5 3-8=-5

-6, —15, —24, -33, ....

uy

a

10-3=7 17-10=7 24-17=7

23,18, 13,8, 3, ...

Uy

=6n+13

d=6

101, 97, 93, 89, .... i

il w5 = 6(15) + 13 =103

U =u1 + (n—1)d

97-101=—4 93 —-97=—4 89 —-93=—4 up =101,

Coou, =101 —4(n—1) .

il w15 = 105 — 4(15) =145

up =105 —4n

d=—4

d

Rl

LU, =8+13(n—1)

Nl

ou, =130+ 6%

31, 36, 41, 46, .... i

36-31=5 41-36=5

U, =up + (n—1)d

U =31+5(n—1)

S

46 —41 =5 U1:31,

e

i w5 =13(15) + 63

ol=

ol

up =ui + (n—1)d

Il

U

Il

-
(12x 27k k=1

or

n

3 12(2)

k=1

1-k

931

Chapter 5 (Sequences and series)

Exercise 5F

185

d 23,4463, .. 43

gl h3

'373

373

..5523#1

3

6%

3

a7

i s= % (2x(3)7) =2+3+41+63

Consecutive terms have a common ratio of 2. the sequence is geometric with

and

u; = 2

+ 102

= 26%

r=3.

Up = ugr"!

wy = 2 x (%)7,,71

Now

S, =2+3+4%+62+..+ (2 x (g)"*l)

Si=%k=1 (2x o

e

11

k=1

@) o k=1X2(3) B

3

k-1

1

1,5, 9,5 iy i 2==2

Ly 4_Z 012

"173

Ly 3_Z 173 1

ji

=

i S5

and

,;12 =

=1+

Consecutive terms have a common ratio of %

1-k % + % + % + 11—6

= 11_5

u; =1

the sequence is geometric with

5

r=3.

Up = ugr™!

un =1 (3)"

i

Now

=105™

S,=1+3+3+5+...+2'7" S

=

f 1,827, 64, .. i

n

k=1

a € e

f

=

1—-k 2°7F

or

i

1

=y 2!

S, =12 +2% 4334+ 4% 1 ... +n® Sp=>

il

f:

kK

S5 =1+8+27+64+125 =225

3

Y 4k=4+8+12

k=l

—9q

4 Y (Bk—5)=—-2+1+4+47

i

{all terms are cubes}

=10

b d

7 > k(k+1)=2+6+12+20+ 30+ 42+ 56 k=1 =168 5

Y10

o

x 2871 =10+ 20 + 40 4+ 80 + 160

=310

6

> (k+1)=24+3+4+5+6+7

il

=27

5 > (11-2k)=9+7+5+3+1

k=t

=25

186 6

Chapter 5 (Sequences and series) u,=3n-1

Exercise S5F

20

+ug+

oup+us

=

... +up

Z(?)k*l)

k=1

24+5+8+11+14+17+20+ 23 +26 + 29 + 32 + 35 +41

F38

+ 50

+ 47

+ 44

+ 56 + 59

+ 53

=610

7

a

Y c=ctetet..tc=cn .

b

—/—/

—

k=1

n times

.

> car =car +cas+ ...+ cay = C((ll +ag

=cy a k=1 [4

| bk)

Z(ak

ik

((Ll

} b2)

t ((LQ

} bl)

.=t

} bn)

(an

= (a1 4 a2+ oo+ an) + (b1 + by + . + by) =Y ar+ Y, b k=1 k=1

8

a

b

k=1

n+n-1)+Mn—-2)+

or

2

2 k=m+1)+n+1)+n+1)4 k=1 =n(n+1)

"

— n(n+1)

2

Y (ak+b)=>ak+ k=1

k=1

=ay

k=1

> b

k=1

k+nb

= ‘m(n; b +nb But

F(n+1)+(n+1)

2

k=1

S

il

_ n(n+1)

ik

¢

4+..+(r-1+n

3

+

2

k=14

{using b}

Y (ak+b) =8n*+11n k=1

“"("2+ D | nb=8n?+11n 2

w-o—nb:&zz—fi—lln %n2+%n+nb=8n2

+11n

gnz + n =8n? +11n Comparing coefficients, we get

g =8

a=16

and

g +b=11

Lip=11 =3

+

+ap)

Chapter 5 (Sequences and series) 9

The sequence of positive odd integers is This is an arithmetic sequence with

Exercise 5F

1, 3, 5, 7, 9, ....

u; =1

and

d = 2.

U, =u1 + (n—1)d u, =14+2(n—1) Uy, =2n

—1

So, the sum of the first n positive odd integers can be represented by n

Sp=>(2k—-1)

L

or

Zi(Zk

1

5

Fon+(2n—3)+(2n—1)

(2n—1)+(2n—3)+(2n—5)+....

1)

o

3

2n

2n

2n

4+ ...

3

1

2n

2n

=nx2n = 2n?

i(?k— 1) =n?

k=1

n

So, S, =3 (2k—1) =n? k=1

10

o k=n(n+ 1)

Now

and

e k2=n(n+

1)(2n+1)

S (k+1)(k+2)= 3 (k*+3k+2)

k=1

k=1

= SR+ Y 3k+ 32 k=1

k=1

N n(n+1)(2n+1)

6 _ n(n+

)6(2n+1)

: n(n+1)(2n+1) 6

k=1

——3§:k+2n

h=1 +3x

n(n+1)

9n(n+1)

Lon

+12_n

6

_nn+1)2n+1)4+9n(n+1) +12n R e

_ n@2n? +3n4+1) +n(9n+9) +12n = _e n@n?+3n+1+9n+9+12)

_N n(2n? 4+ 12n +22) _ 2n(n? + 6n + 11) 6

n(n? + 6n 4 11)

3

6

187

188

Chapter 5 (Sequences and series)

When n=10, but .

when

n =10,

Exercise 5G

10

5 (k+1)(k+2) =6+ 12 + 20 + 30 + 42 + 56 + 72 + 90 + 110 + 132

k=1

=570

n

2

> (k+1)(k+2)= w k=1

20

10(102 4 6(10) + 11)

k=1

(k+1)(k+2) = e

_10(171) 3 =570

1T

a

{from above} )

v

2+6+10+14+18+22+26+30=128

b

S,.= g (ur +u,)


(2K+5)

n = 10.

K=1

2

5

[logab] [FMin[F2(Max|

Sio = 12(7 + 25)

10

=5x32

];1(2/{ +5)=160

=160

b

fj(k

50) = (—49) +

k=1

(—48) + (—47)

This series is arithmetic with and n = 15. .

n

Usipghn ss (Uil

+ ... +

u; = —49,

(=35)

d =1,

),

R

Si5 = 22(—49 + (—35)) =1

-

20

k+3 Z=2+%+3+.‘..+§

k=1

M This

i series

and

n = 20.

Using

e N R arithmetic

is

with

d=

u; =2,

0 HtRedlom] @Rl

)3,

>k=1 (& 20

S, = %(ul +up),

Sn=2(2+3)

[FMin[FMax| 2( [logab] 20

Z(—k+3):135 =\ 2

=135 uy=5 Sn

n=7, =

g(ul

u,=>53 +un)

S7=2(5+53)

=1 x58

K+3

O

—10x X2

6

Vv

> (k—50)=-630 k=1

x(-84)

=—630
0,

n=-19

so

or

18

n=18

So, the bricklayer built 18 layers.

9

a

The number of laps Vicki swims each day can be expressed as an arithmetic sequence

20, 22, 24, 26, ... So,

uy =20

and

d=2.

U, =up + (n—1)d

up =204+2(n—1) U, =2n + 18

il =38

Vicki swims 38 laps on the tenth day. b

ugzo =2(30)+ 18 =78 Vicki swims 78 laps on the final day.

Sp = g(ul + up)

S30 = 22(20 + 78) =15x%x98 = 1470 Vicki swims 1470 laps in total. 10

a

The amount of money the woman deposits each birthday can be expressed as an arithmetic sequence 100, 125, 150, .... So,

u; =100

and

d = 25.

U, =up + (n—1)d

Uy, = 100+ 25(n — 1) L

Up =25+

75

uys = 25(15) + 75 =450 The woman will deposit $450 into her son’s account on his 15th birthday.

b

S, = %(ul +up) S15 = £2(100 + 450)

=10 % 550 = 4125 The woman

will have deposited $4125

over the 15 years.

192 11

Chapter 5 (Sequences and series)

Exercise 5G

The total number of seats in n rows can be expressed as an arithmetic series: 22+23+24+

Row

... 4u, 1 has 22 seats, so u; = 22.

Row 2 has 23 seats, so

d = 1.

S, = g (2u1 + (n —1)d) :g(2> 319.9.

up(r™ —1) r—1

B

160((%)" - 1) % —

= —320((0.5)" — 1) To find n such that

with

S,, > 319.9,

Y; = —320 x (0.5"X~1).

we use a table of values

FeiRedlo] Gk

Y1=-320x (.57 (x)-1) ¥1 ‘

10

319.68

11

319.84

12 EFERER

13 319.96 DA IR

S12 = 319.92,

so

n=12.

12 terms are needed for the sum of the terms to exceed 319.9.

319.921875

(EDIT ) GPH-CON)GPH-PLT

Chapter 5 (Sequences and series) 14

a

Option A:

Exercise SH

205

First year salary = $40 000 Second year salary = $40 000 + 5% x $40 000 = $42 000

Third year salary = $42000 + 5% x $42000 = $44 100 Total earned in three years = $40 000 + $42 000 + $44 100 = $126 100 Option B:

First year salary = $60 000 Second year salary = $60 000 + $1000 = $61 000 Third year salary = $61 000 + $1000 = $62 000

Total earned in three years = $60 000 + $61 000 + $62 000 = $183 000 So, over three years Felicity would earn more under Option B.

b

i

Let A, be the amount of money earned under Option A in the nth year. A, forms a geometric sequence with A; = 40000 and r = 1.05. A, = 40000 x (1.05)""1

i

Let B,, be the amount of money earned under Option B in the nth year. B,, forms an arithmetic sequence with B; = 60000 and d = 1000. B,, = 60000 + 1000(n — 1) = 59000 + 1000n

If A, =

B,,

40000 x (1.05)""! = 59000 + 10007,

We graph A, = 40000 x (1.05)"~!

B, = 59000 + 1000n points of intersection. Since

n >0,

n~

[EXE]:Show coordinates )~( (Z=T 1=40000X

and

on the same set of axes and find their

13.07.

the money earned under Option A will exceed that of Option B after approximately 13.1 years.

n_ 1

:

Sn = £l

for a geometric

r—1

)

= 800000((1.05)" — 1) dollars

Initially Option B is better than Option A, so

Tgp > T4

i

g

;(zul + (n—1)d)

for small values of n.

for an arithmetic

series

series

40000 x ((1.05)™ — 1) Ty=— ) 1.05 — 1 40000 x ((1.05)" — 1) = 0.05 i

g

Tp = %(2(60 000) + 1000(n — 1)) = 600001 + 500n(n — 1)

= 60000n + 50012 — 500n, = 50012 + 59 500 dollars total income

($)

graph 1

graph 2

graph 1 represents 74, graph 2 represents 1.

n (years)

206

Chapter 5 (Sequences and series) ii

Exercise 51

The point P is where T4 meets Tz, which is when

800000((1.05)™ — 1) = 50012 + 59 500n. We graph T4 = 800000(1.05™ — 1) and Ty _= 500n* 2 + 59 500n on the same set of axes and find

[EXE]:Show coordinates

= D

O e

their points of intersection.

Since n >0, P A (22.3, 1580000). iii

1

a

2288661

Option B provides the greater total income for

0 < n < 22 years.

2 _ 3 3 3 03=55+10+1 ™05 + -

3

uy _

(100) 3

U1

3

=

(1500) | 3 Nl |

(15)

(1o5)

10

the series is geometric with

u; = 753

r = 3.1

and

Since we are adding all the terms, it is an infinite geometric series.

b

We need to show that Now

So,let Since

n — oo,

then

0.3 = 1. 3_23 3 3 0.3—E--m+m+...

25+ 25+ Sp==5+ S =

__

!

1—r

3

T

1-(1p)

-1

i 0.3=1 2

a

04=0.444444....

4

_ 4

4

=10+ 100 + 1000 T

is an infinite geometric series with uy =+ 4

and

10

g=1 S = =4

10

b

0.16 =0.161616....

_ 16 | 16% piloil

102 104 106 . e . ) i is an infinite geometric series with

16 up =_ 155

u

bcn i

16 __To0

T

i1

4

o

1=2,35.

and

o 8=

i4

-1

—

04=2

r ==.1

as required

drat

s

—

99

016 =45

100

Chapter 5 (Sequences and series) ¢

Exercise 51

207

0.312=0.312312312.... 312

312

=1t ie T

312

T

is an infinite geometric series with u; = 22 1000

3 0312 = 1% 3

4L

Checking Exercise SH question 9 d:

a

S =

18+ 12+ 8+ % + ... is an infinite geometric series with u; = 18 and r=

18

=

1—

a

) k=1

u1l

1—r

=

189

1-

T

(*3)

=14.175

=54 3

isan infinite and = 18.9

—%.

o 32

v

189—-6.3+2.1—0.7+.... u; geometric series with 8=

1—r

§=

5

b

=1

T =

%

_u

=

B

= 2+ Z+ & +...

isan infinite geometric series with u; =3

and r= 1.

5=11—7r _

s

3

I -1

1

=1

b

k

.

> 6(—2)"=6-6x(2)+6x(£)?—... k=0 up =6 and r=—%

§=11-7r or

6

I SRV

1-(-3) 2 (: 4?) =230

isan infinite geometric series with

208 6

Chapter 5 (Sequences and series)

Exercise 51

Let the terms of the geometric series be

w1,

uy + urr 4+ ugr? =19

and

(1 4r ) =19

"

W=y

Equating (1) and (2),

uyr,

wuir?,

S=

... Ul

=27

1-r

=271 —71)

e (1)

%

... (2)

=27(1—r)

B=(1-n1+r+r?)

B=ttrtri—r—r2—1® Vo143

P=t

-3

Substituting 7 = 3 into (2) gives uw; =27(1 - ) =9 the first term is 9 and the common ratio is %

7

Let the terms of the geometric series be

uyr =8

Loy = %

and

uy,

uir?,

..

2t S:liT:IO

=10-10r

o

- (D

Equating (1) and (2),

uir,

.. (2)

53 =10-10r T

— 5012 8=50r . 50r? —50r +8=0 o

2(25r% — 25r +4) =0 2(5r —1)(5r —4) =0 L

Using 2), either 8

z,

x—2,

a

r=1%

or %

if r=2%,

uw;=10—10(%)=38

if r=2,

u=10—10(3)=2 r:%

u; =8, 217,

or

wu =2,

r:%.

...

=2

The sequence is geometric,

x

LIk

r—2

so(r—2%=x22-T) oa?—dr+4=22%—-Tzx

o2t

—3z—-4=0

(z+1)(z—4)=0 .

b

x=—1lor4d

When z = —1, the sequenceis —1, —3, —9, .... with is divergent so it does not converge.

When

o = 4,

the sequence is 4,2, 1, ...

with

converges. The limiting sum in this case is S =

» =3,

r =1, 4

=8. 1 1—3

and since

andsince

|3|> 1, the series

|$| + ... as required

ground

b

The total time of motion can be written as So,

S, = M

—1,

-

_201-09")

Sn= 1-09 _2(1-09")

where

w3 =2,

[2 + 2(0.9) + 2(0.9)2 +2(0.9)3 +...] — 1 r=0.9

1

Sp = =

1

S, =20(1-0.9") -1 Sp,=20—20x%x09"—1 Sn =19 —20(0.9)" ¢

For the ball to come to rest, n must approach infinity. As

n—o00,

0.9" -0

and so

Sp — 19

(from below)

20 x 0.9" — 0

also.

So, it takes 19 seconds for the ball to come to rest.

Total distance travelled

=h——%h
53%

We try the two values on either side of n = 532,

usz =68 —5(53)

and

and

= -202

usg = —202

is the first term which is less than —200.

3,12, 48, 192, ... 12 e 3

n =53

w54 =68 — 5(54)

= —-197

So,

which are

L

12

48

Consecutive terms have a common ratio of 4. the sequence is geometric with u; =3 and

r =4.

up = ugr”!

L oup =3 x4t

©oug =3 x 48

=196 608 uy = 31

us = —17

cooup +6d=31

.

ouy+14d =17

(1)

{using

u, =u; + (n —1)d}

... (2)

We now solve (1) and (2) simultaneously:

—u; — 6d=-31 up + 14d = —17

.

{multiplying both sides of (1) by —1}

{adding the equations}

8d=—48 s d=-6

ug +6(—6) =31

So in (1),

. uy —36=31 oup

Now

=67

wu, =u; + (n—1)d

up =67 —6(n—1) Uy, =T73—6n

Uzq

=

73—

=-131

6(34)

Check:

6(7) ur =7—3 =31

v

w5 = 73 — 6(15)

=-17

v

n = 54:

Chapter 5 (Sequences and series)

5

a 24,231 221, .. 23%—24=-3

221 -231=-3

the sequence is arithmetic with

Now

Review set 5B

u; =24

and

d = 7%.

u, =u; + (n—1)d

—36=24—3(n—1) —60=—3n+3 3

243

n= g n =381

So, —36 is the 81st term of the sequence.

b ougs =24+34x (—2)


0}

Each vertical support post is 100 cm in height.

b

Volume of tent = area of triangular end x length

= £ x 150 x 100 x 200 cm®

— 1500000 cm®

= 1500000 =+ 100° m® —

¢

piaL

Total area of canvas

iy

3

75cm

= area of two triangular ends + area of two rectangular sides + area of rectangular base

=2 x (4 %150 x 100) + 2 x (200 x 125) + (200 x 150) cm? = 95000 cm?

255

256

Chapter 6 (Measurement)

Exercise 6C.1

"ah

L=

cm

5cm

3cm

e

B

s

Let the sides of the cube be & cm.

V =34.01 cm®

©oxxaxx=34.01

2= 3401

g =l

=

h=58 267

v/34.01

e

The side length is approximately 3.24 cm.

The height is approximately 2.67 cm. ¢

rcm

xcm

xem

Let the height of the rectangular prism be

h em.

e

Let the radius be r cm.

V =43.75 cm® o

xr?x4.6=43.75 2 _

4375

4.6cm

T X 4.6

r=

:i'fG

{as

r >0}

~ 1.74

The radius is approximately 1.74 cm.

6cm 8cm

Let the height of the trapezoidal cross-section be i cm.

h?+1.5% =62

{Pythagoras}

h=4/62—1.52

=1/33.75

{as h >0}

Volume of gold bar = area of cross-section x length

480 = (#) x \/33.75 x length ",

length =

4= =

B335

~12.7 The length of the gold bar is approximately 12.7 cm.

e

Chapter 6 (Measurement)

a

Investigation

The volume of tapered solids

257

V), = length x width x height =10 x 10 x 15 cm®

= 1500 cm®

i

There are n prisms with equal thickness and total height 15 cm. each prism has height

ii

From

LB cm. n

the diagram alongside, the distance from ..

15k

the apex to the base of the kth prism is 2.

cm.

Bk =2 n

=& =

Using similar triangles,

T T

i,

15cm

15

10

o

iii

(E%E>

_ k =

10k n

The volume of the kth prism = area of base x height 15

=T X T X —

n

10k _

10k _

n

n

15 n

~== 1500k> Yok” 3 i

2

o

{using a}

From the 6th row of the spreadsheet, we see that Volume of solid (V) is 1500 cm?®, as calculated in a.

258

Chapter 6 (Measurement) il

Investigation

The volume of tapered solids

Inspecting cell B6, we see that the formula for V), is B3 x B4 x B5, which represents base length x base width x height. Inspecting columns F and H, and checking what each cell reference represents, we see the following logic:

If k2

10

|(—] n

x—

n

2

X 250 x n

Vek? 3 =——cm 3 From the 5th row of the spreadsheet, we see that

Volume of solid (V)

is approximately

785.398 cm?, as calculated in a. Setting

now

m =1

in cell B9, we see in row 12 that the approximate volume of the cone is

785.398 cm® ~ V. Approximate volume (V cm?®)

10 100 1000 10000 100000 iv.

302.378 265.739 262.192 261.839 261.803

The approximate volume appears to approach

2507

==

~ 261.799,

so we expect this to be

the actual volume of the cylinder. This is % of the volume of the corresponding solid with uniform cross-section.

260

Chapter 6 (Measurement)

Exercise 6C.2

2.8cm

V = % (area of base x height) = %(length x width x height)

V= dmd — 4 %7 x (48)° em?®

= 2(2.8 x 1.7 x 2.9) cm® ~ 4.60 em®

~ 463 cm® d


0}

Let the height of the pyramid be & cm.

B + (4V/2)? = 122

h? 432 =144

{Pythagoras}

B? =112 h=+112

Now

B {as

h >0}

V = L(area of base x height)

=18 x8xV112) ecm® 3 ~ 226 cm®

.

4y/2 cm

i

Chapter 6 (Measurement)

2

a

V = volume of rectangular prism

Exercise 6C.2

4+ volume of hemisphere

= length x width x height + 1 x 477?

261

24m

=46x46x12 + 2 x 1 x (%)’

~ 29000 m* bV

= volume of rectangular prism + volume of pyramid

= length x width x height + %(area of base x height)

=6x6x12

=480 m*

+ 2(6x6x4)m’

12m

6m

¢V

= volume of hemisphere + volume of cone

=1 x #mr® + L(area of base x height) =2x7x5 ~ 497 cm®

a

x9cm’

Volume of cylinder = mr?h

o =

i

=

Wl —

W=

Volume of each conical end = % (area of base x height)

W=

() 1 2

X

3

+ & xmx5

2

~0.848 m® Total volume of tanker = volume of cylinder + volume of 2 conical ends

~10.179 + 2 x 0.848 m* ~11.875 m* ~11.9 m*

So, about 11.9 m® of concrete can be held in the tanker.

b

If the ends were hemispheres, the end sections would be as long as the radius of the hemisphere.

. total length=4m+2x0.9m =58m

262

Chapter 6 (Measurement) ¢

Exercise 6C.2

The two hemispherical ends combine to make one sphere with radius 0.9 m.

Volume of sphere = 477

=3 x7x(09)°m’

~ 3.054 m* Volume of 2 conical ends ~ 2 x 0.848 m*

~ 1.696 m®

Difference in volume of ends = volume of sphere — volume of 2 conical ends

~ 3.054 — 1.696 m® ~1.36 m®

So, the tanker could fit about 1.36 m® more concrete if the ends were hemispheres instead of

cones. d

Surface area of cylindrical part of tanker = 27rh

=2x7mx09x4m? ~ 22.62 m?

i

Let the slant height of the cone be s m.

s2=124(0.9)2

s=1/1240.9? 0.9m

{Pythagoras}

{as

s> 0}

=+v1.81m

Surface area of 2 conical ends = 27rs

=2x7x0.9x 1.8l m? ~ 7.61 m? Total surface area of tanker

= surface area of cylindrical part + surface area of 2 conical ends

A 22.62 + 7.61 m? ~ 30.2 m?

So, the surface area of the tanker with conical ends is about 30 m?.

ii

The two hemispherical ends combine to make one sphere with radius 0.9 m. Surface area of sphere = 477>

=4 x 7 x (0.9)? m?

~10.18 m? Total surface area of tanker

= surface area of cylindrical part + surface area of sphere

~ 22.62 + 10.18 m? ~ 32.8 m?

So, the surface area of the tanker with hemispherical ends is about 33 m?. e

The hemispherical ends allow a greater volume to be carried by the tanker. They also allow the length of the vehicle to be shorter. However they have a greater surface area which means they require more steel to manufacture, so they would cost more to produce. This would be a one-off cost however, so for the permanent advantages, the hemispherical design is better.

Chapter 6 (Measurement)

Exercise 6C.2

b

Let the height of the cone be h cm.

V =706 cm®

1 x 7% (12.3)% x h =706 50.43 x 7 x h = 706 h

Let the radius be » m.

V =73.62m?

706

3 xmxr®=73.62

50.43 x

3

~ 4.46

73.62

The height is approximately 4.46 cm.

The radius is approximately 2.60 m. ¢

Let the radius be r cm.

V =203.9 cm®

I xmxr?x6.2=2039 b

o d

.

203.9

3 XX

o A

6.2

203.9 3 XX

6.2

{as

r >0}

~ 5.60


;2

Density

(3% 6—2) cm®

> f 6cm

_{

=4z x 16 cm® ~ 67.0 cm® wh?

ii

V=T(3r—h) T x 32 3

(3x5-3)md

h)

— w = |
0}

=100v2

-,

.

fU

Density of Uranus = I

volume of Uranus

__ 8.681 x 10%° kg ™ 6.83 x 1022 m? ~ 1270 kg m?

100/2m

O

Chapter 6 (Measurement)

1

2

3

a

800 mL = 800 cm®

Exercise 6D

b

12 L = (12 x 1000) cm® = 12000 cm®

¢ 46kL =4.6m?

d

3200 mL = 3200 cm?

a 84cm®=84mL

b

1800 cm® = (1800 = 1000) L =18L

¢ 1.8m?=18kL

d

7154 m® = 7154 kL

3.85x 10" L = 38500 L =38.5 kL

=385 m?

a

V =length x width x height

=34x1.8x21m?

_ Da

21m

=12.852 m?

The tank’s capacity is 12.852 kL.

b V=mh _

=7

5.7\

X

?

X

2.4

m

3

1.8m

3dm

—

=

_f

24m

~61.2 m®

i

The tank’s capacity is approximately 61.2 kL.

V = 1(area of base x heigh)

o =

@

o=

w

e

8

e




x 4.5 m®

2

~ 12200 cm® Approximately 12200 cm® of soup fits in the pot. b

[

57m

‘T

2

—%xwx(m)

‘

Capacity ~ 12200 mL

~ (12200 +1000) L ~12.2L

Approximately 12.2 L of soup fits in the pot.

4.5m

w0 > 8 B

= %(wr2 x h)

@= @

¢

X

L

267

268

Chapter 6 (Measurement)

Exercise 6D

6

dam wall end of

catchment

[25m

area

.

160m

area of trapezium A =

25 4 160

;%r0m

'

5 (

—L

i

(a ;_ b)

=15725 .

170m—>Lf250m

a+b

area of trapezium B = ( 5

) x 170 m?

m?

)

(80 + 160) % 9250 m?

= 30000 m? Total surface area of the reservoir = 15 725 + 30000 m?

= 45725 m?

V = area of cross-section x depth

= 45725 x 13 m* = 594425 m®

The capacity of the reservoir is 594 425 kL.

b

T

T

15em

4.5cm

4.1m

l

&

area of base = 1.2 m?

V =nr?h =7

b

~

.

x (45)2 x 15 om®

V = area of base x height

s

=12x41m s

The capacity of each tin is

=

approximately 954 mL.

¢

Number of tins to be filled from one vat =

Q ~

capacity of vat capacity of one tin

(4.92 x 1000 x 1000) mL 954 mL

4920000 954

=~ 5155.8

So, 5155 tins could be filled from one vat.

Value of one vat of jam = number of tins x cost per tin = 5155 x $3.50

= $18042.50

-

.

The capacity of the mixing vat is 4.92 kL.

4.92KL ~ 954 mL

d

.

3

Chapter 6 (Measurement)

Exercise 6D

269

120 mm

160 mm

160 mm 120 mm

External surface area = 2 x (120 x 88) + 2 x (160 x 88) + 2 x (160 x 120) mm?

= 87680 mm?

b

It is useful to specify the “external” surface area when talking about a container as the external surface area may be different from the internal surface area.

¢

i

The walls of the box are 4 mm thick. the internal length, width, and height of the box are 160 — 2 x 4 =152 mm, and 88 —2 x 4 =80 mm

120 — 2 x 4 = 112 mm, respectively.

internal volume of box = internal length x internal width x internal height

=112 x 152 x 80 mm® = 1361920 mm®

The box can hold 1361920 mm?® of jewellery.

i Capacity of box = (1361920 =+ 10°) mL

{1 cm® = 10° mm®}

=1361.92 mL

iii

Volume of wood used to make the box = total volume of box — internal volume of box = external length x external width x external height — 1361920 mm?® =120

x 160 x 88 — 1361920

= 1689600 — 1361920 mm® = 327680 mm® 9

mm?

10kL = 10m?

Volume of pond = area of base x depth =ar?h

10 =7 x (24)? x h 10

™ x (2.4)2 ~ 0.553 The pond is approximately 0.553 m

(or =~ 55.3 cm)

deep.

{from ¢ i}

270 10

Chapter 6 (Measurement) a

Exercise 6D

The area of the roof is in m?, so we convert 12 mm to metres.

12 mm = (12 +1000) m = 0.012 m The volume of water which fell on the roof = area of roof x depth

=110 x 0.012 m* =1.32m?

b

1.32m® =1.32 kL,

so 1.32 kL of water entered the tank.

¢

The volume added to the tank = area of base x height

=7x22x hm® =47 x hm?®

The volume added to the tank must equal the volume which falls on the roof, so 47 x h=1.32 1.32

="

h= o

.

{dividing both sides by 47}

7T

h~0.105

m

The water level rises by about 10.5 cm.

V = 7r?h

Original tin:

11

=7

7.2\ 2

X (7>

x 15 cm

3

original tin

_9m 5

New tin:

e

(1_0>

2

2

15cm

= %

cm®

xh =22 5

972

=

5 x 52 =7.776 cm

The height of the new tin must be about 7.8 cm.

12

a

V = (arca of base x height) 8.6 2 : =%X7r>

volume of silo

Number of truck loads = ——————

l

volume of container

~

1256.64

l«~4m

736

~ 34.9

So, 35 truck loads are needed to fill the silo.

ACTIVITY 2 1

a

Volume = length x width x height =2rXxT XYy s

b

yem

V=21

The container must hold exactly 1 litre of fluid.

1L = 1000 cm®

2zem

s

222y = 1000

¢

2

22%y=1000 z?y = 500 500 e

Surface area = 2 X (area of longer rectangular ends) + 2 x (area of shorter rectangular ends) -+ area of bottom

=2x

2z xy)+2x

= day + 2xy + 22°

.

A=22"+6xy

(zxy)+2zxw

272

Chapter 6 (Measurement)

3

L4

A

B

C

1

xvalues

yvalues

A values

2

1

500

3002

3

2

125

1508

4

3

5

4

31.25

782

6

5

20

650

7

6

13.888889

572

8

7

10.204082

526.5714286

9

8

7.8125

503

10

9

6.1728395

495.3333333

11

10

55.555556

Review set 6A

1018

5

500

The smallest value of A is ~ 495.33, when When

z = 9.

500

z =9,

6.17cm

y=9—2’\'6.17

The dimensions of the box that your boss desires are shown alongside.

1

a

Arclength= L

X 27r

360 105 360

10 cm

X 21 x 10 cm

~ 18.3 cm

b

¢

Perimeter = 2r + arc length ~2x10+18.3 cm ~ 38.3 cm 0

Area = —

X 77°

360

2 _= 105 555 X ™ x 10 ~ 91.6 cm? 2

0

Area = —

360

x 712

2Ur = 20wy 360

™

2

=30

24rw

rem

360 X

r =108

{as

r >0}

~ 10.4

The radius of the sector is approximately 10.4 cm.

9cem

18cm

Chapter 6 (Measurement) a

hollow top

-

b

and bottom

Review set 6A

273

.

"

10cm

l

Surface area = 4772

=4 x 7 x (5.2)? cm?

Surface area = 27rh =2x7x6x 10 cm?

~ 339.8 cm?

~ 377.0 cm? ¢

Surface area = 7rs + 772

=7 x4x12+7 x 4% cm? ~201.1 cm?

mplN 2.5m

3m

5m

Surface area of shed = area of 2 rectangular ends + area of 4 rectangular sides + area of 2 triangular ends

=2x(3x25)+4x

=7lm? b

(5x25)+2x (3 x3x2)m?

Since the shed is to be painted with two coats of zinc-alum, we need enough zinc-alum to cover

an area of 2 x 71 = 142 m%. The zinc-alum covers 5 m? per litre.

Number of litres of zinc-alum needed =

total area to be painted

-

area covered per litre

142 m? 5m2

L1

=284L

Since the zinc-alum must be purchased in whole litres, we need to purchase 29 L of zinc-alum. Total cost of the zinc-alum = number of litres to be purchased x cost per litre

=29 L x $8.25/L = $239.25

274 5

Chapter 6 (Measurement)

Review set 6A

Total area of netting

=6 x (27rh + 7r?)

2

:6>

3.2m

V = area of end X length

= area of end X height

=1x24x13x32m

f85.3x100m

= 4.992 m®

= 853 om?

~ 4.99 m® ¢V

= volume of external cylinder — volume of internal cylinder =7 R*h

=X

— 7r2h

(E) 2

2

x1.2 — 7mx (%> 2

2

x 1.2 m?

~ 0.452 m®

7

Volume of cone = % (area of base x height)

=1 3 x7x(1.6)*x1.2m?

~3.22 m®

1.2m

‘

Tom has had approximately 3.22 m* of sand delivered. 8

245 L = (245 x 1000) mL = 245000 mL c

total amount of molten iron

Number of spikes made = ——8M——————

amount of iron in each spike

~

245000 mL 15mL

~ 16333.3 So, 16 333 spikes can be made.

9

Volume of sphere= 477° =3 x7x27 cm®

~ 82400 cm® So, the beach ball has volume of approximately 82 400 cm?.

27cm

Chapter 6 (Measurement) 10

a

65L = (65 x 1000) mL

b

65 L = (65+ 1000) kL

= 65000 mL

= 0.065 kL

65000 cm? of petrol is required to

.

fill the tank. 11

a

Review set 6A

0.065 m* of petrol is required to fill the tank.

V = length x width x height

=122x86 x 7 cm® =734.44 cm®

7 cm

The capacity is 734.44 mL.

8.6 cm 12.2 cm

b

V = area of end X length

1 2 = 5 x mr” x length

60 cm 1.4m

=1xmx ()% 140 em®

{14 m = 140 em}

~ 198000 cm® The capacity is approximately 198 000 mL or 198 L. 12

The dimensions of the roof are in m, so we convert 15.4 mm 15.4 mm

= (15.4 +1000)

to metres.

m = 0.0154 m

The volume of water collected by the roof = area of roof x depth

=12 x 5.5 x 0.0154 m* =1.0164 m*

The volume added to the tank = area of base x height

=7 X (4—35)

2

x h m?

.

= 4.62257 x h m® The volume added to the tank must equal the volume which falls on the roof, so

4.6225m x h = 1.0164

10164

T 4.62257 h =~ 0.0684 m So, the level in the tank rises by about 68.4 mm. 13

a

Height of cone = total height of silo — height of cylinder — height of hemisphere

—33-18-1%n 2

_T

1.8m

33m

=0.7m

l

=70 cm Ll.BmJ

275

276

Chapter 6 (Measurement) b

Review set 6A

Let the slant height be s m.

0.8m

52 =(0.7)% + (0.8)*

{Pythagoras}

s =1/(07)2+(0.8)2

sm

~ 1.06

{as

s> 0}

So, the slant height of the cone is approximately 1.06 m.

¢

Surface area of hemisphere = %(surface area of sphere)

= $(4nr?)

=3

x7x(0.8)%) m?

~4.02 m? Surface area of cylinder = 27rh

=2x7x0.8x 1.8 m?

~ 9.05 m?

Surface area of cone = 7rs ~7 x 0.8 x 1.06

~ 2.67 m?

..

total surface area ~ 4.02 + 9.05 + 2.67 m>

~15.7 m?

So, approximately 15.7 m? of steel is used to make the feed silo. d

Volume of hemisphere = %(%777‘3)

=4x4x7x(08)°m?

~1.07 m? Volume of cylinder = 772h

=7x(08)?%*x1.8m3 ~3.62 m*

Volume of cone = & (7r°h) =1 x7x(08)?x0.7m?

~ 0.469 m*

volume of silo = volume of hemisphere + volume of cylinder 4 volume of cone

~ 1.07+ 3.62 + 0.47 m® ~5.16 m* ~52m

So, the silo can hold about 5.2 m® of grain. e

52m®=52kL

So, the capacity of the silo is about 5.2 kL.

Chapter 6 (Measurement)

G40 1

a

Review set 6B

) Arc length = e

360

0

6.18 = = 0=

x 27

x 27 X 4.62

6.18 cm

4.62 cm

6.18 x 360 27 X 4.62

0° ~ 76.6° b

a

0 2 Area:3—>0}

3.60m

~ 3.12

f

T

hem

N

1.8cm

Surface area = area of two triangular ends + area of three rectangular sides

~2x 5 x3.6x312 + 3x6.8x 3.6 cm’ ~ 84.7 cm?

b

74mm

@

Surface area = 27rh + 277>

=27x (%) x 74 + 27 x (2£)” mm? ~ 7110 mm?

< 2.0m

2.2m

1.3m

Let the height of the triangular side with base 2.2 m be d m.

d?+1.12 =2.0

L d=+v202-112

{Pythagoras}

{as d>0}

T

2.0m

dm [1

~ 1.67

Let the height of the triangular side with base 1.3 m be h m.

h* 2 +0.65 2 _=2.0 902

{Pythagoras}

h=1+/2.02-0.652

11m

A

S0

.

{as h >0}

l 0

~ 1.89

l

0.65m

Surface area of pyramid = area of base + area of two triangular sides with base 2.2 m + area of two triangular sides with base 1.3 m

~22x 13 + 2x4x22x167 ~ 8.99 m?

+ 2x1x1.3x1.8) m?

Chapter 6 (Measurement)

Review set 6B

279

The panelling for the gazebo includes 6 interior and 6 exterior triangular panels for the roof, 5 interior and 5 exterior rectangular panels for the walls, and one hexagonal panel for the floor. Total surface area = roof area + wall area + floor area

=2x6x(:x1.2x15)+2x5x

(1.2x0.75)

+6 % (& x 1.2 x 1.04) m?

0751“1

=23.544 m? ~ 23.5 m?

Length of cylinder is

325 mm = 32.5 cm,

and radius of cylinder is

40

_I

4—20 =20 mm = 2 cm.

fe———

Surface area of cylinder

325 mm ———»|

= 2nrh + 2712

= (2x 7 x2x325)+ (2 x 7 x 2%) cm? ~ 434 cm?

So, approximately 434 cm? of bubble wrap is needed to line the cylinder walls.

6

a

V = 13 (area of base x height) =1x7x(44)? x81em’

‘

~ 164 cm®

8.1cm

L—/ 4.4cm

b

3m

V = area of end X length

=(Bx5-Lix2x1)x5m® 4m

B—24>0}

~ 73.3

So, the cruise ship is about 73.3 km on a bearing of 191° from P.

Chapter 7 (Right angled triangle trigonometry)

Suppose the yachts both depart from O.

OAN = 180° — 34° = 146° AOB = 124° — 34° = 90° AAOB

{co-interior angles}

is right angled.

tanf = — 1

I=

0 =tan""(1)1 ~51.8° |

13

-, the bearing ofB from A ~ 360° — 146° — 51.8° ~ 162°

Now,

AB? =112+ 142 .

AB=+/112 + 142

{Pythagoras} {as AB >0}

~ 17.8

So, yacht B is about 17.8 km from yacht A on the bearing of about 162°.

a

i

The projection of [AH] onto the base plane is [EH].

ii

The projection of [BE] onto the base plane is [EF].

iii

The projection of [AG] onto the base plane is [EG].

iv.

The projection of [BH] onto the base plane is [FH].

Exercise 7G

317

318

Chapter 7 (Right angled triangle trigonometry) b

2

Exercise 7G

i

The projection of [RX] onto the base plane is [MR].

ii

The projection of [NX] onto the base plane is [MN].

|

The projection of [AF] onto the base plane is [EF].

a

the required angle is AFE.

il

The projection of [BM] onto the base plane is [FM].

A

B

A

B

the required angle is BMF.

iii

The projection of [AD] onto the base plane is [DE]. the required angle is ADE.

iv

The projection of [BN] onto the base plane is [FN].

the required angle is BNF.

Chapter 7 (Right angled triangle trigonometry)

b

i

Exercise 7G E

The projection of [AB] onto the base plane is [AM].

the required angle is BAM.

B

D‘

c

M

A

ii

319

E

The projection of [BN] onto the base plane is [MN]. the required angle is BNM.

4

B

/

D‘

M

A

iii

W E

The projection of [AE] onto the base plane is [AN]. the required angle is EAN.

D

B

AT

i

The projection of [CF] onto the base plane is [FG].

M

the required angle is CFG.

tanf = ¢

6cm

8

s

f=tan"! (g)

H

0}

Let AGE be a. tana

6

= — V164

tan— « —= tan

1

6 ) (\/m

o=~ 25.1°

The angle is about 25.1°.

C

V164 cm

€

320

Chapter 7 (Right angled triangle trigonometry)

Exercise 7G

The projection of [BX] onto the base plane is [FX]. The required angle is BXF. tan 3 = B

4

o

B=tan™! (%)

.

3~

56.3°

The angle is about 56.3°. iv

The projection of [DX] onto the base plane is [HX].

H {as x>0}

Let DXH be ¢. -

6

= tand V116 _

¢ = tan

—1

¢~ 29.1°

6

(Tm)

The angle is about 29.1°. The projection of [PR] onto the base plane is [RS].

the required angle is PRS. tanf = kil

12

0 =tan~! (%)

0 ~ 33.7° The angle is about 33.7°. The projection of [QU] onto the base plane is [RU].

the required angle is QGR. tana

= E

12

a=tan"!

(%)

cooam33.7° The angle is about 33.7°. The projection of [PU] onto the base plane is [SU].

.. the required angle is PUS. Let SU be x cm. Using Pythagoras in ASTU,

2% =122 + 122 2% =288 x =288

X

6cm m,

a? =107 4 47

e

F

D

Let HX be x cm. Using Pythagoras in AHGX,

z? =116 z=+v116

B

8

the required angle is DXH.

.

A

{as

x> 0}

10em

G

Chapter 7 (Right angled triangle trigonometry) Let PUS be 5. t

.8

anf = =

—1

B = tan

B~ 25.2°

8

(m)

The angle is about 25.2°. iv

The projection of [QM] onto the base plane is [MR]. the required angle is QI\7[R4 Let MR be

cm.

Using Pythagoras in AMUR,

2 =62 + 122 oo 2?2 =180 . z=+180

{as x>0}

Let QMR be 6. 8

tan¢ =

%

-

(bftan

8

-1

(m)

6~ 30.8° The angle is about 30.8°. The projection of [QR] onto the base plane is [MR]. the required angle is MfiQA tanf = L

0.6

6 =tan™' (%) 0 ~ 59.0°

The angle is about 59.0°. The projection of [QU] onto the base plane is [MU].

the required angle is QIAJM. Let MU be z m. Using Pythagoras in AMRU,

2% =0.6% +2.47 2% =612 =+v6.12

o

Let QGM be a. o tana = E=1F,

6.12

=

a = tan

-1 (m) 1

a = 22.0°

The angle is about 22.0°.

{as

>0}

Exercise 7G

321

322

Chapter 7 (Right angled triangle trigonometry) iii

Exercise 7G

The projection of [QN] onto the base plane is [MN]. the required angle is QI?IMA

MN | RU, so MN=RU =24m tan8 =

1

1

o= tan_l(rz)

3 = 22.6° The angle is about 22.6°. d

i

The projection of [AX] onto the base plane is [AM]. the required angle is XAM. Let AM = DM be z cm.

(The base of the figure is a square, so its diagonals [AC] and [BD] perpendicularly bisect each other.)

Using Pythagoras in AAMD,

2% +2? =6?

22% = 36 22 =18

r=+18

{as x>0}

Let XAM be 6. 8 1 V . cosl = — 0 1

@ =cos™! 0 =~64.9°

The angle is about 64.9°. il

The projection of [XY] onto the base plane is [MY].

the required angle is XYM. Let XY be z cm. Using Pythagoras in AXYD,

2 +3% =107 22 =100-9 22 =091

=01

Let XYM be a. - cosa:%

A . a=cos WAk (\/fi) . LoaTLT The angle is about 71.7°.

{as

>0}

Chapter 7 (Right angled triangle trigonometry)

1

a

AC?=AB?+BC? _=8 Q2 +6 @2

{Pythagoras}

B ADJ

=100

HYP

2

a

3

a

10

OPP

8cm

. AC =100 {as AC > 0} . AC=10 The hypotenuse is 10 cm long. bsingzfl:fizé

5

C HYP

HYP

b

10

dtang:flzfizg

5

ADJ

sin8° ~0.139

¢

b

HYP 8m

6em

A

ccosazfl:i:é

cosh9° ~0.515

tan76°

-

ADJ

7m

Trm

om

[

OPP

oz

cos64° = =

{cosf

HYP

_ADJ = fi}

8 xcosb4’ =
0}

=v2074 m

"

2074

ch

-

A

B

tanf = .

7 V2074

6 = tan 17 ( m)

~ 8.74° So, the angle of depression from A to B is approximately 8.74°. 12

We draw the isosceles triangle cross-section of the cone as shown. Let the radius of the cone be r cm.

1Sy

rem

tan17.5° = — 18 x tan17.5° =r .

.

17.5°

r~568

V= 1mr’h

~ 3 X x 5.68% x 18 cm® ~ 607 cm® ~ 607 mL

~ 0.607 L The cone has a capacity of about 0.607 L. 13

a

The projection of [AC] onto the base plane is [BC]. the required angle is ACB.

m

A

-

tanf = s

6

8cm

A =tan"! (%)

0 ~ 53.1° The angle is about 53.1°.

c fi

B

Chapter 7 (Right angled triangle trigonometry) b

Review set 7B

327

The projection of [AD] onto the base plane is [BD].

the required angle is ADB. Let DB be & cm, and the centre of the circular base be O.

OB =OD

=3 cm

{both radii of circle}

8cm

Using Pythagoras in ABOD,

z? =32+ 32

5B

a2 =18

L x=+18

{as z>0}

Let ADB be a.

A

©otana na = NG ——8 -.

— 18 (\/fi) a=tan

8cm

a = 62.1°

The angle is about 62.1°.

C

fi

—

D

1

Let the unknown side be x cm. r2 2 122 _=5"+11 {Pythagoras}

z°

V18cm

HYP xcm

5om

=146

/‘ B

OPP

z -=146

{as

= >0}

Tom

Ll

ADJ

sinf = OpP _ 5 HYP

2

,

146

a

cosf = 2D

HYP

2(;111:11 A

ADJ

4.5cm

e

146

C

5.0cm HYP

¢

tanf = Oileks

)

5

ADJ

OPP

11

2.2

i

sin26 26° = —~ A

=~

il

cos26° = ADI

45 5 0.90

il

tan26° = OFF ¢ 22

b

-

,

HYP

sin26° ~ 0.44,

ADJ

0 044

5.0

4.5

0.49

cos26° ~ 0.90,

tan26° ~ 0.49

328

Chapter 7 (Right angled triangle trigonometry)

3

a

b

HYP

OPP 4m

Review set 7B

7m

ADJ

.

10cm

13cm HYP

ADJ

sinf = 2

{sin@zfl}

7

cosg = 20

HYP

{co 9=fl}

13

. O =sin"t (%)

HYP

—1(10 0 =cos™'(19)

~ 34.8°

¢

= 39.7°

tang =22

{tané’:fi}

6.2

ADJ 6.2m

ADJ

0 = tan~! (4—3) ~ 36.0° 4.5m OPP

fsinf= 2t}

4 sin23° =LAB

HYP

HYP

A

B

. AB= sin2 23°

47 mm

~ 120 mm

tan23° = 2

g

C

{tanf = %}

AC

ADJ

& tan 23°

. AC=

~ 111 mm

5

2% +19? = 322

z? =32° - 19°

{Pythagoras}

M 32

r? = 663

x =663

{as

32

HYP

0 =~ 53.6°

sina = 32

32

.

{sinezfl}

a=sin! (%)

oo

36.4°

HYP

19cm [

K

{cos@zfl}

i, 0 =cos™'(33)

o

= >0}

~ 25.7

cosf =2

OPP

Trcm

L

Chapter 7 (Right angled triangle trigonometry) 6

g

Review set 7B

The diagonals of a rhombus bisect each other at right angles.

W

So,

tang

013 = 1

_o

{tan = ADJ}

g =tan"! (%)

7.5cm

‘:‘

~2 124° So the larger angle of the rhombus is about 124°.

7

ABC = 90° {angle in a semi-circle} AABC is right angled at B. cos32° = 43

AC

AC — "

or = "

4.3 cos32°

243

cos32° 4.3

Lr=—

2 X cos 32°

~ 2.54

The radius is approximately 2.54 cm. Let the height of the building be h m.

tan20° = — "

x 4+ 80

h = (z + 80) tan 20° Also

tan23° = h

xr

tan23° =

(x + 80) tan 20° x

2 tan 23° = x tan 20° + 80 tan 20°

.

z(tan23° — tan 20°) = 80 tan 20° s "

80 tan 20° tan23°

~ 481.25

h ~ (481.25 4 80) tan 20° m ~ 204 m

The building is about 204 m tall. 9

a

A

B

D

H

6 cm /7

329

8 cm

G

F 4 cm

AAHG

is right angled at H.

AHG = 90°

— tan 20°

330

Chapter 7 (Right angled triangle trigonometry)

b

E

F

Review set 7B

Consider the base of the prism. Let FH be z cm. Using Pythagoras, z% =42 + &°

N

4cm

o2 =80

Scm

=80 ADFH tanf an

{as x>0}

is right angled at H.

=

T —

6 =— tan

D

tan—1 ( (-680)

6em

9i33'9 So, 10

DFH ~ 33.9°.

Suppose Aaron starts at S, travels to O, and finishes at F.

FON; = 360° — 303° =57°

{angles at a point}

OSN, = 360° — 213° = 147° . N;OS = 180° — 147° = 33°

{angles at a point} {co-interior angles}

. FOS = 57° +33° = 90°

AFOS is right angled at O. tanf = 25 3

6 =tan™"' (%2) ~ 39.8° -, the bearing of F from S ~ 213° 4 39.8° ~ 253° Now,

2 =2.5%4 32

r=1/2524+32

{Pythagoras}

{as x>0}

~ 3.91

So, Aaron is about 3.91 km on a bearing of about 253° from his starting point. 1

Let the height of the pyramid be /& m, and the diagonal of the base of the pyramid be 2z m.

Consider the base of the pyramid.

2%+ 2% =122

2% =144 L2t =T2

L:\/fi

{Pythagoras}

{as

z > 0}

12m

v

V80 cm

Chapter 7 (Right angled triangle trigonometry)

tan40° = V7 h

T

VT2

tan 40°

~ 10.1

Volume of pyramid = % x (area of base) x height

~ 3 x12x12x101m’ ~ 485 m®

12

a

The projection of [BH] onto the base plane is [FH].

the required angle is BHF. Let FH be x cm. Using Pythagoras in AFEH,

2% = 5%+67 2t =61 x=\/6_1

{as = >0}

Let BHF be 0 tanf = NG

2

0 = tan 12 (fi) 0 ~ 14.4°

The angle is about 14.4°. b

The projection of [CM] onto the base plane is [GM].

the required angle is CMG. Let GM be « cm. Using Pythagoras in AGHM,

2? =52+3?

Loa?=34

Cx=+34 Let CMG be o tancy—i

V34

a == tan I (N)

a =~18.9°

The angle is about 18.9°.

{as z>0}

Review set 7B

331

332

Chapter 7 (Right angled triangle trigonometry) ¢

Review set 7B

The projection of [XM] onto the base plane is [MY].

B

l"

the required angle is XMY.

..

tanff =2

2

A

o

o. Ba2L8° B=tn(f)

’

o

a

25

i In AAMH,

6cm

OPP

sin65° =-—

{sinf = —}

AH

AH=

G

L M H

The angle is about 21.8°.

13

X

HYP

2

sin 65°

AH =~ 27.6 cm

i In AAMH,

tan65° = =2 AM

{tang = 2Py ADJ

25

"~ tan65°

AM ~ 11.66 cm

.

{altitude of isosceles triangle bisects the base}

CM ~ 11.66 cm . AC~2x11.66 . AC=23.3 cm .

and

50

{from a ii}

tan 65°

" tan65° .

i

Let the height of the prism, AE, be h cm length of the prism, EH, be x cm.

In AADC,

&

50

2 _

rEes

222 =

JdG

L

N

tan 65°

E

)2

(tan 65°

(

hcm

{Pythagoras}

AD? +CD? = AC? 9

and the

)

—

c


0}

2

4.092 4 9.862 — 8.42

2 % 4.09 x 9.86

¢ Area of ABCD = 1 x DB x BC x sinDBC ~ § % 4.00 x 9.86 x sin 57.5° ~17.0 m?

Chapter 8 (Non-right angled triangle trigonometry)

9

a By the cosine rule:

7 =22+ 6% -2 x 2 x 6 x cos60°

436 — 122 % (3) s 49=2"

coat—6r—13=0

b

Lo

4 6EV6 2 — 2(1)(

+

-1

6em

7em

3)

_6::\/@

e,

-2

>0,

so

Exercise 8C

rem

=3+V22 But

10

a

x =3+

By the cosine rule:

V22

2 2, Q2 0 11" =2*+8" —2x x x 8 X cosT0 o121 = 2% + 64 — 162 X cos 70°

vem

11em

oo 2?2 = (16c0s70°%)z — 57 =0 Using technology, But

b

x>0,

so

= ~ —5.29 or 10.8.

8cm

x~10.8.

By the cosine rule: 52 =32 4 2% — 2 x 3 x x x cos 120°

5em

3em

— 6z x (—1) 25=9+2%

s

a2 +3r—-16=0

—3+4/3%2

Trcm

—4(1)(-16)

2

But

¢

>0,

so

v =—

+ VB3 o7

By the cosine rule:

52 = 2% + (22)% — 2 x & X 27 X cos60°

5cm

ZCI m

25 = 2% + 42® — 42’ x (3) = 5% — 222 = 322

2

2 _

2x cm

25 3

r=4/8

{as x>0}

z = 2.89 11

By the cosine rule:

25 =22 + 36 — 122 cos 40° . 22— (12c0s40°)z +11=0

Using technology,

o

52 =22 462 — 2 x & x 6 x cos40°

= ~ 1.41

or

7.78.

5cm

xcm 6cm

343

344

Chapter 8 (Non-right angled triangle trigonometry)

12

Exercise 8C

Let CAB be a and DAC be . .

In AABC, by the cosine rule:

72 4122 — 92

cosa = ———— 2x7x12

Q=

—1(112 (m)

COs

2 2 _ cos3 = D

In ADAC, by the cosine rule:

12

2x8x12

. B=cosT(EE) A

ges

Now in ADAB,

B

8em

DAB =« + 3

= Cos —1(112 (1—68) —+ cos —1(87 (1—92) ~111.2°

. By the cosine rule:

2 Q 2 2 o BD” ~ 8% 4 7% —2 x 8 x 7 x cos 111.2

D

BD~

/82 +72—2x8x

BD~

124

7 xcos111.2°

[BD] is about 12.4 cm long. 13

In ABCD, by the cosine rule:

BC? =52 462 —2 x5 x 6 x cos 130°

. BC=+/52+62—-2x5x6xcos130° . BC~9.98

{as BC> 0}

In AABC, by the cosine rule:

BC2=821+92 -2 x8x9xcosf ©. 9.98% ~ 64 + 81 — 144 cos6 .

145 — 9.982

cosf R —04 —

144

L drxalo (145

L

9.98 2 )

144

O =T1.6° 14

A

Let the distance from P to C be = cm. In AABP,

~o

cosPAB =

524102 62 o X't i

e

__ 89

= 7100

- PAC =60° —cos™! (&)

rlem

{since AABC is equilateral}

- PAC ~ 32.87° Now, in AAPC, by the cosine rule:

5% — 2 x 10 x 5 x cos PAC 22 =1+0%

.

2 2 2.87° x cos3 x 55 104 2% za /10 T ~6.40

So, P is about 6.40 cm from C.

‘

{as

= > 0}

5

'

10cm

h

Chapter 8 (Non-right angled triangle trigonometry)

Exercise 8D.1

1,2,3

7cm

5

sinA

We notice that

_

sinB

a

b

_

sinC

or equivalently,

a

sin A

EXERCISE 8D.1 1

a

Using the sine rule, x

_

sin48°

23

sin37°

r =

23 x sin48°

———

sin 37°

x~28.4

b

Using the sine rule, x

11

sin 115°

sin 48° xTr=

11 xsin115°

r~13.4

sin 48°

115°

482

1l1cm

rem

sin B

sin C'

345

346

Chapter 8 (Non-right angled triangle trigonometry)

¢

Using the sine rule, T

48

sin 51°

sin 80°

4.8 X sin51°

r=

sin 80°

x km

z~3.79 d

Exercise 8D.1

The unknown angle is

180° — 100° — 45° = 35°

{angles in a triangle}

Using the sine rule, @

.

sin100°

6

sin35° 6 x sin 100° oL BIC D sin 35°

g

x~10.3

e

The unknown angle is

k6m 180° — 58° — 55° =67°

Using the sine rule, T

o

sin67°

4

sin55°

"~

58°

{angles in a triangle}

xcm

P

55°

4 x sin67°

sinB5°

r~4.49

f

The unknown angle is Using the sine rule, T

_

sin31°

T

9

180° — 108° — 31° =41°

{angles in a triangle}

108°

sin41° 9 x sin 31°

sin4l°

r =~ 7.07 2

a

Using the sine rule, a

18

sin63°

sin49° x sin 63° = 18 Lt O} sin 49°

a~21.3 cm

b

The unknown angle is Using the sine rule,

b

sin73°

x4

sin25°

Pl

34 x sin 73° "sin25°

b~ 76.9 cm

£ i

18cm

180° — 25° — 82° =73°

34 cm

{angles in a triangle}

Chapter 8 (Non-right angled triangle trigonometry)

¢

The unknown angle is ~ 180° — 48° — 21°

347

Exercise 8D.1

{angles in a triangle}

=111°

Using the sine rule, c

6.4

sin 48°

sin 111°

c—

A

6.4 X sin48°

T

sint11°

c~5.09 cm 3

a

BAC=180° =T74°

48°

c

— 58° — 48°

6.4om

21°

{angles in a triangle}

A

Using the sine rule,

-

AB

=

sin 58°

7

—

and

sin 48°

. AB — 7 >< sin 29°

XY

sin43° =

19 >

SIn

R

60

—

x ~ 9.85

b

Using the sine rule, 7°

7 mm 9mm

_ sin58°

sinz®

58°

i SO

9

o 58 7 X sin

T

=

e

sIn

r~41.3

9

O

(7 x sin58°) T

Chapter 8 (Non-right angled triangle trigonometry)

349

Exercise 8D.2

Using the sine rule,


< %

254

40

Chapter 10 (Probability)

b

Review set 10A

445

P(they are the same colour) = P(2 reds) + P(2 yellows) + P(2 blues)

= (ExfH)+(xH)+(HxH) 9+16+25 144 50 144 25

=72