Mathematics Core Topics for the IB Diploma HL 1 Worked Solutions, for use with Mathematics Analysis and Approaches HL & Mathematics Applications and Intepretation HL, [1 ed.] 9781925489859

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HAESE

MATHENMATICS

(M58

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

for use with

IB Diploma Programme Bradley Steventon Charlotte Frost

Joseph Small Michael Mampusti

SNOILNTOS

@)

daNd

Mathematicsilr

MATHEMATICS: Bradley Steventon Charlotte Frost Joseph Small Michael Mampusti

CORE TOPICS HL WORKED

SOLUTIONS

B.Ma.Sc. B.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

978-1-925489-85-9

© Haese & Harris Publications 2019

Published by Haese Mathematics.

152 Richmond Road, Marleston, First Edition

SA 5033, AUSTRALIA

2019

Artwork by Brian Houston and Charlotte Frost. Typeset in Australia by Charlotte Frost and 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 HL. 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.

BS

e-mail: web:

[email protected] www.haesemathematics.com

CF

JS

MM

TABLE OF CONTENTS

Chapter 1 Chapter 2

Chapter 3 Chapter 4

Chapter 5 Chapter 6

Chapter 7 Chapter 8 Chapter 9

Chapter 10 Chapter 11

Chapter 12 Chapter 13 Chapter 14

Chapter 15 Chapter 16

Chapter 17

STRAIGHT LINES SETS AND VENN DIAGRAMS SURDS AND EXPONENTS EQUATIONS SEQUENCES AND SERIES MEASUREMENT RIGHT ANGLED TRIANGLE TRIGONOMETRY THE UNIT CIRCLE AND RADIAN MEASURE NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY POINTS IN SPACE PROBABILITY SAMPLING AND DATA STATISTICS QUADRATIC FUNCTIONS FUNCTIONS TRANSFORMATIONS OF FUNCTIONS TRIGONOMETRIC FUNCTIONS

45 70 97 135 229 281 331 371 418 456 523 543 616 689 745 790

Chapter 1

STRAIGHT LINES

1

a

y=3x+7

has gradient

y = —2x — 5

m =3

has gradient

and y-intercept

m = —2

c = 7.

and y-intercept

¢ = —5.

¢

y=32x— 4 has gradient m =2

d

y= h; Z_ Zx+ 2 has gradient m = £ and y-intercept ¢ = 2.

¢

_2x-3 y==—F5—

1 =357~ 31

and y-intercept ¢ = —3.

¢iy =y= —3.

ek Sient 7m =3 g hasgradient and y-intercept

fy=2"5%_23_ y=—5— =5 3 ¢ hasgradient hasgradient mm =—2 = —g and and y-i y-intercept t c=3 ¢= . 2

a

The equation of the lineis

y —1=3(z —4) Ly—1=3zx—-12

y=3r—11

b

The equation of the line is Sy

y — 5= —2(z — (—3)) —5=-2(x+3)

L

¢

The equation of the line is

y—5=-2r—-6 y=-2x—-1

y — (—3) = (z — 4)

L y+3=qz-1 Ly=

3

d

The equation of the line is

a

The line passes through i

.

the gradient is

80—90

%w —4

y = —%z + 4.

(0, 90)

and

(1, 80),

so

= —10.

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

¢

100

80 &

balance was $90.

10

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

20

The account runs out of money when

y =0

~10z 490 =0

10z =90 =9

The account will run out of money after 9 years.

by T

%

s

3

x (years

4

6

Chapter 1 (Straight lines)

4

a

The line passes through )

)

so the gradient is

b

Exercise 1A

(0, 46)

0—46

0

S0t

a

(1820, 0),

Yy

23

)

.

510"

The gradient is —£% and the y-intercept is 46, so the equation is y = 7%x

5

and

When

t=0,

+ 46.

1820

>

H =150+ 120(0) =150

The helicopter took off from a height of 150 m. b

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

¢

When

t=2,

H=150+120(2) =390

The helicopter is 390 m above sea level after 2 minutes.

d

When the helicopter is 650 m above sea level,

H = 650

150 + 120t = 650 120t = 500

— 500 _

T 120 =45

41

The helicopter is 650 m above sea level after 4% minutes, or 4 minutes 10 seconds.

6

a

y=—4z+6

. dr4+y=6

b

{adding 4z to both sides}

y=>5x—3

—br+y=-3 br—y=3

c

y=-3z+32 o dy=-3z+5 . 3r+4y=5

d

a

y=-br+2

3z+Ty=-2 Ty=-3z—2 LYy=— gx — %

¢

{multiplying both sides by 5} {subtracting 3z from both sides} {multiplying both sides by —1}

br+y=2

L b

{multiplying both sides by 4} {adding 3z to both sides}

y= %z — % 5y=3z—1 —3x+5y=—1 3r—b5y=1

7

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

{subtracting 5z from both sides} {subtracting 3z from both sides} {dividing both sides by 7}

2xr—y=6

—y=-22+6

{subtracting 2z from both sides}

y=2r—6

{multiplying both sides by —1}

Chapter 1 (Straight lines) d

3z—-13y=-4

—13y=-3z—4

y=Zx+ 8

Exercise 1A

{subtracting 3z from both sides}

5

{dividing both sides by —13}

ar+by=d by =—azx+d y=

—%x +%l

which has the form

The gradient of the line is 9

a

y = mz + ¢

7% )

Since the line has gradient —4, the general form of its equation is 4z +y =d

Using the point

(1, 2),

the equationis

4z +y=4(1)+2

whichis b

4z +y =6.

Since the line has gradient %, the general form of its equation is = — 2y =d

Using the point

(3, —5),

the equationis

=z — 2y =3 — 2(—5)

which is ¢

= —2y = 13.

Since the line has gradient —%, the general form of its equation is 5z + 3y = d

Using the point

(—2, 6), the equation is

5z + 3y = 5(—2) + 3(6)

which is

d

5z + 3y = 8.

Since the line has gradient Z, the general form of its equation is

Using the point

(—1, —4),

the equationis

7z —6y =7(—1) —6(—4)

which is 10

a

The line has gradient

311 (_ 12) = % =2,

", the equation of the line is

7z — 6y = d

7z — 6y = 17. and passes through the point

A(—2, 1).

y — 1 =2(z — (—2)) Ly—1=2z+4

L

b

The line has gradient

s = %

", the equation of the lineis

The line has gradient

*.

2-(=5) 3-(—2)

the equation of the line is

= —1,

and passes through the point

y—2=—(z L

¢

y=2x+5

A(7, 2).

—7)

Y—2=—x+7 Ly=—x+9 = g,

and passes through the point

y — (=5) = %(z - (-2))

y+5==%(z+2)

y+5==%o+ 3

y=3T-% _

7

M(—2,

—5).

and passes through the point R(5, —1). 8 |

ol

y — (—1) =

lat

the equation of the line is

= ——,

+ —

-,

E _7(_51)) . %

ol

The line has gradient

8

oluon

|

d

Exercise 1A

Il

Chapter 1 (Straight lines)




v

So, the a-intercept is 2. For When

6z 4 4y = —36: =0,

4y=-36 L y=-9

So, the y-intercept is —9. When

y =0,

6x=—-36

Lor=—6 So, the z-intercept is —6.

For When

7Tx 44y = 42: z =0,

4y =42

Loy=24 21

So, the y-intercept is 2. When

y =0,

T7z=42

=6 So, the z-intercept is 6.

>

z

Exercise 1B

13

14 3

Chapter 1 (Straight lines) ay= b

7%z+2 i

When

Exercise 1B

has gradient z =8,

m = 7%

and y-intercept

we have

¢ = 2.

c

y=-30®)+2 =—6+2 =—4 v

So, ii

(8, —4)

When

So,

x =1,

(1, 3)

When

So,

4

a

For

we have

does not lie on the line.

z = —2,

we have

y=—3(-2)+2 NI

iii

does lie on the line.

(-2, %)

v

does lie on the line.

2z —3y=18:

When

z =0,

—3y=18 L

y=—6

So, the y-intercept is —6. When

y =0,

2z=18

=9 So, the z-intercept is 9.

b

|

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

¢

If (=3, ¢)

(3, —4)

into

i

vV

does lie on the line.

lies on the line then

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

2(—3) —3c =18 —6—-3c=18 —3c=24 c=-8

(7, —2)

into

does not lie on the line.

Chapter 1 (Straight lines)

5

a

x serves of nigiri at $4.50 each and y serves o 4.5z + 9y =45

b

When

z =4,

4.5(4)+9y =45

Exercise 1C

sashimi at $9 each adds up to a total of $45.

¢

When

18 +9y =45 9y =27

y=1,

4.52+49(1) =45 .

45x4+9=45 . 4.5 =36 =38

Ly=3

Hiroko bought 3 serves of sashimi.

d

1

Hiroko bought 8 serves of nigiri.

4y

5Nd.5z + 9y = 45

a

The midpoint M of [AB] is

b

The gradient of [AB] is

¢

The gradient of the perpendicular bisector is —1, negative reciprocal of the gradient of [AB].

d

(3—;5, 1—;7)

or (4,4).

which is

2BG.T)

’577; = g =3

\

the

1 The perpendicular bisector has gradient —3 and passes through (4, 4).

its equation is

z + 3y =4+ 3(4) z + 3y = 16.

15

M(4,4)]

A(3,1)0’:

;

16 ~ Chapter 1 (Straight lines)

2

Exercise 1C

a The midpoint M of [AB] is (fl The gradient of [AB] is

4-2

——

M) v (3, 3).

= —

1-5

74

= 0)

the gradient of the perpendicular bisector is 2. the equation of the perpendicular

bisector is whichis

X(s, 2)

2z —y = 2(3) — 2z —y=3.

The midpoint M of [AB] is

(71; 5, %) 3-5

The gradient of [AB] is

or (2,4).

-2

“—1)

1

6

3

the gradient of the perpendlcular bisector is 3. the equation of the perpendicular

bisector is whichis

3z —y = 3(2) — 3z —y=2.

The midpoint P of [MN] is (%

*3; 1) or (4, —1).

= 4—4 — 1

The gradient of [MN] is 13 2—-6

the gradient of the perpendicular bisector is 1. the equation of the perpendicular

bisectoris whichis

M(6,-3)

=z —y=4—(-1) z —y =5.

The midpoint P of [MN] is The gradient of [MN] is

(7+§71), w) T

et

-1-7

-8

or (3,4). 2

P(:;Zj‘

the gradient of the perpendicular bisector is 2. the equation of the perpendicular

bisector is whichis

\1.\;(7, 2)

2z —y = 2(3) — 2z —y=2.

The midpoint M of [OP] ( is The gradient of [OP] is

-0

)

or

(g, O).

=0.

So, [OP] is horizontal, and the perpendicular bisector is

the vertical line passing through

N(-L6)

(%, 0).

the equation of the perpendicular bisector is = = %

Chapter 1 (Straight lines)

f The midpoint M of [AB] is (“;1),@) 2

.

——0. = =3—4 _3 4 The gradient of [AB] is -1-3 "

)

Exercise 1C

)

-

the gradient of the perpendicular bisector is —3. the equation of the perpendicular

bisector is which is or

a

4z + 3y = 4(1) +3(3) 4z + 3y = 2 8z + 6y = 35.

The midpoint M of [PQ] is

L

.

The gradient of [PQ] is

4-(-3) _

The gradient of [AC] is

11

(—i, 5),

or

)

1

3—(—4)

the gradient of the perpendicular bisector is —1. . the equation of the perpendicular bisector is whichis

x4y = —% + % z+4+y=0

or ¢ 5

B

a

b

2=—(-2)

Vv

D:

-1=—-(1)

= —z.

Vv

i3z-2y+1=0

il

2y =3z +1 . y=3z+1

The equation of the perpendicular bisector is

2z + 3y = 2(3) + 3(5) 2z +3y =21

or a

The midpoint between the hospitals is

or

gradient —2.

has gradient £ which is

6

The perpendicular bisector has

2x+3y—21=0.

(_32}

2,

22} 4)

AY

(-3, 1).

o

B(2,4)

3

b The gradient of [AB] is - E:g =2 the gradient of the perpendicular bisector is 7%‘

T3

%

the equation of the perpendicular bisector is

52 + 6y = 5(—3) +6(1) which is 5z + 6y = I or

A2 v

10z +12y=7

The perpendicular bisector of the line joining the two hospitals divides the town into two regions. An ambulance crew should be sent from A to locations below this line, and from B to locations above this line.

7

a

The midpoint of [AB] is

The gradient of [AB] is

(zl ;zz,

i ;yz)

£2—Y

2 — 1

the gradient of the perpendicular bisector is

—

the equation of the perpendicular bisector is

b

(z2

—x1)z + (y2 — 1)y = (22

(z2

—z1)z + (Y2 — Y1)y= o

(IZ

zl)z

(y2

y1)y—

(

2

o

—

I

Y2—un

l’l)xl;g62 F(y2 211

.2

2_

p Y2 2y1

ef2 +yi)2) =. (2(

1

.2

+ 9 1)

We can find the perpendicular bisector of any two points substituting in the values of =1, x2, y1, and ys.

l/l)yléry2

A(zy, y1)

and

B(zs, y2)

by

Chapter 1 (Straight lines)

19

i The midpoint of [AB] is (IT+4 2_;5) or (5 1) =1

the gradient of the perpendicular bisector is —1. the equation of the perpendicular bisector is z + Y whichis z+y

Bil The midpoint T, (142 of [AC] is (T’ The gradient of [AC] is

-1-2

NI

The gradient of [AB] is

5—2

+

.

.

e

a

O

&

Exercise 1C

24(=1) e ) or (3, 4). 5 =-3

the gradient of the perpendicular bisector is %

the equation of the perpendicular bisector is whichis

ili

The midpoint of [BC] is The gradient of [BC] is

(#,

-1-5

5+;71)>

= — 3y = % — 3(%) = —3y=0.

or (3,2).

=3

the gradient of the perpendicular bisector is —%.

the equation of the perpendicular bisector is whichis b

= + 3y = 3 + 3(2) z+ 3y =

Ay

The perpendicular bisectors all intersect at

(5, 2).

A, B, and C are all equidistant from this point. ¢

9

a

The perpendicular bisectors of each pair of points will meet at a single point. As the three points are equidistant from the point of intersection, a circle centred at the point of intersection that passes through one of them will pass through all of them.

—645 e of [PQ] is. ( (841 ;i The midpoint ) or (-3, -13). prs )

L

The gradient of [PQ] is

5—(=6) _ 11 =8

9

the gradient of the perpendicular bisector is —7.9

the equation of the perpendicular bisector is which is

9z + 11y = 9(—%) +11(—3) 9z + 11y = —37.

Chapter 1 (Straight lines)

Exercise 1D

il The midpoint of [PR] is

.

(_8 iy _6+T(_2))

o2

The gradient of [PR] is

or (=2, —4).

(6) 1

T-(®)

3

the gradient of the perpendicular bisector is —3.

the equation of the perpendicular bisector is which is

iil The midpoint of [QR] is (# .

.

The gradient of [QR] is

—2-5

fre

3z +y = —10.

or (52024).

2)) 5”* >T g =

3z +y = 3(—2) + (—4)

7

the gradient of the perpendicular bisector is % the equation of the perpendicular bisector is which is

3z — Ty = 3(%) — 7(

)

wolee

20

3z — 7y = —3.

We draw the graphs of y =3z +2 and y =x — 2 on the same set of axes. The graphs meet at the point (-2, —4). the solution is

= = —2,

y=

—4.

Check:

Substituting these values into:

o

y=23z+2

o

y=x—2

gives gives

—4=3(—2)+2 -4=-2-2

v

Chapter 1 (Straight lines)

Exercise ID

We draw the graphs of y = —4x +1 y =3z — 6 on the same set of axes.

The graphs meet at the point the solutionis

z =1,

21

and

(1, —3).

y=—3.

Check: Substituting these values into:

o

y=—4dz+1

e

y=3r—6

gives gives

-3=—-4(1)+1 -3=3(1)-6

We draw the graphs of y =2x —5

y =42 +4

on the same set of axes.

The graphs meet at the point the solutionis

v

z =6,

and

(6, 7).

y=717.

Check: Substituting these values into:

o

y=2r—5

gives

o y=1z+4

7=2(6)—5

Vv

gives 7=1(6)+4

v

We draw the graphs of y =2 —1 and 2x + 3y = 12 on the same set of axes.

The graphs meet at the point the solutionis

z =3,

(3, 2).

y=2.

Check: Substituting these values into: e y=x—1 gives 2=3-1

o

2z +3y

v =12 gives 2(3) +3(2) =12

v

Chapter 1 (Straight lines)

Exercise 1D

We draw the graphs of x +3y =9 and x — 2y =4 on the same set of axes. The graphs meet at the point (6, 1). the solutionis

=z =6,

y=1.

Check: Substituting these values into:

o o

z+3y=9 z—2y=4

gives 6+3(1)=9 gives 6-2(1)=4

the solution is = =2,

y= —12.

Check: Substituting these values into:

e

3z —2y=230 gives 3(2) —2(-12)=30

o

4dxr+y=—4

gives

42)+(-12)=—-4

a

y=z+2

e

20 4+3y =21

.. (2

(D)

Substituting (1) into (2) gives

2z + 3(z +2) =21

2z + 3z +6 =21 5z =15 Lox=3

Substituting

= =3

into (1) gives

y=3+2 y=>5

The solutionis

Check:

(1) 2)

= =3,

y=>5.

5=3+2 V 23)+3()=6+15=21

Vv Vv

We draw the graphs of 3z — 2y = 30 and 4x +y = —4 on the same set of axes. The graphs meet at the point (2, —12).

|y

22

v Vv

Chapter 1 (Straight lines) b

y=2:-3

..

dz—3y="7

.. (2)

Substituting (1) into (2) gives

4z —3(2z —3) =7 LA

-6z +9="7 —2r = -2 =],

Substituting

= =1

into (1) gives

y =2(1)—3 Ly=-1

The solutionis

Check:

(1) Q@

z =1,

y=—1.

—1=2(1)-3=2-3 « 41)-3(-1)=4+3=7 v

¢ Sz+3y=19

.. (1)

y=6—2x

.

(2)

Substituting (2) into (1) gives

5z + 3(6 — 2z) = 19 .

5+ 18 —6x =19

—r=1 Lx=-1

Substituting

z = —1

into (2) gives

y=6—2(—1) Ly=2~8

The solution is

Check: d

(1)

=z = —1,

y=8.

5(-1)+3(8)=-5+24=19

2)

8=6-2(-1)=6+2

z=y—-3 br—2y=9

v

«

e (1) .. (Q2)

Substituting (1) into (2) gives

5(y —3) —2y =9 hy—15—-2y=9 L 3y=24

Substituting

y = 8

The solutionis

Check:

(1) ()

into (1) gives

=z =5,

*=8—3

y=38.

5=8-3 « 5(5)—-28)=25-16=9

Exercise ID

23

24

Chapter 1 (Straight lines)

e 3rtdy=-13 T=8y—2

Exercise 1D

... (1) e

Substituting (2) into (1) gives

Substituting

y = 7%

3(8y —2) +4y = —13

2y — 644y =—13

into (2) gives

= = 8(7%) -2 L

r=-2-2

Lox=—4 The solution is

Check:

f

= = —4,

y=—%.

(1)

3(-4)+4(—%)=-12-1=-13

@

-4=8(—3)-2=—2-2

o=-by—2

e

Tr+4y=-10

... (2)

v

v

(D)

Substituting (1) into (2) gives

7(—5y —2) +4y = —10

—35y — 14 + 4y = —10 —3ly =4

y=—3r

Substituting

into (1) gives

y = —5¢

= = —5(—=5) — 2

=2 T = —11—}

y = —=.

The solution is @ = —14%,

Check:

g

(1)

—14

@

7(-14)

y=%x+5

e

3x+4y=5

.. (2)

5(-4)—2=2-2

2

g

+4

(D)

Substituting (1) into (2) gives

3z + 4(%35 + 5) =5 L

3x+2x+20=5 br=-15

.

05

TN

Substituting = = —3

into (1) gives

y=1(-3)+5

=-3%+5

y=33 The solution is

Check:

= = —3,

y = 33.

(1)

33=3(-3)+5=-3+5=1

v

@

3(-3)+4(33) =—9+4(3)=-9+14=5

v

Chapter 1 (Straight lines)

dr—5y=-24

Exercise 1D

.. (2 —3y — by =—-24

8y=-2 y =

Substituting y = 3 into (1) gives

= —3(3) r=-2%

The solution is = = -2, Check:

y=3. v

(1)

—23=-3(3)=-%

@

4(-23)-503)=4(-%)

i 3z+Ty=6

-15=-9-15=-24

v

..(1) ..

r=%y—-1

Substituting (2) into (1) gives

3(5y —1) + 7y 5y —3+Ty 12y

y=1%

Substituting

y = 2

into (2) gives

« = g(%) ~1 T = % -1

The solution is = =1,

y=3.

r=t

Check: () 3(3)+7(})=1+%=6

v

@ i=3(}-1=9-1

3r—y=5

:

{ 4z +z =9 The coefficients of y are the same size but opposite in sign. We add the LHSs and the RHSs to get an equation which contains x only. 3r—y=5

de+y=9 Adding,

Tz

= =2

The solutionis

Check:

In(2):

..

=14 LT

Substituting

.. ()

=2

into (1) gives

= =2,

3(2) —y =5 . 6—y=5 —y=—1

y=1.

4(2)+1=8+1=9

25

26

Chapter 1 (Straight lines)

b

Exercise 1D

5r —2y =17 3x+2y=7 The coefficients of y are the same size but opposite in sign. We add the LHSs and the RHSs to get an equation which contains z only. br—2y=17

... (1)

3r+2y=7

.2

Adding,

8x

=24 Lr=3

Substituting

= =3

into (1) gives

5(3) — 2y =17

15— 2y =17 —2y=2 Ly=-1

The solutionis

Check: ¢

=3,

In(2):

3zx+y=16 Tr—2y=7

y=—1.

3(3)+2(-1)=9-2=7

V

.. (1) ..(2)

To make the coefficients of y the same size but opposite in sign, we multiply (1) by 2.

6z +2y =32

{(1)x 2}

Tx—2y="7 Adding,

13z

=39

=3

Substituting

= =3

into (1) gives

3(3) +y = 16

91 y=16 L oy=7

The solutionis

Check: d

=z =3,

In(2):

y="17.

73)—2(7)=21-14=7

3z—-Ty=-27 —6z+5y =18

V

... (1) ... (2)

To make the coefficients of x the same size but opposite in sign, we multiply (1) by 2.

6z — 14y = —54 —6z + 5y =18

Adding,

{(1) x 2}

9y = —36

Ly=4

Substituting

y =4

into (1) gives

3z — 7(4) = —27 3r —28 = 27

3r=1

The solution is

Check:

Tn(2):

x = 13

L e=1 y =4

—6(3)+5(4)=-2+20=18

Chapter 1 (Straight lines) e

Jx—Ty=-8

.. (1)

9z + 11y =16

... (2)

Exercise ID

27

To make the coefficients of x the same size but opposite in sign, we multiply (1) by —3.

-9z + 21y =24

{(1) x -3}

9z + 11y = 16

Adding,

32y — 40 y=13

Substituting y = 13

into (1) gives

3z — 7(13) = -8

L 3r—-7(3)=-8 L3-8 .

The solution is = =4,

Check:

In(2):

fdr+3y=14 Sr—dy=23

=-38

3z = %

soar=1

y =14

9(§)+11(1})

=%+11(3)=3+F=%=16

v

.. (1) .. (2

To make the coefficients of y the same size but opposite in sign, we multiply (1) by 4 and 2)

by

@ by

3.

160 +12y =56 {(1)x 4} 9z —12y =69

Adding,

25z

{(2) x 3}

=125

=5

Substituting

z =5

into (1) gives

4(5) + 3y = 14

20+ 3y =14

o 3y=—6 The solution is

Check: g

= =15,

In (2):

y=—2.

L

y=-2

3(5) —4(-2)=15+8=23

2xr—3y=6

..(1)

br—4dy=1

... (2)

To make the coefficients of y the same size but opposite in sign, we multiply (1) by —4 and (2) by 3. 8z + 12y =—-24 {(1)x —4} 152 — 12y =3 {(2) x 3} Adding,

Tz

=-21 Sox=-3

Substituting

= = —3

into (1) gives

2(—3) —3y =6 —6—-3y==6

3y =12

The solutionis

Check:

In (2):

= = -3,

y=—4.

Ly=-4

5(—3)—4(-4)=-15+16=1

v

Chapter 1 (Straight lines)

h 4z+2=-23 br—Ty=-5

Exercise 1D

.. (1) .. (2)

To make the coefficients of x the same size but opposite in sign, we multiply (1) by 5 and (2) by —4.

20z 410y =—115 —20z+28y =20 Adding,

{(1) x 5} {(2) x -4}

38y = —95

Substituting y = —2%

into (1) gives

4z +2(—23) = 23

Az +2(-5)=-23 ©odr—5=-23 4r = —18

r=—41

The solution is @ = —43,

Check:

dz—Ty=9 Sr—8y=-2

5(—43)

-7(-23)

=5(-2)-7(-3)=—2+L=-L=-5

v

.. (1 .. (2

To make the coefficients of = the same size but opposite in sign, we multiply (1) by 5 and (2)

by —4.

.

20z — 35y =45

{(1) x5}

200432y =8

Adding,

{(2) x —4}

-3y =53 y=

Substituting y = —17%

—17%

into (1) gives

4z —7(-173) =9

dr—7(-%)=9 dx + %

=9

dr = 7% r=-282

Check:

In(2):

5(—28%)

y = —173.

—8(—-172)

=5(—5%)

—8(—%2) =

—430 + 41




2 —5=3T _2_ 4

2

2

4

2 _ brt+x=1

4, _2 ' —3zx=%

241 Fr =3 1 T+

2\2 _ 2

232

(-5)°"=3+(-3)

-3+

2\2 _ 2,

4_ 10

(z-3)y=3+5=% /10 g

_ . 2 T—3=I4/

2 _ 4 V10

rT—3==x73 _

2 4

1

5

12

xr

3

32% +1= b 2,1

_ _5

T+

g5=—32

2. T+ 2.5

5.3T _1 =

—3

5y2 _

1

5\2

T4+ 3e+ (3 =-35+%) 5)2 _ 1,25 _ 13 (+8)=—3+% =% 5 __ 4

/13

T+HE=%4/3 5

_

1

V13

T+g=x%% o= 54 Vi3 =76T7% az® +br+c=0 b 2+S=0 a a

P+

24 a

x

lro_c a

b\

¢




b\

x4+

a

c

— ) +(2a)

=

241

1y2 _

b2 — dac 4a2

b

b2 — dac

z+%7__

4a? —b =+ /b2 — 4ac

rT=—

1

(5)° =5+

e+t

1y2

1,

(z+75) T TT

Y10

T=35x

345 =—2

Exercise 4C.2

1y2

(1)

1 _ 21

510 = 10

21 1 15 = %4/ 100 41

421

10 = *710 —

_ 1

- T=T1

4

T

V21

105

106

1

Chapter 4 (Equations)

a

Exercise 4C.3

b

22-42-3=0 has

a=1,

b=—-4,

_ () L= T =

o—

4::\/%

¢=

224+62+7=0 has

-3

£ V(=92 —4(1)(=3) 2(1)

+1=0

whichhas

a=1,

T =

=

a4 b=—-4,

¢=1

.

.

S o

4::2\/§

o

f

b=—-4,

¢c=2

LT

has

h c=—-1

has

¢c=-3

2+ 28

4

xr =

2::2\/7

T =

1::\/7

4 2

a=—-1,

b=4,

e )

L

=

2(—1) —4::\/@

r=——Y" -2

=

c=6

\/42 — 4(-1)(6)

T

2(3)

6

b=-2,

—2?2+4c+6=0

_ —(=5) £ V(=5)2 —4B)(-1)

LoT=

a=2,

xr =

.Z':2::\/§

\/fi

2

—(=2) + /(=2)? — 4(2)(=3) 2(2)

2

5+

——

4@

4::2\/5

b=-5,

—4::2\/5

222-22-3=0

2

a=3,

—4(1)(-1)

x=—2::\/5

4::\/§

=

/42

¢=—1

. xi—Q(l)

L=

2

322 -52—-1=0 has

4+

b=4,

L= 2

2

2(1)

=

a=1,

—4::m

_= (V2

LT

dr—1=0

whichhas

V(=92 -4

22 —4r+2=0

)

P’ 4z =1

4+/12

a=1,

76::2\/5

x:73::\/§

d

.Z':2::\/§

has

g

“

a4z

_ (=)

e

—64++/8 2

.=

\/62 —4(1)(7)

2

2’ +1=4dz

b

.

2

c

xr

=6+

=

EI=NF2) == x/?

.

.

2 4::2fi

c=7

b=6,

a=1,

74::72\/fi

Z=2::\/E

Chapter 4 (Equations)

= r

=

T

=

=TT —

—4(=2)(-2)

2(—2)

::\/E

—4

7+4/33

b

(z+2)(z—1)=2-3z

LAt —r+2r-2=2-3z ot tdr—4=0 whichhas e

Exercise 4C.3

a=1,

—4+ 4+

b=4,

/4% 2

_

¢c=—4

2z +1)?2=3—=x

L4’ t4r+1=3-z o4z 450 -2=0 which has

L —4(1)(—4)

a =4,

—54

b=5,

4/52 — 4(4)(-2)

=~/

2(4)

2(1) 1:74::\/3_2

T =

2

¢= -2

-5+

=

\/fi

8

—4+ 42 — L=

2

c,

:L‘=—2::2\/§

d

(z—22=1+z ot

(Bz+1)2 =22

92?46z +1=—2z 482 4+1=0 o922

—dr+4=1+=x

2?—5z+3=0 whichhas

a=1,

_ (=5

f

whichhas

¢=3

/(52— 4B 2(1)

5413

o=

b= -5,

[

f

(z+3)(2z+1)=9 2 +x4+62+3=9

oy

a=2, (==

2(2) -7+

\/W

4

=

8

-8+

8

2\/7

74::\/7

9

2e+3)(2c-3)=2 —9=2

42’ —2-9=0 ¢=—6

72 — 4(2)(—6)

r= o=

b=7,

T

—8++/28

4P

2%+ T2 —-6=0

whichhas

.

¢c=1

209)

r = .

b=8,

_ —8+,/82-4(9)(1)

r=

2

a =9,

whichhas

a =4,

b=-1,

¢=-9

—(=1) £ V/(=1)* — 4(4)(=9)

2(4)

1++/145

8

107

108

Chapter 4 (Equations) z—1

S

Exercise 4C.4 h

=2x+1

2—x

1

r——-—=1 T

Lr—1=Q2e+1)(2-1x)

sLort-l=2

Lr—l=dr—-2%+2—x 222 —22-3=0

whichhas

SPE

T r = r = r =

a=2,

b=-2,

st —x—1=0 whichhas a=1,

c=-3

L —(-1) +

—(=2) +/(=2)? —4(2)(=3)

B

2::\/%

i

ey

L

g 1EVE =

b=-1,

/CDE =2

¢=—1

(=)

2(1)

2

4 2427 4 147

2

i

2w—-L1=3 T

22?7 —1 =3z 22

—3c—-1=0

which has

a =2,

b=-3,

¢c=-1

—(=3) £ /(=3)2 —4(2)(-1) 2(2)

1

22—Tc+9=0 a

has

a=1,

A=0b>—4dac = (=7)2 = 4(1)(9)

b=-7,

¢=9 b

=13
0, but 13 is not a square, there are 2 distinct irrational roots.

Chapter 4 (Equations) 2

422 —42+1=0 a

has

a=4,

b=-4,

Exercise 4C.4

109

c=1

A=0b>—4dac

b

Since

A =0,

there is one repeated root.

= (—4)> —4(4)(1) =0

—b+


0, but 76 is not a square, there are 2 distinct irrational roots.

f 1622 —8z+1=0

A =b? —dac

Since

522 +42-3=0

=176

22+z+5=0 has

and 1 is a square, there

=42 —4(5)(-3)

Since A > 0, and 16 is a square, there are 2 distinct rational roots.

e

> 0,

A =b* —dac

4(3)(-1)

=16

A

are 2 distinct rational roots.

d

A =b* —4ac

=22

¢c=2

there are no real roots.

has

a =16,

b=-8,

c=1

A =b? —dac

= (-8)° — 4(16)(1) =0

Since

A =0,

there is one repeated root.

110

Chapter 4 (Equations)

Exercise 4C.4

202 —Tx —5=0

622 —5r —6=0 has

a =6,

b=

-5,

¢=—6

has

A =b% — dac

a=2,

A =0

= (=5)* — 4(6)(—6)

b=-7,

¢=-5

—dac

= (=7)* —4(2)(-5)

=169

=89

Since A > 0, and 169 is a square, there are 2 distinct rational roots which can be

Since A > 0, but 89 is not a square, there are 2 distinct irrational roots.

322 +4x+1=0

622 —47x —8=0

found by factorisation.

c=1

b=4,

a =3,

has

has

a=6,

b= —47,

¢c= -8

A =1 — dac =4%—4(3)(1) =4

A =" —dac

Since A > 0, and 4 is a square, there are 2 distinct rational roots which can be found by factorisation.

Since A > 0, and 2401 is a square, there are 2 distinct rational roots which can be found by factorisation.

422 —3x +2=0

822 +22-3=0

has

a=4,

A =

= (—47)% - 4(6)(~8) = 2401

b=-3,

¢c=2

has

— dac

A =

=-23

A < 0,

b=2,

¢=-3

— dac

=22 —4(8)(-3)

= (-3 —4(4)(2) Since

a =8,

=100

there are no real roots.

Since A > 0, and 100 is a square, there are 2 distinct rational roots which can be

found by factorisation.

22 4+4z+m=0

has

a=1,

b=4,

c=m

A = b% — dac

=47 —4(1)(m) =16 —4m i

For a repeated root,

ii

For no real roots,

A>0 16 —4m >0

A0 9—-8m >0 —8m

For no real roots,

A +p

are of the form

and

pqui.

Chapter 4 (Equations)

Exercise 4D

a 2*+924+14=0

aX2 +bX+c=0

Using technology, r=—-2or -7

|

EJDEIR3([CLEAR]

b 2?—8z+16=0 Using technology,

[REPEAT

a8

= =4

aX2? +bX+c=£)

s

a

< -¢ I

1

EJDEIRI3([CLEAR]

¢

4’ +2-8=0 Using technology, r~1.29 or —1.54

aX2 +bX+c=0

a

C

et Des)Forn) aX? +bX+c=0 x1C I

-T2 +2=0

Roots

Press: [EXIT] EJDEIR[3[CLEAR]

g

Real

a8

aX2? +bX+c=£)

a

PN

) e lrn)

aX2 +bX+c=0

Xl[ 2L

Using technology, x ~ 1.39 or 0.360 E|DEIR[3([CLEAR]

[REPEAT

0.3696.

Chapter 4 (Equations)

3.8z + 2.12% = 52.6 2122 +3.80 — 526 =0 Using technology, z ~ 4.18 or —5.99

B

Efbefom]

aX? +bX+c=9 L

[Ffes)

2.1

B

3.8 I

2?4+ 5r+20+12=0 24Tz 4+12=0

Using technology, r=-3o0r —4

(x —1)(z+9) =5z 2?2 +8—9—5x=0

(7] Real aX2 +bX+c=0 b

119

FHifefon) [@)Fea

o

aX2 +bX+c=0 X2l

-5.9

4.181121809

-52.6

(SOLVE]RRIARCLEARICEDIT]

z(x+5)+2(x+6)=0

Exercise 4D

c

aX2 +bX+c=0

(37c)Rel

sl

7 I

[SOLVE) BEYERECLEARIEDIT] ath] Deglforn])

aX2? +bX+c=0

a

22 +32-9=0

[HathDeglForm1) (d7c]Real

aX? +bX+c=0

b

i;[ —A.SEJ

Using technology, z ~ 1.85 or —4.85

3z(z+2)—5(x—3)=18

322 + 6z — 5z +15—-18=0 3z +2-3=0 Using technology, r ~ 0.847 or —1.18

2x(r —6) =2 — 25

2% — 122 —x+25=0 222 — 132 +25=0

(7] Real aX2 +bX+c=0 b

1

c

aX2 +bX+c=0

sl

-1+437 6

(SOLVE)FEEBCLERR) Heth)Dedforn]) (d7c)Rea)

aX2 +bX+c=0 a

L

b

2

[HathDeglforn1) (d7c)Real = -

a c

No

-13

it Desllorn) (d7e))

Using technology, x=3,0,o0r

Real

Roots

Press: [EXIT]

Using technology, there are no real solutions.

22 —92 =0

(GFe)Re)

ot es)flor) (d7e)fes)

aX3 +bX2 +c‘)’(+d=0

aX3 +bX2 +cX+d=0

X1

—3

X2

0

X3

-3

EJDEIE3

2?22 +4=0

Using technology, r~—1.13

Hath) DeglFornT) aX3 +bX2 +cX+d=0

a

b

c

[HathDegForn1) (d7c)Rea) aX3 +bX2 +cX+d=0

x1[

e

-1.130395435

120

Chapter 4 (Equations) ¢

28 —22

-z

Exercise 4D

+24=0

Fetifedforn]) (JFc)fFes)

aXs3 +bX2 +c§+d=0

Using technology, r=3,2,or

5

—4

1

j

—23+2=2c—2? 2422

—204+2=0

Using technology,

FHibedforn] [dF)Re)

aX3 +bX2 +cX+d=0

Xl{fl X2 2 X3 -

-1

=L

24

[SOLVE) RRIAM CLEARICEDIT) d

B

[REPEAT) B

[Hat Degl Norm1) aX3 +bX2 +cz(+d=0 a c

[tDegforn]) ({Fc)Res)

aX3 +bX2 +cX+d=0

x1C I

= =1 DELETE

e

2% 42?2 =3z —1

2% 422 —3x+1=0 Using technology, r =0.5,~0.618, or —1.62

[HathDegforn]) [97c)Rea)

aX3 +bX2 +cl)’(+d=0

a

C

2

B

EhD:

(&) Read

aX3 +bX2 +cX+d=0


i

B HathDesNorn]) (d/cReal aoX2+a1 X3+ - -+as=0 143407 X2

-3.17]

1.340778787

Chapter 4 (Equations) 2t =222+

1=0

B

aoX4+a1

Using technology, z=1or

Ffifedform]) [d)Fesd a0

X3+

- -+as=0

al

a2

C-

-1

0

Exercise 4D

121

HetfiBeforn]) (J7c)Fes)

a3

-2

-

o

aoX4+a1

X3+

Xl[‘] X2

-1]

x2

- -+as=0

x2

(SOLVE)(ANEI3([CLEAR]

2t -2 +322 -2+6=0

B

WetfiDedforn] (dFc)Red)

aoX4+a1 X3+

Using technology, there are no real solutions.

a0

C-

al

- -+asa=0 a2

1

3

[Hath Deg florm1)

a3

>

agE=s

-1

Press: DELETE

zt -z =522 +3 't =522 —2-3=0

PIENEN3

B [EatiRadfiorn]) (dFc)Resl ao X4 +a1 X3+ - -+as=0 a0

C-

Using technology, T~ 243 or —2.27

[EXIT]

al

a2

a3

-

PINEY3

[HathRed forn]) (d7c)Real aoX4+a1 X3+- - -+tas=0

Xl[ x2l -2.267]

2.432693049

[REPEAT]

zt 4223 + 3 = x(92 + 14) 2t ot +22°

—92?

+22%

4+ 3

922 +

14

— 142 +3=0

Using technology,

B

HetRedfom] (dFc)Red)

aoX4+a1 X3+

r ~ 2.82,0.192,

a0

C-

—1.65, or —3.36

al

- -+aa=0

2

a2

-8

a3

>

-14

[HathRed Forn]) aoX*+a1 X3+ X1

X2|

0.1017

X3|

-1.651

X4l

-3.369.

2.81944095

[REPEAT]

z(z? —1) =2z

B

EHfedfom] [@)Re)

BB forn] (J/c)Fes)

aXs +bX2 +cl)’(+d=0

B —r—22=0

aX3 +bX2 +cX+d=0 X1 X2

=

22 —3x=0

- -+as=0

0

X3L-1.732

Using technology, r=0,~1.73,or

b

—1.73

(x—2)(z+1)=2> 22 —r—-2-2%=0 —ad

-z

Using technology, z~ —0.811

(SOLVE)NANEI3(CL

B

Eatfledlorn]) (d7cIReal

aX® +bX2 +cl)’(+d=0

a

(dZe)Read

c

aX3 +bX2 +cX+d=0 X1 I

-2=0 [REPEAT]

-0.8105357138

122 6

Chapter 4 (Equations) a

Exercise 4E

Za P

pdi

z—3

5

x?

zxa? -2z —3)=22%(r—3)

{multiplying both sides by

z*(z —3)}

% — 20+6 = 22° — 622

2% — 62 +20-6=0 b

23— 62> +22-6=0

B Eaialien] TR

.

aX3 +bX2 +cX+d=0

Using technology,

a

7~ 5.83

b

SR

BB

@)

aX3 +bX2 +cX+d=0

¢

10

R

5.83346916

1

a

2 —5=x+1

b

22—-5—-2—-1=0 Ltz

—6=

Wegraph

y =22 -5

on the same set of axes.

and

y =2 +1

_EXE]:Show coordinates

Yot

(z+2)(z—3)=0 Lox=-2o0r3

INTSECT

The graphs intersect at (3, 4). the solutions are 2

a

Wegraph

y=2

and

y=5—+/

the same set of axes.

on

b

We graph

y = B

xT

(—2,

—1)

and

x = —2 or 3. and

y =3z +4

on

the same set of axes.

B [EXE]:Show coordinates Yi=x ¥2=5-(Jx)

[EXE]:Show coordinates

YI=2.x Y2=3x+4 T1

INTSECT ¥=3.208712163

The graphs intersect at (3.21, 3.21). the solution is

x ~ 3.21.

INTSECT

=-1.16227766

The graphs intersect at and

(0.387,

—1.16)

5.16)

the solutions are

0.387.

(—1.72,

= ~ —1.72

or

Chapter 4 (Equations)

¢

We graph y =3% same set of axes.

and

y =15

on the

d

We graph y =5x2""1 on the same set of axes.

[EXE]:Show coordinates

Y1=3"(x) vo=15

(2.46, 15).

y=22+5

The graphs intersect at

and y =4+

..

f

on the same set of axes.

and

on the same set of axes.

[EXE]:Show coordinates

V1=(x+1) (-3):gy

V2=4 (x+3]

INTSECK

ON=7.304883905

-

°

y =%+

(3.56, 2.56).

x ~ 1.52 or 2.83.

4 xr

v¥=1.302775638

The graphs intersectat (—1.30, 1.30) and

(2.83, 13.0).

the solutions are

o

INTSECT

X=-1.302775

The graphs intersect at (1.52, 7.30) and

We graph

(5.17, 90).

y = (z+1)(z—3)

y =+/x + 3

7y

X=1.518184411

a

X

the solution is z ~ 5.17.

We graph

[EXE]:Show coordinates

Y1=x2+5 Y2=4x+(Sx)

INTSECT

¥=0I

ON=15

the solution is = ~ 2.46.

Wegraph

y =90

=6x2"(x~1)

The graphs intersect at

e

and

123

[EXE]:Show coordinates

7y

X=2.464973521

Exercise 4E

.

b

the solutions are

We graph

= ~ —1.30 or 3.56.

y = 2% — 2%,

[EXE]:Show coordinates

[EXE]:Show coordinates

Y1=27) =% (4)

T

X=-1.587401062

The z-intercept is = —1.59. the solution is = ~ —1.59.

The z-intercepts are ~ —0.861, and 16. the solutions are r~

¢

Wegraph

y=z+

¢z +4.

d We graph

[EXE]:Show coordinates YI=x+(Ax)+4 13]

—0.861,

1.24, or 16.

y =222 —r + 1

[EXE]:Show coordinates ¥Y1=2x2-(J(x+1)) 13¥

X=-0.571950556

The z-intercept is ~ —2.62. the solution is

1.24,

= ~ —2.62.

|

The z-intercepts are ~ —0.572 and 0.821 . ..

the solutions are z ~ —0.572 or 0.821.

124

Chapter 4 (Equations)

e

We graph

Exercise 4E

y=az*+277% —30.

f We graph

[EXE]:Show coordinates

[l

Y{=x"(8)+2~(-2)-30¥

4

y=

1

z2+5

x24+1°

[EXE]:Show coordinates

B

rl: 2.1(xZ+8))

(11

(RF+1

= X=-2.242084731-3:.8=0

T

e

the solutions are

e

I

The z-intercepts are ~ —0.577 and 0.577.

x ~ —2.24 or 2.34.

the solutions are

z~ —0.577 or 0.577.

4

a

i Wegraph y =1 B

y = 222 — 122+ 11

and

on the same set of axes.

ii We graph y = —7

[EXE]:Show coordinates

B

VI=2x7-12x+11

y = 22? — 122 + 11

on the same set of axes.

[EXE]:Show coordinates

YI=2x7-12x+11

INTSECT

The

graphs intersect at

(5, 1).

the solutions are

ili

We graph y = —10

INTSECT

(1,

1)

and

The graphs intersect (touch) at (3, —7).

.

= =1 or 5.

y = 222 — 122 + 11

the solution is = = 3.

and

on the same set of axes.

The graphs do not intersect. there are no real solutions.

b

2% — 120 +11=k

202 120+ (11 —k) =0

has

a=2,

b=-12,

c=11—k

A = b — 4ac

= (—12)? —4(2)(11 — k) =144 — 88 + 8k = 8k + 56

i

For two real solutions,

and

ii

A>0

8k + 56 >0 8k > —56 k>-—T7 which agrees with our results in a.

For exactly one real

solution,

.

A =0

8k+56=0 .8k = —56 oo k=-T7

iii

For no real solutions,

A—dac

=17~

—(—1)2

_

= Since

a=6,

b=—-1,

c=-2 b

46)(-2) L

A > 0, and 49 is a square, there are 2 distinct rational roots.

i~

o

_;:—A

nivE B

T T147 e o

Lrz=For -5 Lax=%2or—1 So there are 2 distinct rational roots

expected.

8

202 -52x+4=0

has

a=2,

b=-5

A =b% — 4ac

= (=5 - 4(2)(4) =-7 Since

A < 0,

there are no real roots.

c=4

as

Chapter 4 (Equations)

9

222 —3z+m =0

has

a=2,

b=-3,

Review set 4A

127

c=m

A = b — 4ac

= (-3) — 4(2)(m) =9—-8m

a

For a repeated root,

b

For two distinct real roots,

A= 9—8m = —8m

10

A>0 9—8m >0

= -9

m

=

¢

—8m

%

m

For no real roots,

.

A

%

%

For all quadratic equations with roots 4 and —3,

the sum of the roots = & + (—3) = -3

and the product of the roots = (3)(—3) = —3 So, we have

MoJB_5 a

and

2

The simplest solutionis

E=—§. a

a=2,

2

b=5,

c¢= —3.

all quadratic equations with roots % and —3 are of the form

k(222 +52—3) =0, 1

kER,

k#0.

If a and 3 are the roots of

222 — 3z = 4,

a+p

af

(-3) 3

=—-—~=-

35

and

4 =—=-2 )

For the quadratic equation with roots 1

1

a

B

and

o

and the product of the roots = (—) @

1

B+ a

af

= @

3

—a

-2—92

-2

-4

-3 So, we have

all =8 a

The simplest solution is

4

_Eul and

E=—l=—2. a

a=4,

the quadratic equation is

b=3,

2

222 — 3z — 4 = 0,

%, 1

the sum of the roots = — + — =

i

which can be rewritten as

4

c=-2.

4z + 3z -2 =0.

1

-

B8

then

128

Chapter 4 (Equations)

Review set 4A

12 ka?+ (1-3k)z+(k—6)=0

has

are the roots, then

If a and —1

«

a=Fk,

b=1-3k

c=k—6

— £ qEs5

a(—l)

k

a

«a

_k=6 k

—-k=k-6 = —6 —2k k=3

For

k=3,

the equation becomes

3z*+ (1 —3(3))z+(3—6)=0

32 -8z —3=0

.&3z+1)(z—3) =0

)

So,

13

a

k=3

and the two roots of the equation are

223 -322-924+10=0

8 Gl

.

=3,

r

1, or —2

a

G0k b

2

1

or3

and 3.

aX3 +bX2 +cX+d=0

Using technology, 5

7§

z=—3

-3

c

-9

8 Weilelies] [0k

aX3 +bX2 +cX+d=0

X1

d

xz{ X3

1 _i

PN

b

32% = a(7x — 2) 308 — 722 _ 9 =

..‘

—

32 — 722+ 22 =0

Hath Deg ore]

et e farn) (07}

aX3 +bX2 +cz(+d=0 a8 _ a

=

-7

3

aX3® +bX2 +cX+d=0 h

c

iflo.aaag

Using technology,

z=2,%0r0




6

14

i)

GEs

aoX4+a1 X3+ - -+aa=0 Xl[mi]

x2l-6.91

1.835003664

Chapter 4 (Equations)

14

a

We graph

y =10x 27!

and

on the same set of axes. B

y =35

b We graph

Review set 4B

y = /z

and

y:éfl

on

xT

the same set of axes.

[EXE]:Show coordinates

129

[EXE]:Show coordinates

Yi={x

Y2=(4.x)-1

fl INTSECT

The graphs intersect at

INTSECT

%

(2.81, 35).

The graphs intersect at (1.73, 1.31).

the solution is = ~ 2.81.

¢

We graph

=1.314606212

X=1.728163201

the solution is

= ~ 1.73.

y =% — Yz +5.

[EXE]:Show coordinates V1=x"(3)-(3{x)+6 13y

The z-intercept is ~ —1.84. the solution is =z ~ —1.84.

1

a

—722=0

b

—42® = 128

¢ (e-V3)7°=16

x:}:i%

2%=

=0

_

Lo

5

3

r—V3==+V16 r—V3==44

T=4""%

r=v3+4

oo VI NE)

z=-3 2

a

T

4

:%

b

81 — 1 r==+4/5

2 =—18 r = ¥/Z18

¢

(z—1)"2=4 (x—1)2=1

- V6

T =x3

\/[%

r—1==+

xi__“Sl

r—1=

]

::%

r=1%+

%

z=3or3

3

a

P=p q

. p=0,

b ¢#0

2xz

Th

202 =0,

y#0

z2=0,

y#0

=0

or

2=0,

¢ y#0

5

TwT

—5=

which is impossible

there are no solutions.

b

22% —b5r =0

322 - 120 =0 o 3x(zx—4)=0

ox(2r—5)=0 %

ox=0or

=0

22 -7z 4+6=0

2’ 4+4=—dz

d

(x=1)(z—-6)=0

a4t

o x=1 or

dr+4=0

o (@+2)%=0

6

Lo

—4r—12=0

(z+2)(x—6)=0 o

x=-2

a

1:2

or

6

13 =2

b

xT

sor?=9

¢

a

3zt —x=2 o3t —r—-2=0 Bzr2)(z—-1)=0

wax=-3%orl

a=1, e

b ¢=3

has

/52

—4(1)(3)

)

—54

—5::\/E r=_2=VvV>

.=

U

2?2 —6r+4=0 oa?—6r=—4

2% — 6z + (—3)> = —4+ (-3)? (x—3)2=5 T —

a=3,

s

2

a

3z2+11z-2=0

b=5,

2(1)

7

-3=0

x

224+5x+3=0 has

=3-227

b

b=11,

11+

¢=-2

/112 — 4(3)(-2)

—2(3) —11++/145

6

222 +8x =1 2?44z

=

=::\/§ $:3:Z\/5

Sk

6

2

2z +3)(z—1)=0 La=-2orl

=243

3p—1=2 Lo

oz

2%+

soz2—9=0 (z+3)(x—3)=0 LT

or 2

Low=—3

n m

5

(Bz+5)(z—2)=0

.

v | sk sl S5

s

322 —2—10=0

f

2 —12=4x

5

e

=-2

Do

L

4

or

I f

¢

x—4=0

3z=0or

.

2r—5=0

or

wx=0

Il |

a

Review set 4B

Do

L

Chapter 4 (Equations)

8

130

Chapter 4 (Equations)

¢

Review set 4B

131

42% — 52 =6

a? -3z =3

P Sot (-5 =3+ (-

@-§F=3+ -y= r- %=,/ :L‘—g—::

8

il

V64

zfgzzz% x:%::%

les—fiorfg

a

b

22-8x+16=0 has a=1, b=-8,

2:2—2z-5=0

¢=16

has

A =b* —dac

A =0,

-5

=41

there is one repeated root.

¢ 322 +52+3=0 a =3,

¢=

= (-1* - 4(2)(-5)

=0

has

b=-1,

A =b* —dac

= (—8)* —4(1)(16) Since

a =2,

b=5,

Since A > 0, but 41 is not a square, there are 2 distinct irrational roots.

c=3

A =1 — dac

=52 —4(3)(3) =—11 Since

a

A < 0,

there are no real roots.

i 2z(z+4)=8(z+k) When

k=29,

2z(z +4) =8(x +9)

22% + 8z = 8z + 72 222 =72 z? =36

il

2z(z+4)=8(x+k) When

k=9,

2z(z+4)=8(x+9)

We graph y = 2z(x +4) and y=38(z+9) on the same set of axes. B

[EXE]:Show coordinates

Y1§2z(x+4) Y2: (x+9)

145y

I=::\/%

I=::6

The graphs intersect at

(6, 120).

the solutions are

(—6, 24)

= = —6 or 6.

and

132

Chapter 4 (Equations)

b

Review set 4B

2z(z+4)=8x+k)

2%

c

+ 8z =8z + 8k co22% =8k ©oa? =4k

2% =4k

. x=+V4k Z=::2\/E

i

10

The equation

z? =4k

has two real solutions if & > 0.

i

The equation

22 = 4k

has one real solution if k& = 0.

ili

The equation

x? =4k

has no real solutions if % < 0.

ax® +br+c=0, i,

a#0 —b+

B

The sum of the

/b2

— dac

2a

. solutions

=

b+ /02 —dac —2~ V" —dac

T

=——,

a#0

a

has

2a

2a

b

202+ kr+12=0

|

2a

%

—b— /b2 —4 — —b= Vb7 dac

a=2

Ifa and 3« are the roots, then

b=k,

c=12

o+ 3a = b

—g

a

da

L

_E 2

=

- k=-8a c

and

e (1)

a(3a) = iy

12

3a% =6 a?=2

.

Using (1), o= ,g Ry

Substituting

« =

3

k.

}

into (2) gives

k)2

(—§>

=2

2

£64 N, k* =128

k=482

When k = 8V, and

a:fsf:f\/fi 3a = —3v2

So, the two roots are

—v/2

and

When k = —8V3, and

—3+/2.

So, the two roots are

a:J*ggfi) 3a =3V2 V2

and

3v/2.

)

Chapter 4 (Equations) 12

If p and q are the roots of 422 —3x —3 =0,

then

=2 3 Aty

p+q=——2

For all quadratic equations with roots p* and ¢?, the sum of the roots = p* + ¢*

Review set 4B

and the product of the roots = p¢®

= (p+9)* - 3p%¢ - 3p¢® =(p+a)® - 3(pa)(p+q)

— @7 _ 27

“uti

-3 3 =03 = 4

and

e

133

= (pg)®

=(-§)® _27 64

21

— 135 ~ 64 So, we have

b

135

a

64

—— = —

27

and

a

a=64,

The simplest solution is

Ly 64

b= —135, c= —27.

all quadratic equations with roots p* and ¢* are of the form k(6422 — 1352 —27) =0, 13

a

k€R,

z® — 15z = 222

k#O0. Wt Dediorn]

L

R

2® —22% — 150 =0

Using technology,

[Hath]DeglForn]) aX3 +bX2 +cX+d=0

SR

B

X1 X2

]

X3

-3

xr=2>5,0,o0r —3 DELETE

b

23422 - 6x+7=0

REPEAT]

o) Des)farml) _(d7e)es)

ot Oelforn) (7

Xl[EE:E

-8

X2L

0.4485.

2.390541979

REPEAT]

b

We graph y =2*

and y =9—2/

the same set of axes.

same set of axes. B

the

4.929142304

REPEAT]

PN

14

0.8144

[EXE]:Show coordinates

[EXE]:Show coordinates

Y1=x"(3) ¥2=9-2(x)

5

INTSECT

5 X=1.845270174f

The graphs intersect at (2.81, 7). the solution is = ~ 2.81.

13¥

5

INTSECT

-

'=6.2831855661

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

on

134

Chapter 4 (Equations) ¢

We graph

2

Review set 4B

y:%fx/:c

3.

[EXE]:Show coordinates

The z-intercepts are ~ —2.15 the solutions are

and 3.58.

x ~ —2.15

or 3.58.

Chapter 5

SEQUENCES AND SERIES

1

a

4,13, 22,31

b

N A AA

45, 39, 33, 27 A

+9 +9 +9

2

3

A A

2,6,18, A

—6 —6 —6

2,3,5, 7,11, 13,17,

a

¢

x3

A

x3

A

54

d

x3

96, 48, 24, 12 AN

2

A

2

u=3

b

us=11

¢

ujo =29

b

u

{the 10th prime number}

4,7,10,13, 16, ... a

We start with 4 and add 3 each

¢

ug=us+3+3+3

time.

=4,

us=13

wu,=2n+5

a b

u;=2(1)+5

=9

=11

=13

° ° °

n

0

2345

>

u,=3n-2 a

u1=3(1)—2

b

=l

6

ug =2(4)+5

°

6 4 2 0

ug =2(3)+5

AU

8

5

up =2(2)+5

=

14 12 10

U5=3(5)—2


which are n = 7691 uzgor = 13(7691) + 10 = 99993 and wurgg2 = 13(7692) + 10 = 100 006 So, the first term to exceed 100000 is

a

wu7g92 = 100 006.

Upg1 = Un + 7 Up41

— Up

=

b

7

So, consecutive terms differ by 7. the sequence is arithmetic.

¢ Let

and

U, =uy + (n—1)d VL.

U200

=

—12+

.

ugee = 1381

199(7)

w,=1000=—12+7(n—1) Tn—7=1012 7n = 1019 n =145%

but » must be an integer, so 1000 is not a member of the sequence.

a

32,k 3 Since the terms are consecutive,

k—32 =3 —k

{equating differences}

2k =35

k=173

b

k7,10 Since the terms are consecutive,

7

7

k=1 0—7 k=3 k=4

{equating differences}

n = 7692:

143

144

Chapter 5 (Sequences and series) ¢

k,

2k—1,

Exercise 5B.1

13

Since the terms are consecutive,

2k —1—k =13 — (2k —1)

{equating differences}

k—1=14-2k 3k=15 k=5

.

d

k

2k+1,

8-k

Since the terms are consecutive,

2k +1—k=8—k

— (2k+1)

{equating differences}

k+1=7-3k . 4k=6 e

2k+7,

3k+5,

L h=$-3

5k —4

Since the terms are consecutive,

3k+5—(2k+7) =5k —4 — (3k+5)

{equating differences}

k—2=2k-9 k=T f

2k+18,

—2—k,

2k+2

Since the terms are consecutive,

—2—k—(2k+18)

=2k +2— (-2 —k)

{equating differences}

—3k—20=3k+4

—6k =24 L k=-4 g

k k% Kk*+6 Since the terms are consecutive,

k? —k =k? +6 —k?

kK —k—-6=0

{equating differences}

(k+2)(k—=3)=0 wk=-2o0r3 h

5,k k*-38 Since the terms are consecutive,

k —5=k®

ook —26-3=0

—8 — k

{equating differences}

(k+1)(k—-3)=0 wk=-1or3

9

10k+1, a

2k,

4k* -5

Since the terms are consecutive,

2k — (10k + 1) = 4k* —5 — 2k

2%k —10k—1=4k* —5— 2k ;. —8k—1=4k*>—-5—2k

s 4K +6k—4=0 202k +3k—2)=0 L2k —1)(k+2)=0

1 k=_5

or —2

{equating differences}

Chapter 5 (Sequences and series)

b

For k=1,

Exercise 5B.1

the common difference d = 2(%) — (10(3) + 1) =1-6

=5 For k= —2,

the common difference d = 2(—2) — (10(—2) +1)

= —4—(~19) =15

10

up =41

.

ug =77

.

up+6d=41

.. (1)

ou +12d =77

... (2)

{using

u, =u; + (n —1)d}

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

—u; —

6d=—41

{multiplying both sides of (1) by —1}

uy +12d =77

.

6d =36 . d=6

So, in (1):

{adding the equations} uy + 6(6) =41

Check:

—1 wr =6(7)

w436 =41 cooup

Now

=5

=42—-1

wu, =uj+

(n—1)d

=41

U, =5+6(n—1) L

upz = 6(13) —1 =78-1

up,=6n-1

=77

U5 = —2

upp =—123

cooup +4d = =2

-

wp+1ld=-128

Vv

a1

{using

V

w, =uy + (n —1)d}

.. ()

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

—up —

4d =2

{multiplying both sides of (1) by —1}

up +11d = —123

7d=—103 L d=-} So, in (1):

{adding the equations}

up +4(-3)=-2

Check:

Loup—6=-2

Now

U5:7%(5)+1—21

4800.

Using a graphics calculator with Bl

we view a table of values:

_HatiRadforn]) (b/dRel

e

(»I3)Y:(x—1) 12

13

14

15

[FORULH|

1683.56

2916 8748

5050.660155 EDIT J GPHECONIGPHEPLT

The first term to exceed 4800 is

¢

Y; =4 x (v/3) (X — 1),

w14 = 29161/3 ~ 5050.

12,6,3,1.5, ... The sequence is geometric with

u; =12

and

r = 3.1

ol

10

Exercise 5C

u, —=12 x ()1yn—1 cooup=12x21 We need to find n such that

u,, < 0.0001.

Using a graphics calculator with

Y7 = 12 x 2(1

— X),

we view a table of values:

B HetRedforn] [b/dRes) Y1=1%>

{20% = 0.2}

= 1280 So, after 3 years the value of the lathe is €1280.

w4, = ug (

100 z)‘“l

100+7r/

168

2

Chapter 5 (Sequences and series)

a

us=upx(1-d)?°

= 110000 x (0.75)° ~ 26103.52

Exercise SE.5

{25% = 0.25}

b

The depreciation = €110 000 — €26 103.52

b

The depreciation = ¥87 500 — ¥30013

= €83896.48

So, after 5 years the value of the tractor is

€26103.52.

3

a

uz=upx(1-d)?

= 87500 x (0.7)>

{30% = 0.3}

=30012.50

= ¥57487

So, after 3 years the value of the laptop is

¥30013. 4

U4:u0>0,

n~

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.

g

¥2-590

:

isoss ks 0003

198

Chapter 5 (Sequences and series)

d

i

S,= LLII) "

Exercise SH

for a geometric series

I, — 40000 % (1L05)" 1.05—-1

- w.05

il

1)

S, = g(2u1 + (n—1)d)

Tp = 2(2(60 000) + 1000(n — 1))

= 60000n + 500n(n — 1) = 600001 + 500n° — 500n

.

.= 500n° 2 + 59 500n dollars

= 800000((1.05)" — 1) dollars e

i

for an arithmetic series

[Initially Option B is better than Option A, so Tp > T, for small values of n.

total income ($)

graph 1 graph 2

graph 1 represents 74, graph 2 represents 1.

n (years)

il

The point P is where 7'y meets Tz, which is when

800000((1.05)™ — 1) = 500n2 + 59 500n. i

graph

TBA—= 5QOn

s00 000(1‘05

=

TA

e

1)

and

2 +459 500n' on the same set of axes and find

[EXE]:Show coordinates

VI=80000x(1.05"(0-1) || e e B

their points of intersection.

INTSECEK

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

15

Ds2288661

Option B provides the greater total income for

=l577367.$97

0 < n < 22 years.

a A3=A,x103—-R

= ($8000 x (1.03)2 —1.03R— R) x 1.03— R

= $8000 x (1.03)® — (1.03)*R — 1.03R — R

b Ag = $8000 x (1.03)® — (1.03)"R — (1.03)°R — (1.03)°R — (1.03)*R — (1.03)°R —(1.03°R-1.03R— R ¢

Ag=0

©. $8000 x (1.03)% = R(1 + 1.03 4 (1.03)% + (1.03)® + (1.03)* + ... + (1.03)") i

. $8000 x (1.03)° ~

1% ((1.03)8 — 1)

s _

=R (T)

i

g (r 1)

{using S, = 7}

~$8000 x (1.03)% x 0.03 R (1.03)8 — 1 = $1139.65

d

Notice in ¢ that P =8000 r=20.03

and

m=8

So, each repayment is

R =

and

(1.03)% = (14-0.03)% = (1+7)"

P(1+7r)m xr

(1+r)m—1

dollars.

Chapter 5 (Sequences and series)

3

3

27(m>7(moo

o

Erool

gl

(3

e

Exercise 51

il

[o:

()

10

the series is geometric with u; = %

and r = .

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

b

We need to show that

Now

0.3 = 3.

03=35+ 3+ 35+

n — oo,

Since

35+

Sp=3+

So,let

S =

then

+

B

1—r

3 -

_1 =3

10

1- (1)

0.3 =13 2

a

04=0.444444....

.4

_ 4,

1

as required b

4

0.16 =0.161616....

is an infinite geometric series with u; == 4

and

-

5=

.

_ =4

=%

1 r = ==.

=

up = 155

and r =

u o8 =—L 1—r

16 _ _ T

T

e :

0.312=0.312312312.... 312

- L100

99

o1 s

= 03=4%

¢

312

=S40 103 b 106

312

g 109

is an infinite geometric series with

16

. e I . . . is an infinite geometric series with 1 16

1-r4

__T1 .

16

— 16

=16t 100 T 000 + -

312 u; = 1606

and

r = 1000+

99

5.

199

200

Chapter 5 (Sequences and series)

3

Checking Exercise SH question 8 d:

L

a

18+124+8+

1—36 + ....

Exercise 51 S =

is an infinite

189-6.34+2.1—0.7+....

r=—1.E

r=2. e

1—r

)

18 5

i

1-

=2

5

2+ 2+ & +....

a 1;1 T

5=

(=Y

"

)

W

)

;

and r = 1.

is an infinite geometric series with u; =3

1—7r

3

1

_

B

=1 b

189 e

=14.175

=54

S

1-1

T

=6—6x(§)+6x($)*—...

];06 (—g)

.

.

.

i

.

.

isan infinite geometric series with

r=—%

and

5=

2\2

2

ok

e

up =6

18—

6

=

Bl 1-(-2

30

=7 6

(42

(=43)

Let the terms of the geometric series be uy +urr

4+ ugr? =19 19

“= i Equating (1) and (2),

uy,

uyr,

and

ur(l+r+r?) =19 .

18.9

=

g—_"m

_w =

isan infinite

u;

series with

geometric

and

u; = 18

geometric series with

b

wyr?, S=

...

lu—lr =27

wn = 271 — 1)

W 19

T

27(1—r)

L-@1-r1+r+r?

Bottr4+r?—r—r>—13 ;—57’:177"3

=25

,=1

Substituting r = 2 into (2) gives u; =27(1—3) =9 the first term is 9 and the common ratio is %

@

and

Chapter 5 (Sequences and series) 7

Let the terms of the geometric series be

and

ulr:%

8 up =— ! 5r

.

Ul

S=

©oup

M

uir,

uir?,

201

..

=10

1—r

(1

Equating (1) and (2),

wuy,

Exercise 51

. Q)

=10—10r

35r _10-10r 8 = 50r — 501>

o 50r? —50r +8 =0 2(25r% — 25r 4+4) =0

2(5r —1)(5br —4) =0 r=1

Using (2),

or#

uw =10—10(3) =8

if r=

up =10-10(3) =2 either

wu; =8,

r=1

or

w=2

r=4%

Let the terms of the geometric series be S3=uy

+wur +ur?

uy,

uir,

uyr?,

and

=21

..

S=

Cu(l+r+r?) =21 o

Equating (1) and (2),

21

=—-7

.. (

14+r+72 21

M

64

——=F1-r

T+7r 472

5(

)

21x3 = =(1-r)(1+r+7r?) G64 _

g2

2

_p—r2_y

3

63 _1_ .3 Gg=1-r 1 L3 A’I"—64

So,

Sy =wuy

+urr

+uir® 2 + urr’ 3 + ugr

4

)=% __

64

X

wlee

Substituting r = 41 into (2) gives uy = & (1

=

=1

= 16.

=u (1 +r+r24+r3 40

1601+ 3+(3 + 3+ )

_=763 9

a

Total time of motion

=1+ (90% x 1) + (90%x 1) + (90% x 90% x 1) + (90% x 90% x 1) + (90% x 90% x 90% x 1) + .... =1+40.9+0.9+ (0.9)% + (0.9)% + (0.9)* + .... =1+2(0.9) +2(0.9)% + 2(0.9)° + ...

as required

ground

202

Chapter 5 (Sequences and series)

b

Exercise 51

The total time of motion can be written as So,

S, = M

—1,

-7

where

uy =2,

[2 4+ 2(0.9) 4+ 2(0.9)? +2(0.9)3 +...] — 1 r=0.9

g, 200

1-09

g, —20-08m 0.1

S, =20(1—0.9") —1 S, =20-20x09"—1 S, =19 — 20(0.9)" ¢

For the ball to come to rest, n must approach infinity. As n— o0, 0.9" — 0 and so 20 x 0.9 — 0 also. Sp — 19

(from below)

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

10

Total distance travelled

—h+2(2)h+2(3) (E)h+ ..

b @R[ 3+ @) ()]

()

But

o e

7h =490,

s

so h=70.

The ball was dropped from a height of 70 cm.

11

a

09= 3%+ 135+ 1055 + -~

which is geometric with

uy = %

and

r =15

S

09=9=-""=1 -1

b

9

un*fi


0}

V48.17

Surface area = (8.2)% + 4 x ($ x 8.2 x V48.17) m?

~ 181 m?

i

o 4.1m

236

Chapter 6 (Measurement)

;

3

Exercise 6B.1

+«—219ecm——>

81lcm

34cm v

a

219¢m

The area of the top surface is the same as the

bottom surface. This area is a trapezium, so total area of top and bottom surfaces

81cm

=2x

[ b

(75+219)

x 81 cm?

= 23814 cm? (219 —75) =144 cm

-

22 =812 + 144>

{Pythagoras}

2% = 27297 =/27297

81lcm

{as z > 0}

75cm

So, the four sides are:

xn A 8lcm area

area

=81 x 34 cm? = 2754 cm? ¢

219cm

=219 x 34 cm? = 7446 cm?

[ V27297

75cm area

area

=75 x 34 cm? = 2550 cm?

— /27297 x 34 cm? ~ 5617 cm?

The surface area = area of top and bottom surfaces + area of four sides

A 23814 + 2550 + 2754 + 7446 + 5617 cm?

~ 42181 cm? ~ 42181+ 10000 m? ~ 4.2181 m?

*.

timber cost ~ 4.2181 m? x €128/m?

~ €539.92 ~ €540

em

{using a and b}

Chapter 6 (Measurement) 4

a

Exercise 6B.1

25cm

16cm

90 cm

The surfaces of the steps are: side

back

front of step

48 cm

25cm 75¢cm

90cm

area

90cm

area

=25 x 16 + 50 x 16 + 75 x 16 cm? = 2400 cm?

area

=90 x 48 cm? = 4320 cm?

=90 x 16 cm? = 1440 cm?

base

top of step

75¢em

90 cm

90 cm

area

area

=90 x 25 cm? = 2250 cm?

=90 x 75 cm? = 6750 cm?

. the surface area = area of two sides + area of back + area of three front of steps + area of three top of steps + area of base

= 2 x 2400 + 4320 + 3 x 1440 + 3 x 2250 + 6750 cm?

= 26940 cm®

237

238

Chapter 6 (Measurement)

Exercise 6B.1

25m

b

The sides of the pool are: side

deep end side

25m

.

12m

=(18+11)X25m2 841

12m

o

1.1m

area

shallow end side

[ ]

area

area

=12x 1.12m m?

=12x1.8Em m? =21.6 m

= 36.25 m?

=132 m

The base of the pool is:

22 =252+ (0.7)> coox=1/252+(0.7)2

12m

= v625.49

rm

area = v/625.49 x 12 m?

~ 300 m?

-,

the surface area = area of two sides + area of deep end side + area of shallow end side + area of base

~ 2 x 36.25 + 21.6 + 13.2 + 300 m? ~ 407 m* The sides are:

38cm

AL

38cm

8.5cm

area = 8.5 x 38 cm? =323 cm?

1.1m

{Pythagoras} {as x>0}

Chapter 6 (Measurement)

Investigation 1

Archimedes and the sphere

239

area of triangle = § x 8.5 x 7.35 cm?

The ends are:

= 31.2375 em? -, -,

area of end = 6 x 31.2375 cm? = 187.425 cm?

total surface area = 6 x sides + 2 x ends

=6 x 323 + 2 x 187.425 cm? = 2312.85 cm? ~ 2310 cm? 6

a

Surface area of prism

=2xz(r+2)+2

e

x z(2z) + 2 x 2z(x + 2) cm?

= 2z(x + 2) + 22(2x) + 4o (x + 2) cm? =222 + 4z + 42% + 42% 4 8z cm? = (102® + 12z) cm? b

2 om

triangular side

zcm

(z+2)em

Let the height of the triangular sides be h cm.

h2 4+ (%)

2

= g2

{Pythagoras}

22

T cm

S [1

Fom

h=

x2,I

_ = 4

_ V3 2 Surface area of pyramid = area of base + area of four triangular sides

=zxx+4x 2

2 I(\z/gw) om?

= 2% + V32% cm? = (1+v3)2? em?

1

Curved surface area of the cylinder = 2 x 7 x radius x height =2XTXTX2r = 47r?

2z cm

cm?

{as

h >0}

240

Chapter 6 (Measurement)

2

Investigation 1

Archimedes and the sphere

a

Curved surface area of the slice = 2 x 7 x radius x height =2X7TXrxAy =27rAy

b

In the diagram below,

2 +y? =r? a2 =12

{Pythagoras} g2

Using g similar triangles, gles,

— Bs Ay = —— rAs =rAy

d

The curved surface area of the slice is a rectangle with length 27z and width As. -, the surface area of the sphere from this slice = 27z x As

=2mrAy 3

{using ¢}

Each thin slice of thickness Ay contributes 27rAy to the surface area of the sphere and the surface area of the curved surface of the cylinder which just contains it. So, the surface areas are therefore equal as we sum these contributions over the height 2r.

-,

the surface area of the sphere = the curved surface area of the cylinder =27r X 2r = 4nr?

Chapter 6 (Measurement)

Exercise 6B.2

EXERCISE 6B.2 e ;T;cm

v

b

8cm

4.5km

A = 2nrh 4 2nr?

=2xTXx8x12+2x 7 x 8 ~ 1005.3 cm?

The diameter

d = 4.5 km,

so the radius

r = %

= 2.25 km.

A = 4nr?

=4 x 7 x (2.25)?

~ 63.6 km? hollow

6cm

7

10 cm

|

The cone is hollow at the top, so we only have the curved surface.

The diameter

A=nmrs

.

so the radius

=mx6x10

~ 188.5 cm?

d = 14 m, 14

r = ik

7 m.

A =7rs +mr?

=7 x7Tx 1847 x 72 ~ 549.8

hollow top only

m?

f

3cm

12cm

L—

20cm —J

The diameter .

so the radius

4

+ area of flat end

d = 20 cm, 2

r = ?0 =10 cm.

The cylinder is hollow at the top, so we

only have the curved circular end. A = 2rrh + 7r?

=2x7x10x ~ 1068.1 cm?

A = area of curved surface

surface and one

1247 x 10

=1 xdmr? 4 or?

=2x7mx3%+mx3? ~ 84.8 cm?

241

242

Chapter 6 (Measurement)

Exercise 6B.2

b

8cm

4.6m

A = 2nrh + 2nr? =2Xx

A = 47r?

=4 x 7 x (2.3)? ~ 66.5 m?

7T Xx8x36+2xmx8?

~ 2210 cm?

d

A=mnrs+nr

2

Let the slant height of the cone be s cm.

=7 x 38 x 86 + 7 x 382 ~ 14800 mm?

Now

5% =

(1.6)% + (1.2)?

coos=1/(1.6)24(1.2)2

=1

=l A = 7wrs + wr?

=7 x1.2x2+7x (12)

~ 12.1 cm?

{Pythagoras}

s2 =22 452

s =1/22 +52

=29

{as s> 0} 5m

1

~ 5.39

b

Area of canvas = 7rs + 71> =7 x2x V2947 x 22 ~ 46.4 m?

F—Zm*‘

approximately 46.4 m? of canvas is required to make the tent. ¢

Cost of canvas = area of canvas

x cost per m>

~ 46.4 m? x $18/m?

~ $835.24

{Pythagoras} {as s> 0}

Chapter 6 (Measurement)

4

a

The diameter d =8 m, so the radius

T

r = g =4

m.

6m

Area of base = 772 =7 x 42

i

~50.3 m®

b

Exercise 6B.2

4m-f

Cost of lining the base = area of base x cost per m?

~ 50.3 m* x $23.20/m? ~ $1166.16

¢

Area of curved wall = 27rh

=2X7TXx4x%x6

~150.8 m? d

Cost of lining the curved wall = area of wall x cost per m?

~150.8 m* x $18.50/m? ~ $2789.73

e

Total cost of the lining = cost of lining the base + cost of lining the curved wall

~ $1166.16 + $2789.73 ~ $3960 5

Area of top = b

360

{using b and d}

10cm

X mr?

=50 % ™ 10 _ 95

2

‘

~ 82.9 cm?

5cm

Area of curved surface = %

_ 5955 =

x 27rh

X2x7x10x5

~ 82.9 cm? Area of two flat surfaces = 2 x length x width =2x10x5

= 100 cm?

Total surface area to be iced = area of top + area of curved surface + area of two flat surfaces

A~ 82.9 + 82.9 + 100 cm? ~ 266 cm”

6

a

The radius is r, so the diameter is 2r.

Since the height is the same as the diameter, the height is also 2r. Surface area = 27rh =2XTXTX2r

= 4772

=y

hollow top

and bottom

243

244

Chapter 6 (Measurement)

b

Exercise 6B.2

The surface area is 91.6 m?

. Arr? =91.6

;2 9L 47

91.6

r=4/"

{as

r >0}

~ 2.70

h=2x270

{h=2r}

~ 5.40 So, the height of the cylinder is approximately 5.40 m.

7

a

=T

r

hollow

Surface area = 7rs X7rX3r

/

= 3nr?

3r

b

i

The surface area is 21.2 cm?

- 3mr? =21.2

Lo,

!

2y

212 3

Y oo

.

21.2 e

{as

r >0}

r=1.50

3r~3x1.50 s &~ 4.50

So, the slant height of the cone is approximately 4.50 cm. ii

Let the height of the cone be h cm.

h? 412 = (3r)2

{Pythagoras}

h? 472 = 9r?

hcm

h* = 82

h=372

~ /8% (1.50)2

{as

h >0}

{from b i}

~4.24

So, the height of the cone is approximately 4.24 cm.

Chapter 6 (Measurement) Surface area = 2mrh + 2712

Exercise 6B.2

T

=2XTXEX2+2XT Xz = d7a? + 2mwa?

2zm

= 6mz? cm?

emol Surface area = £ x d7r?® + 77

rem

= 27172 4 7r?

= 37r? cm?

Let the slant height of the cone be s cm.

s* = 2% + (22)?

{Pythagoras}

= 52

s =5z

{as

s> 0}

Surface area = 7rs + w12

=7 xzXxVbr+7xaz? =\/§7rx2+7rz2

= m2? (V5 + 1) em? Surface area of a sphere = 4772

The surface area is 647 cm? 47r? = 647

.

=16

r=4

{as r>0}

The radius of the sphere is 4 cm. Surface area of a solid cylinder = 27rh + 2772 The radius is 6.3 cm and the surface area is 1243 cm?

2x7x6.3xh+2x7mx6.3%=1243 12.67h + 79.387 = 1243

12.6mh = 1243 — 79.387

j_ 1243 — 79.387 -

12.67

h =~ 25.1

The height of the cylinder is approximately 25.1 cm. Surface area of a cone = 7rs + 772

The slant height is 143 mm and the surface area is 60 000 mm?

7 x 7 x 1434+ 7 x ¥ = 60000 1437r + ar? = 60000 7r? + 1437r — 60000 = 0

Using technology,

r ~84.1

or

—227

but

>0

The radius of the cone is approximately 84.1 mm.

..

r~84.1.

|

245

246 10

Chapter 6 (Measurement)

Exercise 6B.2

a

4

Surface area

1.2m

= surface area of hemispherical top + surface area of cylindrical base

¥

=1xdmr® + (2nrh +mr?) =2x7mx(1.2)? + 2x7x1.2x28+7x ~ 34.7 m?

2.8lm

b

Surface area

‘

= surface area of half cylindrical top + surface area of rectangular prism base

2m

om

=1 x (27rh + 27r?) F(2x6x2+2x12%x2+4+12x%6)

6m

=7 x3x12+7x3%

~ 285.4 m?

T

&3

4 (24+48+72)

Surface area = surface area of hemispherical top +

surface area of conical base

= % x 47

6cm

=2x7x22

+

7rs

+ Tx2x6

~ 62.8 cm? 11

Surface area of a sphere = 4772

The surface area is ~ 7.618 x 10° km?

4mr? 2 7.618 x 10° r

2

7.618 x 109 4

TrR

7.618 x 109

{as

>0}

~ 24600 The radius of Neptune is approximately 24 600 km. 12

a

Arc length AB = 3%0 x 2 x 7 x radius of sector 0

=—

360

_

A

X 27s

h

Oms

T 180

b

Arc length AB = circumference of base circle O

180

= 2mr

9=

360r S

‘

B

(1.2)

Chapter 6 (Measurement) ¢

Exercise 6C.1

Surface area of cone = area of sector + area of base circle 0

.

=_— x7s?> + 7>

{using a and b}

360

360r s

=

360

X

TS

2

2

r

g

= 7rs + mr?

Ny

’

2.3cm

2.6cm

——

_+_ 5cm

4.2cm

V' = length x width x height

V' = area of cross-section x length . 3

=4.2% 26 x 2.3 cm®

B

=25.116 cm®

L

c

d 13.8cm

2 o

O —f 13.9cm

g

—120cm—=

S0cm

i

V' = area of cross-section x length

=

I&Q Cn-‘l

(45 Xx base base xx heiheight ght) ) xx len length g

V = mr2h — 7 x (82)2 x 13.9 om®

_= (5(1 x86x13.8) x12.9 cm 3

= 765.486 cm® e

~ 2940 cm?

80cm

V' = area of cross-section x length

= (22 x 7 x (0.8)°) x2m® ~3.13 m®

f

{80 cm = 0.8 m}

Let the prism have height ~ cm.

h*+5% =13 RZ 495

13cm 24

cm

10cm

Cop2h

..

s

Volume = area of cross-section x length

= (3 x10x12) x 24 em® = 1440 cm?

=169

—

{Pythagoras}

13cm

144

h=12

{as

h >0}

5cm

O

247

248

2

Chapter 6 (Measurement)

a

Exercise 6C.1

Let the equal length sides be x mm.

3z =900 x =300

.

So, the end is made up of six 300 mm by 300 mm squares. V = area of end X length

rmm

900 mm

= (6 x 300 x 300) x 1200 mm? = 648000000 mm?®

|

=

V = area of end X length

= (area of large circle

46m

=(rx1.22 -7 x 0.8%) x 4.6 m*

~11.6 m®

6cm 10cm

7

3cm

7cm

V = area of end x length = (area of large rectangle — area of small rectangle) x length

=(7x10—-3x6) x 3 cm®

= 156 cm? a

The diameter

d =1m

so the radius

r = % =0.5m.

So, the external radius of a pipe is 0.5 m. b

internal radius = external radius — width of concrete

. r=05-0.05m =0.45m

So, the internal radius of a pipe is 0.45 m. c

r=0.8m

— area of small circle) x length

= (mR? — 7r?) x length

3

1200 mm

Volume of concrete necessary to make one pipe = volume of whole cylinder — volume of hollow section

=7 x(0.5)%x25 — mx (0.45)% x 2.5 m® ~0.373 m®

R=12m

Chapter 6 (Measurement) 4

a

Depth of floor = 120 mm

—120+ 10

Exercise 6C.1

249

0.12m=

100 m

3~

6.5m

=0.12m

9.2m

Volume of concrete = length x width x depth

=92x65x0.12m’ =7.176 m*

b

7.176 m® of concrete is needed. Since concrete 7.2 m? of concrete will need to be ordered.

is only supplied in multiples of 0.2 m?,

Cost of concrete = volume of concrete x cost per m*

=72m* x $135/m® =$972

So, it will cost $972 to concrete the floor.

5

a

b

The diameter of the small circle )

12

radius

r = ol

so the

6 m.

So the radius of the large circle tm

d = 12,

R=7r+1=7m.

Surface area of concrete = area of large circle — area of small circle =7R? — 7r? =7

x 7 —7x62

~ 40.8 m? ¢

Volume = surface area of concrete x depth

~40.8 x 0.1 m® ~4.08 m®

{as 10 cm = 0.1 m}

approximately 4.08 m? of concrete is needed for the path. Length of piping required = 1000 km

= 1000 x 1000 m = 1000000 m Internal diameter of piping = 13 mm

= (13+ 10+ 100) m =0.013 m internal radius = 0.013 =2 m = 0.0065 m External radius of piping = = = = =

internal 0.0065 0.0065 0.0065 0.0085

radius + wall m + 2 mm m + (2 + 10 ~ 94.248 m3

250

Chapter 6 (Measurement)

Exercise 6C.1

Weight of plastic required = volume of plastic required x weight of plastic per cubic metre ~ 94.248 m® x 0.86 tonnes/m* ~ 81.1 tonnes

7

a

Volume of garden = length x width x depth

=86x24x0.15m*

=3.096 m®

{15ecm=0.15 m}

Volume of trailer = length x width x height

=22x18x(06-02)m*

{60cm=0.6m,

=1.584 m®

Number of trailer loads required =

20 cm=0.2m}

volume of soil required

volume of soil per trailer load

~3.096 m? T 1.584m3 ~ 1.95

So, T will need 2 trailer loads of soil.

b

Cost of soil = number of loads X cost per load =2 x $87.30 = $174.60

¢

i

Area of garden = length x width =8.6 x 2.4 m?

Number of loads of bark needed area of garden _ "

= 20.64 m?

area covered by one load

~20.64 m? T 1im? ~ 1.88

So, I will need 2 trailer loads of bark.

il

Cost of bark = number of loads X cost per load

=2 x$47.95 = $95.90 d

Total cost of establishing garden = cost of soil + cost of bark

= $174.60 + $95.90 = $270.50 8

a

Let the triangular prism have height h cm.

h* 4+ 752 = 125%

h? + 5625 = 15625 h? = 10000 h =100

{Pythagoras}

{as h >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® =15m’

125cm

75cm

Chapter 6 (Measurement)

¢

Exercise 6C.1

251

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

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

T

P

hcm

5cm

3cm

5x3xh=40 h— 40

om?

T cem

Let the sides of the cube be z cm.

V = 3401 cm?

Sz Xxaoxx=34.01 ooa®=34.01

oz =3401

s

h=2%~267

~ 3.24

The side length is approximately 3.24 cm.

The height is approximately 2.67 cm. ¢

rem

rem

Let the height of the rectangular prism be

om0

e

Let the radius be r cm.

V =43.75 cm® T xr?x4.6=43.75 o

L

T

4375

ax46

r=a

4.6cm

43.75

e

{as

r >0}

~ 1.74

The radius is approximately 1.74 cm.

10 6cm

8cm

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

h* 4+ 1.5% = 62

{Pythagoras}

h=1+/62—1.52

=V33.75

{as

h >0}

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

480 — (5;—8) x \/33.75 x length 480

length = ——

¢

13./3375

~12.7

The length of the gold bar is approximately 12.7 cm.

15cm

252

Chapter 6 (Measurement)

10

>k

1

Investigation 2

Volume formulae

100

10(1041)

il

- 2

102

>k

100(100+1)

1002

10000

k=1

100

2

_ 110 © 200

10100 "~ 20000

L

__

=1 =055 1000

i

101 __ =101 — 0.505

>k

1000(1000+1)

10002

1000000

£=L

10 000

:

2

iv

1001000

k=1

k

100002

_

= —

100000 000 100010000 "~ 200000000

~ 2000000

= 4358 = 0.5005 b

As

n—oo,

10

2

k=1

a

k2

103

10001 __ = 10001 — (.50005 __

2L

.

100

S k%

100041)(2(10)+1) 6

2310

i

1000

100(100+1)(2(100)+1)

k=1

e

1003

=21231 -0.385

1000

e

m

10003

==

= 20301 20301 __— (.33835 __

1000(10004-1)(2(1000)+1 6

1000000000 2003001000

S

k=1

K2

6

__

__

>k

k=1

=

100003

=

10 000(10 000+1)(2(10000)+1)

"~ 6000000000

200030001 = 200080001 , ).333 38

n

3

10000

1000 000 000 000 2000300010 000 "~ 6000000000000

2003001 = 200300 1 ~ (0.33383

n— oo,

6

1000000 2030100 6000 000

"~ 6000

> k2 k=1

10000(10 000+1) 2

2

5= _ %1

b

As

a

There are n prisms with equal thickness and total height h. the thickness of each prism is fi. n

Chapter 6 (Measurement) b

Investigation 2

Volume formulae

253

From the diagram alongside, the distance from .

.

hk

the apex to the base of the kth prism is —. n

hk o

X,

.

L

hk n

_L

=% = %

Using similar triangles,

h

x

X

-

T

_F

-

n

=t n

¢

The kth prism will have *.

length = width = x—k,

and height

n

E n

the volume of the kth prism = length x width x height zk

—

n

_xk _

X =X n

h n

_h (zk)2 -

n

n

The volume of the pyramid can be approximated by the sum of the volume of all n prisms.

2

n

S

o

(RN &

Na b

o Vay =1

oT)

=X =1

2hk2_2hk§1

Thm=oh

=5

n

d

As n—oo,

> K

B=L— n

1

{from2 b}

n

as n— oo,

> K

a?h =L n

— 2%hx =1

x base area x height

As the number of prisms in our approximation approaches infinity, the volume is given by : 1 1,27 52°h = 3 X base area x height.

a

There are n cylinders of equal thickness and total height h. the height of each cylinder is E n

Let Ry, be the radius of the kth cylinder.

From the diagram alongside, the distance from the apex to the base of the kth cylinder .

hk

is —. Using similar triangles,

—&o = %

Ry

&

T

nk

.

n

% L

L

h

™\ |

O

254

Chapter 6 (Measurement) b

Investigation 2

Volume formulae

The volume of the kth cylinder = base area x height h

=T xRZx = n

h (rk)2 —ho(rk n

n

The volume of the cone can be approximated by the sum of the volumes of all n cones. So,

Soh (k& k2 g_:lk Vx> ’flfl-(’n) B Zm’th3 = 7r2h ’n3 k=1 k=1

2

n

> K

¢ As n— oo,

k:n13

— %

{from2 b}

n

> K

as n— oo,

mr?h =L n

— mr2hx

The volume of the cone is given by

5

a

=1

x base area x height

2mr?h = % x base area x height.

There are n solids of uniform cross-section with equal thickness and total height h. the height of each solid of uniform o)

cross-section

1S

—. n

Let A, be the base area of the kth solid of uniform cross-section.

The base of the kth solid of uniform cross-section is a reduction of the base of the .

:

tapered solid with scale factor

n-a(l)

k

—. n

n

b

The volume of the kth solid of uniform cross-section = base area x height

-4 (E)

2

Lafty n


0}

l O

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

~22x13 + 2x3x22%x1.67 + 2x 1 x13x1.89m? ~ 8.99 m?

277

278 4

Chapter 6 (Measurement)

Review set 6B

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(2x1.2x15)+2x5x(1.2x0.75)

+6x (3 x1.2x 1.04) m?

= 23.544 m? ~23.5 m? 5

Length of cylinder is 325 mm = 32.5 cm, and radius of cylinder is . Surface area of cylinder

I fe— 325 mm ——>|

4—20 =20 mm = 2 cm.

= 27rh 4 272

=(2x7x2x325)+ (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=

% (area of base x height)

=3 x7x(44)° x 81 em®

‘ 8.1cm

~ 164 cm®

L—/ 4.4cm

b

3m

V = area of end x length

=(Bx5-1x2x1)x5m’

T

=24 x5 m? =120 m®

5m 5m

¢

V=

%m“n’ 43 X

273

~ 10300 mm®

3

;

]

40

TR

Chapter 6 (Measurement)

7

a

The front face of the letter F is made up of 3 rectangles.

Review set 6B

0.5m

Volume of letter F

V/A

= area of front face x length

0.4m

=1.04x 0.5 m® =052 m?

0-3m 04m

= (1.7x04+04 x 0.4+ 0.5 x 0.4) x 0.5 m*

1L7m

Frank will need 0.52 m® of plastic. b

Surface area of letter F = 2 x area of front face + 2 x area of side face + area of top

e

horizontal fa


(5684755500 = 10%) cm?® 5684755.5 cm® (5684 755.5 + 1000) L 5684.7555 L 5680 L

The bench has a storage capacity of approximately 5680 L.

9

10kL =10 m? Volume of cylindrical drum = 10 m® arh =10

7 xr?x3=10 '4

""2

o=

=

E

3

1o

3m

{as

r >0}

~ 1.03

So, the radius of the drum is approximately 1.03 m. 10

a

279

Surface area of the Sun ~ 47r?

~ 47 x (6.955 x 10%)? m?

~ 6.08 x 108 m?

The Sun’s surface area is approximately 6.08 x 10'® m?.

280

Chapter 6 (Measurement)

b

Review set 6B

Volume of the Sun ~ 4mr? ~ 4w x (6.955 x 10%)* m?

~ 141 x 10%" m®

The Sun’s volume is approximately

11

a

1.41 x 1027 m?®. 3

Volume of hemispherical top = § x 37r =27 x 3% em =187

3

cm®

~ 56.5 cm®

b

Volume of cone-shaped base = % x area of base x height

=Lixmr’h

=17 %32 x h cm?

= 37h cm® Now, volume of cone = 3 x volume of hemisphere

3rh = § x 187 3h=9 h=3 So, the cone-shaped base has height 3 cm.
3cm

Sos=V18

scm

o

3\/5

{Pythagoras}

{as s> 0}

Outer surface area of spinning top = surface area of hemispherical top + surface area of cone-shaped base =%>0}

x ~ 2.65

tand = 2

{tan9=m}

x

i 0 =

tan

3cm ADJ

—1(_2 0}

~ 854

So, the finishing point is about 854 km from the starting point. b

tanf = 460

720

0 = tan~" (450 0 ~ 32.6° So, the bearing of the finishing point from the starting point is about 236° — 32.6° &~ 203°.

312 10

Chapter 7 (Right angled triangle trigonometry)

Exercise 7E

Suppose the cruise ship first sails from P to Q, then from QtoS.

NQP = 180° — 112° = 68° PQS = 360° — 68° —202° =90° APQS

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

is right angled at Q.

tanf = %

6= tan~! (-12) 0~ 179.3° -,

the bearing of the cruise ship from P ~ 112° + 79.3° ~ 191°

Now,

PS?=13.6% +72?

{Pythagoras}

PS =1/13.62+722

{as PS> 0}

~73.3 So, the cruise ship is about 73.3 km on a bearing of 191° from P. 11

Suppose the yachts both depart from O.

OAN = 180° — 34° = 146°

{co-interior angles}

AOB = 124° — 34° = 90° AAOB

is right angled.

tanf = — 1

6 =tan™'(1}) ~51.8° -,

the bearing of B from A ~ 360° — 146° — 51.8°

Now,

~ 162°

AB?=11% 4 142

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°. 12

Suppose the eagle starts at Q and flies to R, and the eagle’s nest is at P.

NPQ = 360° — 293°

{angles at a point}

=67°

QPR = 67° +23° =90° APQR is right angled at P.

N

Chapter 7 (Right angled triangle trigonometry)

Exercise 7F

a cosf = £

5

B

O:COS’I(%)

..

0 ~66.4°

The eagle flew on a bearing of

= 180° — § — NPQ

{co-interior angles}

~ 180° — 66.4° — 67°

~ 046.6° b

In APQR,

PR?+2%=52

{Pythagoras}

. PR=+/52-22

R

{as PR >0}

. PR=+21 km

Let T be the point directly south of R and directly east of P.

In APRT,

RPT = 90° — 23°

Sin67° = %

=67°

{complementary angles}

d = /21 x sin67°

~ 4.22

The eagle is now approximately 4.22 km north of its nest.

1

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].

d km

S

313

314

Chapter 7 (Right angled triangle trigonometry) b

2

Exercise 7F

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 7F E

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

the required angle is BAM.

B

D‘

c

M

A

ii

315

E

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

4

B

/

D‘ A

iii

QW E

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

D

AT

i

M

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

B

M

the required angle is CFG.

tanf = ¢

6cm

8

s

f=tan"! (%)

H

0}

Let AGE be a. tana

6

= — V164

_

o = tan

-1(_s6 (\/m>

o=~ 25.1°

The angle is about 25.1°.

c

V164 cm

E

316

Chapter 7 (Right angled triangle trigonometry)

Exercise 7F

The projection of [BX] onto the base plane is [FX]. The required angle is BXF. 6

t an 3 =2 7

flztan_l(%)

.

3 ~56.3°

The angle is about 56.3°. iv

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

Let HX be z cm. Using Pythagoras in AHGX,

D 6cm m,

a? = 10% 4 47 z? =116 z=+v116

H {as x> 0}

Let DXH be ¢. B

i

_

6

tanqfi——\/m 6 -1 = ¢7tan (fi)

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

the required angle is PRS. tanf = 3

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

= kel

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 z =288

B

8

the required angle is DXH.

.

A

{as

= > 0}

10em

X

G

F

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

8 o

tan 8 =

fi=tan_1(\/§?8) B~ 25.2°

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,

22 =62 + 122 oo 2?2 =180 . z=+180

{as x>0}

Let QMR be 6. 8

tan¢ =

%

N

(}5 =

tan

8

-1

(m)

- $~308°

The angle is about 30.8°. The projection of [QR] onto the base plane is [MR]. the required angle is MfiQA tanf = ok 0.6

w g =tan"" (55) o 6 ~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 = m. Using Pythagoras in AMRU,

2% =0.6% +2.42 oozt =612 =612

Let QfiM be a. otana

1

= —

6.12

o = tan

11 ( _6'12)

o~ 22.0° The angle is about 22.0°.

{as

z >0}

Exercise 7F

317

318

Chapter 7 (Right angled triangle trigonometry) iii

Exercise 7F

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

MN| RU,

so MN

=RU

=24

m

1

tan3 = o1

L

B= tan_l(fi)

B

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

z? =18

z=+18

{as x>0}

Let XAM be 6. 8 1 V . cosf = — 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,

z? + 3% =10° 22 =100—9 22 =091

=01

Let XYM be a. - cosa:%

gDty G(\/fi) T . a=cos LoarTLT The angle is about 71.7°.

{as

>0}

Chapter 7 (Right angled triangle trigonometry) L

Let the equal sides of the pyramid be z. The projection of [AX] onto the base plane is [AM], where M is the point directly below A on the base plane.

Review set 7A A T

the required angle is AXM.

Consider the pentagonal base of the pyramid. The angle at the centre of the pentagon is 360°.

{angles at a point} —~

X

o

shix = 25 = 72° Y

o

TVX = 2= = 36° =

In ATMX,

sin36° = —— MX

MX=_——2%

_

2 X sin 36°

A

—z . cosAXM = 2Xsin367

In AAXM,

x

=T

L

z

= cos™!

(;) 2 X sin 36°

T 2 xsin36° .

AXM

~ 31.7°

X

2

M

2 X sin 36°

So, the angle between [AX] and the base plane is approximately 31.7°.

i

1

a

AC?=AB>+BC? _

=8

Q2

=100 . .

{Pythagoras}

B

2

+6

ADIJ

8cm A

C

The hypotenuse is 10 cm long. bsinezflzfizé 2

6cm

AC > 0}

{as

AC =100 AC=10

HYP

oPP

10

S ¢C059:fl:fi:é

°

HYP

a

10

°

b

HYP Sm

=

rm

ADJ o

c0564°:§ 8 X cos64° =z

T

ADJ

{cos@zm}

ADJ

OPP

ADJ

OPP

dtamg:flzfiz%

7m 9m HYP

.

) OPP {sme:m}

o ox® =sin—

(%)

51nw0:5 7

8

319

320

Chapter 7 (Right angled triangle trigonometry)


Also

tan23° = -

tan23°

—

x

( 4+ 80) tan 20° T

. ztan23° = xtan 20° + 80 tan 20°

z(tan 23° — tan 20°) = 80 tan 20° = "

80 tan 20° tan23°

~ 481.25 o

h~(481.25 + 80) tan 20° m

~~ 204 m

The building is about 204 m tall.

AAHG

6 cm

H

/.7

8 cm

G

4cm

is right angled at H.

AHG = 90°

— tan 20°

Chapter 7 (Right angled triangle trigonometry)

b

E

F

Consider the base of the prism.

4em

Let FH be = cm. : Pythagoras, Using

327

2 _ 42, Q2 z°=4%48

o a2 =380

s

H

Review set 7B

z=+80

{as

x>0}

ADFH is right angled at H. tanf

=

—

M=% 0=

6

tan_l

D (%)

6cm

9 ~ 33.9° So, DEH ~ 33.9°. 9

H

L

V80em

F

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

-

FON; = 360° — 303° =57°

{angles at a point}

OSNj = 360° — 213° = 147°

{angles at a point}

N;OS = 180° — 147° = 33°

{co-interior angles}

" FOS = 57° +33° = 90°

AFOS is right angled at O. tanf = 25 3

0 =tan""(%2) ~39.8° .. the bearing ofF from S ~ 213° + 39.8° ~ 253°

Now,

{Pythagoras}

% =25%+32

z=14/252+32

{as x>0}

~3.91 So, Aaron is about 3.91 km on a bearing of about 253° from his starting point. 10

Suppose Amelia and Kristos both depart from point S. Kristos jogs = km to point K, and Amelia

jogs

(z+2) km

to point A.

ASK = 164° — 74° = 90° AAKS is right angled at S.

SAN = 180° — 74°

{co-interior angles}

=106°

SAK = 180° — 106° — 44° = 30°

{co-interior angles}

328

Chapter 7 (Right angled triangle trigonometry)

In AAKS,

Also,

Review set 7B

tan30° =

sin30° =

So, the joggers are approximately 5.46 km apart at this time. 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.

22 4 2? =122

{Pythagoras}

. 227 =144 2P =12

L x=+72

{as z>0}

tan40° = @

L

h

"

VT

tan40°

~ 10.1

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

V72m

~ 3 x12x12x10.1 m® ~ 485 m® 12

a

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

. the required angle is BHF. Let FH be x cm.

B

A

X

”

2em

Using Pythagoras in AFEH, coa?

G E

=61

z =61

{as

= >0}

M

H

Chapter 7 (Right angled triangle trigonometry)

Let BHF be 6.

tanf = ——2 =

V61

0=tan” 8

-1

0~ 14.4°

(k) 2

The angle is about 14.4°. The projection of [CM] onto the base plane is [GM].

the required angle is CMG. Let GM be = cm. Using Pythagoras in AGHM, z? =52 + 32

coat=34 L

{as

x=+34

= >0}

Let CMG be a. .

’

tanoz*i

V34

_2_ a —= tan -1 (\/3—4)

o=~ 18.9° The angle is about 18.9°. The projection of [XM] onto the base plane is [MY].

the required angle is XMY. 2

tan8 = z

.

B=tan"'(2) [ =21.8°

The angle is about 21.8°.

13

i In AAMH,

sin65° = 2AH

. ) AH=_2 " sin65° .

AH =~ 27.6 cm

{sinf= v} HYP

Review set 7B

329

330

Chapter 7 (Right angled triangle trigonometry)

il In AAMH,

tan65° — —> AM

AM ="

j .

{tang = 2P ADJ

2 tan65°

AM ~ 11.66 cm

CM =~ 11.66 cm .

and

—

AC

A

B -(

{from a i}

2 sin 65° tan 65° 50

tan 65°

hecm

.

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

JdG

L

{Pythagoras}

AD?+ CD? = AC?

In AADC,

E

2

50

=(tan65°)

it

2 = (o) (

o

@t

2

tan 65°

2

\fan 650)

=

e

(D

(i)

tan 65° 7 -

Loxr=

2

or~16.49

{Pythagoras}

AE? +EH? = AH?

In AAEH,

2,

2 b+

C

T

A

{from a ii}

—2

AC=2x

{altitude of isosceles triangle bisects the base}

AC =2 x11.66 cm AC=23.3 cm

s

b CH=AH=

Review set 7B

50

e

(m)

[

2

2

25 )2 (sin65°

~

2

25

N (

B

.

)2

{using (1)}

\sin65°

0}

Tcm

H

Chapter 8

THE UNIT CIRCLE AND RADIAN MEASURE EXERCISE 8A a

E

90° = (90 X L) 180

radians

= 7 radians

60° = (60 X L) 80

radians

= 7 radians

30° = (30 x 1&5) radians = ¢ radians

9° = (9 x %)

b

18° = (18 x &;)

radians

= {; radians

radians

135° =

(135 x &)

radians

= 35 radians

225° = (225 x 1&;) radians = 52 radians

i 360° = (360 x &) radians 180 = 27 radians

315° = (315 x &) radians = IZ radians

36° = (36 x &) radians 180 = £ radians

270° = (270 x &) radians = X radians

i 720° = (720 x

80;) = 47 radians

radians

540° = (540 x

;) radians

= 3 radians

80° = (80 x &;) radians = 4= radians

230° = (230 x 1%;) radians = B2 radians

36.7° = (36.7 x %5) radians ~ 0.641 radians

317.9° = (317.9 x %) radians ~ 5.55 radians

137.2° = (137.2 x {&5) radians ~ 2.39 radians

219.6° = (219.6 x %) radians ~ 3.83 radians

396.7° = (396.7 x &) radians ~ 6.92 radians

= 36°

%= (% x 10)° = 135°

F=(Gx2)°

¥ = (i x 0y° =108°

%= (< 129)° =10°

332

Chapter 8 (The unit circle and radian measure)

s fi=(Hx5)

h 5= (55

=18°

i %= (X 4

=27°

)°

=210°

Exercise 8B

P 3=(3x12)° =225°

a 2radians= (2 x 182)° ~ 114.59°

b 1.53 radians = (1.53 x 182)° ~ 87.66°

¢ 0.867 radians= (0.867 x 180)° ~ 49.68°

d 3.179 radians= (3.179 x 180)° ~ 182.14°

e 5.267 radians= (5.267 x 122)° ~ 301.78°

s+ B o ] [0 [ [0 5 [ 5 o [0 n b

o3

[+13 w5 [« [# [# ¥ ¥ [ls 180 | 210 | 240 | 270 | 300 | 330

EXERCISE 8B 1

a

&

arclength=6r =1x7 =Tcm 7cm

b

arc length = 6r =4x3 =12

cm 3cm

¢

arc length = 6r

25 % 6.2 ~13.0 m 6.2m

Chapter 8 (The unit circle and radian measure)

Exercise 8B

area = 10r°

=1x3x2? =6cm? 2cm

area = 10r°

=1x15x8? =48

cm?

8cm

area = 10r?

=3 xZx56 ~ 8.21 cm? 5.6 cm

arc length = Or

=1

area = 10r®

x9

=ixItx9?

~49.5 cm

~ 223 cm?

arc length = 6r

area = 20r?

=4.67 x 4.93

=1 % 4.67 x 4.93

~ 23.0 cm

~ 56.8 cm?

l=06r

2.95m

13m

6cm

/

295 =60x43

-

.

Scm

l=0r

6=0x8 ==

.

8.4cm

]

0 ~0.686°

S

jem

b

6

8

area= %07"2

co30=1x6x

.

g

.

9=06°

B

30x2

100

area = 10r°

=150.75%x8

B =24

cm?

0=0.75¢°

1=06r

- 84=0x5 -

6= o

5

area = 10r?

=1x1.68x5 =21 cm?

102

333

334

Chapter 8 (The unit circle and radian measure)
-~

Let the smaller semi-circle have radius r cm.

In APTR, we have:

P

b,om T

Thus (6+7)% =62+ (12—7)2 364 12r + 12 =36 + 144 — 24r + 12

(6 +r)em (12-r)em

o 36r =144

R

Lor=4

Chapter 8 (The unit circle and radian measure)

b cos(TPR) = & = 0.6 TPR = cos™(0.6) ~ 0.927°

Exercise 8B

337

PRT = I - 0.927° ~ 0.644°

i area A =area APTR — (area sector PQT + area sector RQS)

~ (8 x 6) — (3(0.927)(6%) + 1(0.644)(4%)) ~ 24 —16.69 — 5.15

~2.16 cm? ii

area B = area of quarter circle — area of semi-circles — area A

~ im(12?) — in(6?) — iw(4%) — 2.16 ~ 36m — 187 — 8w — 2.16 ~ 10m — 2.16 ~ 29.3 cm?

sinZ =

1

10 —ry

{in AOAB}

1__"n

2

10 —7ry

10-r

=2r

3ry =10 T1=13—0

the largest circle has radius % cm.

So, in successive circles, radii are reduced by a factor of 3.

10 Ty _=g,

_ 10 I3 =g,

10 T4 _=gy,

and so on.

Thus the total area of the circles = 7T7‘12 + 7r7’22 + 777"32 + 7rr42 + ...

=7 (B + (B + @+ @)+ )

mx (B [1+ @)+ @) +3) +]

=mx

10 x (

= 257 ypits?

1

N

- >

9

.

)

.

.

.

{infinite geometric series with uy =1,

r = %}

338

Chapter 8 (The unit circle and radian measure) ¢

Exercise 8C

area of sector = %07"2 [ =5x%x10 2 — 50w _—h3

.

-,

.

.

total area of circles

fraction of sector occupied by circles = ————— area of sector

257

R

~

50

e

Il

3

(G

oei

o|i

positive |

positive |

positive

negative |

positive | negative

negative | negative | positive positive | negative | negative

2

e

sinf, cosf, and tan@ are all positive in quadrant 1

e

only sinf is positive in quadrant 2

e e

only tanf is positive in quadrant 3 only cosf is positive in quadrant 4.

EXERCISE 8C 1

2

Since any point P on the unit circle has coordinates

a

i

A(cos26° sin26°), B(cos146°, sin146°), C(cos199°, sin199°)

il A(0.899, 0.438), B(—0.829, 0.559), C(—0.946, —0.326)

(cos6, sinf),

then:

Chapter 8 (The unit circle and radian measure)

b

1 A(cos123° sin123°), B(cos251°, sin251°), C(cos(—35°), sin(—35°))

il A(—0.545, 0.839), C(0.819, —0.574)

3

Exercise 8C

a

B(—0.326, —0.946),

i L0707

i 2 ~0.866

e

“Mmsliflllllflflll s

us

s

3

5

4 0° < 0 0

cos@zffi\/%T in

=

C

T

cos? 0 +sin26 =1

Now

cos? 0 +sin26 =1

cos® 0 + A cos® 0 = 1

cos® 0+ g cos®0 =1 13 . 29 5 cos =1

25 24 Feosf=1

cos” 29_ 0 = 59

cos’0 =2

cosf = A L3 t5

(:059:::\/i1—3

But 6 is in quadrant 1 where cosf sinf are positive. G6 E i3, 752 sin cosfifx/fi

G

and

But 6 is in quadrant 2 where negative and sinf is positive. cosf —_3 = s,

gineg=4%4 smefs

cosf

is

354

Chapter 8 (The unit circle and radian measure)

S

¢ tang =300 _ cos

A

0

Exercise 8F

d tanf—=

T

C

cos?0 +sin?0 =1 Now . 29 0 _=1 144 2 s?0+ )coco M cos?

=1 cos? +sin?6

Now

= 1

cos® 0 + 2 cos’

9 2 . .169 costl=1 S

19—4c0520:1

. 20 soocosl

cos29=%

.

But @ is in quadrant 3 where cosf sin@ are both negative. A 0 3 n_ cosf = T sinf = Tl in

tanf = 227

6

= k

cos 6

S

0

sinf = kcos@

Now .

2 =_ 155

= +% cosf

cosl9=::\/—31_4

3

6

cos

. 12 cos ¢ = —== f) sin

sin@:-@cosa

-

=L

But . 6 is in quadrant ; . 4 where ; positive and sin@ is negative. i . 12 sind=—33 =33,5 coocosf

and

cosf

A

cos?@ +sin?6 =1 ocos?O+ k2cos? =1

(B2 +1)cos?f =1 .

cosf =

+1

VEZ2+1

But 6 is in quadrant 3 where k is positive. (zost9=_—1

1

a

cosf

sinf =

and

sinf

are both negative, and

tanf

—k

tanf=4

b

cosf =0.83

Using technology,

Using technology,

tan™!(4) ~ 75.96°

cos~1(0.83) ~ 33.90°

75.96°

o

0=

.

O0=T76.0°

is positive

or

180° + 75.96°

.

0 ~33.90°

or

360° — 33.90°

or

256°

s

60=33.9°

or

326.1°

is

Chapter 8 (The unit circle and radian measure) ¢

ing =2 sinf=¢

d

Exercise 8F

cosf =0

cos~1(0) = 90°

Using technology, sin ™! (%) ~ 36.87°

.

. e

0~ 36.87°

or

180° — 36.87°

0 =~36.9°

or

143.1°

tanf =6.67

f

Using technology,

o

0=90°

or

360° —90°

.

6=090°

or

270°

cost oo =5

Using technology,

tan~1(6.67) ~ 81.47°

cos™! (&) ~ 83.24°

7117 . .

2

a

or or

0 ~81.47° 0~81.5°

180° + 81.47° 261.5°

_13 tanf =

o . b

or 360° —83.24° or 276.8°

0~ 83.24° 0~=832°

cosf = %

Using technology,

Using technology,

tan~!(3) ~ 0.322

cos™!(2) ~ 1.128

But

0

3

1

B

or its supplement

C ~62.1°

or

(180 —62.1)°

C=~62.1°

or

117.9°

a

Let BAC be 2°. .

C

.

Using the sine rule,

8cm

11cm

in

Sot

z°

8

A

= 3¢ ) 11

inz®

B

in45°

— 8 x sin45°

o

11

$_Sin_1(8XSin45O)

-

11

z ~ 30.9 BAC is approximately 30.9°. b

Let ABC be z°.

A

i

Using the sine rule,

LR c

sinz°

=

23

sin 42° 32

sin g© — 23 % sind2° 32 o — sin—1 (23 X sin42°)

B

32cm

32

T~ 28.7 ABC is approximately 28.7°.


180°

which is impossible in a triangle.

{sine rule}

- inABC = 22:1xXsin38° ’

-

.

16.5

in

389

. ABC =sin™! (%)

- ABC~55.5°

sinACB _ sin 18° )

16.5cm

c

124.5° + 38° = 162.5°

B

which is < 180°.

)

T

SinACB

or its supplement

or 124.5°

both of which are possible as S

22.1cm

or 180° — 55.5°

. ABC~55.5° f

A

{sine rule}

A

— 4.3 X sin 18° 3.1 3

~

-, ACB =sin""! (%)

. ACB~25.4° px

ACB =~ 25.4°

a

Using the sine rule,

A

or its supplement

c

or 180° — 25.4° or

154.6°

both of which are possible as

&

O

sin z®

—

. o .oy

154.6° + 18° = 172.6°

= =

3.1km

B

which is < 180°.

sin 48°

Txsin4g® —mm

c.osinz® ~ 1.04

’

5m

7m

z°

48°

But sinz® is always between —1 and 1 (inclusive), so we cannot solve for = and the question cannot be solved. b

This means that it is impossible to draw a real diagram with the dimensions Mr Whiffen has given.

Chapter 9 (Non-right angled triangle trigonometry)

5

a

i

Using the sine rule,

sin ACB _ sin30° —

.

-

=~

7

in

30°

sinACB = X500 L

=

C in

LETh

30°

. ACB =sin"? (X%)

- ACB ~22.9°

)

389

cm

Tem

B

.

or its supplement

or 180° — 22.9°

or 157.1°

- ACB~22.9°

But

A

g

=

Exercise 9C.2

so this case is impossible.

157.1 4+ 30 > 180,

- ACB ~ 22.9° ii b

BAC ~ 180° — 30° —22.9°

BAC = 127°

{angles in a triangle}

Area of triangle ABC = 3bcsin A

A § X9 %7 xsinl27° ~ 25.1 cm?

6

Let the angle opposite the 9.8 cm side be 6. Using the sine rule,

sinf _ sin75°

98

9

9.8 X sin 75°

sinf =

. But

9.8cm

9

sinf ~ 1.05

—1 0,

PQ ~ 2.98

or

—0.453

so PQ~2.98.

So, the yachts are about 2.98 km apart. ¢

sinlPQ _ sin53°

Using the sine rule in ALPQ,

.

2.1

=~

- sinlPQ= =

LPQ

2.4 21xsin53° Xsin53” 21 2.4 . _1/2.1

= sin

LPQ ~ 44.3°

xsin53° 1(—X = 2.4

)

The bearing of the Queen Maria from the Porpoise is 223° — 44.3° ~ 179°.

11

8cm

U

a


8000

~ 97.403° area of the property = area ABCD + area ABED + area AABE

~ 1 x50 x 100 x sin83° + 1 x 60 x 106.213 x sin67.145° + 1 x 50 x 80 x sin 97.403°

~ 7400 m? 18

Let the location of the Chinese restaurant be C and the location of the fish and chip shop be F, such

that CF = 7 km. The area enclosed by the larger circle shows the free delivery region for the Chinese restaurant and the area enclosed by the smaller circle shows the free delivery region for the fish and chip shop.

i

Chapter 9 (Non-right angled triangle trigonometry)

Exercise 9D

399

The region that receives free delivery from both locations is the intersection of these two regions. So, we need to find the shaded area.

Using the cosine rule in AAFC:

2,72

47 -3

cosf =

ACB =20

and

and

2x5xT7

6 =cos™*

Now AAFC and ABFC are congruent.

g2

(%)

2,

cosa = et

a=cos!

AFB=2a

In the larger circle, area of AABC = % x5 x5 x sin26 = 2—25 X sin (2 cos™! (%))

~ 8.6161 km? X m X 52

2cos ™!

(%)

X 7 X 52

360

~ 9.5063 km? Now, area of segment of larger circle = area of sector ABC — area of AABC

~ 9.5063 — 8.6161 km?

~ 0.8902 km? In the smaller circle, area of AABF

= % X 3 X 3 x sin 2«

=2 xsin (2cos™" (33))

~ 4.3743 km? Also, area of sector ABF = 2a

X T X

2cos™! (%)

360

32

x

— 5

2x3x7

{SSS}

Also, area of sector ABC = 20

72

X 32

~ 6.0025 km? Now, area of segment of smaller circle = area of sector ABF — area of AABF

~ 6.0025 — 4.3743 km? ~ 1.6282 km?

So, shaded area = area of segment of larger circle + area of segment of smaller circle

~ 0.8902 + 1.6282 km® ~ 2.52 km?

the region that receives free delivery from both locations is about 2.52 km?.

(%)

g2

400

Chapter 9 (Non-right angled triangle trigonometry)

19

A

Exercise 9D

The perimeter is 36 m.

a+btc=36 L b=3-a—c

.. (1)

The area is 30v/3 m?. b

¢

Lacsin60° = 30v3 1acx (é) =30V3 %

C

@

B

ac = 30V3

c.oac=120

)

c Using the cosine rule:

b = a® + ¢® — 2accos 60°

(36 —a—c)*=a*+c* —2acx (3) —ac o (36—a—c)(36—a—c)=a’+c?

1296 — 36a — 36¢ — 36a + &% + ac — 36¢ + ac + £ = a + £ — ac 1296 — 72a — 72c + 3ac =0

432 —24a — 24c+ac=0

.

{using (2)}

4322 (%O) —2dct (1—30) c=0 o4i120-0

432 - B8 C

940

o552 — 2880 c

. 24¢% — 552¢ 42880 =0 . 24(c? —23¢+120) =0 .

24(c—8)(c—15)=0 .

.

Substituting o

Substituting

When

¢=8

¢ =8

¢ =15

and

.

.

.

.

into (2) gives

into (2) gives

a=15,

120

a = -

120

a = ==

b=36—15-8

=

c=8or

5

8

{using (1)}

=13

When

¢=15

and

a=8,

b=36—8—15 =13

{using (1)}

15

{using (1)}

Chapter 9 (Non-right angled triangle trigonometry) Using the sine rule: sinB

_

b

sin 60°

_

13

Also, using the sine rule:

sinC

sin 60°

sinC'

c

13

15

sinC'

{When

P

V3o z_

8}

8

fi

13

~

sinC

Lf

8

{when

15

¢=15}

=13sinC

sinc = B8 26

sinC = 4—\/5 13

C =sin!

(43 C = sin =1 (—1‘3£>

(15263>

C ~ 87.8°

and

C ~ 32.2°

A=180°-B-C {angles in a triangle} A~

sinC

13

. 4/3=13sinC

and

Exercise 9D

A=180°-B—-C {angles in a triangle} .

A~

180° —60° — 87.8°

Ar322°

180° — 60° — 32.2°

A~ 87.8° So the remaining two angles of the garden are approximately 32.2° and 87.8°. 20

Suppose Sam and Markus are z m and y m from the tree respectively, and the tree is A m high. N

In APST,

tan25°

h == ) _h "

tan25°

= 2.145h

In APMT,

tan15° =2 oy

.

Y

h

tan 15°

~ 3.732h

But

STM = 65°

and

100% = 2% + ¢ — 22y cos 65°

{equal alternate angles} {cosine rule}

10000 ~ (2.145h)% + (3.732h)% — 2 x (2.145)(3.732)h? cos 65° 10000 ~ 11.762 h? h? ~ 850.17 h ~29.2

So, the tree is about 29.2 m high.

401

402

Chapter 9 (Non-right angled triangle trigonometry)

Exercise 9D

T

b

3km

Consider the following net of the curved surface of the cone: In order for AB to be as short as possible, [AB]

must be a straight line. The circumference of the cone’s base is equal to the arc length of the sector. Ox3=2rx2

L

30 =4dnw

_ 4rm

0="%

Using the cosine rule in AABT,

AB? =32+ (3)>—2x3x 3cos &

. AB=4/9+2-9(-1) .

AB = v15.75

{as AB>0}

~ 3.97 km

The length of the path from A to B is approximately 3.97 km.

b

T

The path from A to B is horizontal at the point D where [DT] is perpendicular to [AB].

Using the sine rule in ATAB,

A

sinABT _ sin3 3

-

:

ABT

3km

fi

/1575

1.5km D

.

km——>

B

V15.75

=

B

sin-1

W

3sin sz

7

- ABT =~ 40.89°

Now in ABDT,

cos40.89° ~ % .

BD~1.13

km

The length of the part of the path from approximately 1.13 km.

B to the point where

the path is horizontal is

Chapter 9 (Non-right angled triangle trigonometry)

Exercise 9D

403

sin%:é e

T

"

6r

=it6

O _

11

7= (%) g =4sin"' (1)

3r

rv

:

Area of sector of larger circle =

=

We can divide the figure into two parts, the sector of the larger circle, and the unshaded section of the smaller circle.

x 6 x (3r)%

{radius = 3r

for larger circle}

or?

Nlo

b

0:

Area of unshaded section of smaller circle = 772 — area of shaded section =7l —m Total area = area of sector of larger circle + area of unshaded section of smaller circle

=20 +mrP -7

.. (1)

The figure can also be divided into two isosceles triangles and a sector of the smaller circle as shown. Let each of the base angles of the isosceles triangles be a.

a+a+ g =7

{angles in a triangle}

2a=7r—€

Now,

2

(-9

angle of sector of smaller circle = 27 — 2«

—7T+5

{angles at a point}

0

Total area = area of two isosceles triangles + area of sector of smaller circle 0

=2x (£(2 x3rx3rxsing 2) +3 l>0}

\| Ho+35 —9sing

r~1467

{using 0 =4sin"'(})

from a}

The radius of the smaller circle is approximately 1.467 cm.

ACTIVITY 1

The surface area of the lune is 21 of the total surface area T

of the sphere. Sx’ 0 =

i

X 471'7'2

2

=2r%0

2

a

A’, B/, and C’ are diametrically opposite to A, B, and C respectively.

So, arc AB is diametrically opposite arc A'B’, arc AC is diametrically opposite arc A'C/, and arc BC is diametrically opposite arc B'C’. Thus, AB = A’B’, AC = A’C’, and BC = B'C’ so that triangles ABC and A’B’C’ are congruent. {SSS} b

We can find the surface area of the sphere by adding the surface area of each lune twice, then subtracting the areas of the spherical triangles ABC and A’B’C’ that have been added more than once. Both triangles are counted 2 times more than necessary, so we subtract the area of 242 =4 spherical triangles. surface area of the sphere = 4772 = 25, o + 25p, 5 + 2S¢, 4 — 44


0 La+fB+y—m>0

Review set 9A

405

{A r>0}

Lat+fB4y>w the angle sum of a spherical triangle is greater than 180°. 4

Suppose two spherical triangles on a given sphere are similar. the angles of the triangles must be the same. From 2 ¢, two spherical triangles on a given sphere with the same angles must have the same area. the triangles have the same size and shape, and are therefore congruent. It is therefore not possible for two spherical triangles on a given sphere to be similar but not congruent.

1

a

b 7km 9cm

Gem

8km

Area = § X 6 x 9 x sin83°

Area = £ X 7 x 8 x sin30°

~ 26.8 cm?

=28x1

= 14 km?
0}

z ~ 10.5

The remaining side is about 10.5 cm in length.

7cm

406

Chapter 9 (Non-right angled triangle trigonometry) b

Review set 9A

Let the remaining side have length z m.

6m

By the cosine rule:

2% =624 6.8% — 2 x 6 x 6.8 x cos 130°

rm

{as x>0}

. x=1/62+6.82-2x6x68xcos130°

6.8m

orx~11.6 The remaining side is about 11.6 m in length. L

a

Using the sine rule, T

8

sin52°

snd3° 8 X sin52°

8em

zem

T Teimaz T~

b

9.24

.

:

Using the cosine rule,

102 + 122 — 112

cosf = ik il 2x10x

12

102 4 122 — 112 — 117 127 0 = cogt (LT

10m

2% 10 x 12

6 _= cos

—1/123

(%)

9 ~ 59.9°

¢

. 12m

Using the cosine rule,

2?2 =42 +9% -2 x4 x 9 x cos37°

o=\ +92_2x4x9xcos37°

{as z>0}

4

-

Jem

ox~6.28 rcm

5

By the cosine rule:

DB? =72 +112 — 2 x 7 x 11 X cos 110° .

{as

11 xcos110° DB=+/72+112-2x7x ~ 14.922 cm

DB > 0}

Total area = area AABD + area ABCD

A

X 7 x 11 xsin110° + § x 16 x 14.922 x sin 40°

~ 113 cm?

6

. By the cosine rule:

4122 — 172 92 cosf =iet 2x9x12

9 = cos™!

2 4 192 _ 172

97 +127

2x9x12

(=64) 0 _= cos eoq—1 (W 0~ 107.2°

So, area of triangle &~ 1 x 9 x 12 x sin107.2° ~ 51.6 cm?

177

.

Chapter 9 (Non-right angled triangle trigonometry)

a

By the cosine rule:

407

72 = 22 4+ 82 — 2 x x x 8 X cos 60° o

49 = 2% + 64— 163 x %

8cm

coa?—8x+15=0

7cm

(@=3)(z—5)=0

s

-

.

or

x=3

b

rcm

There are two possible values for x, so Kady can draw two triangles:

b

Review set 9A

8cm 7cm 3cm

So, Kady’s response should be that she needs information to know which triangle to draw. a

more

By the cosine rule:

217 = 2% + 3% — 2 x = x 3 x cos 120°

o441 =22 49— 6z x (-1

a? +32—432=0 b

r= _

P

X50m\

M

o

s L

0

2lcm

—344/32 — 4(1)(—-432) — -

2(1)

=3+

V1737

2

3193

= =3

But >0, ¢

so v =-5+34%~193

Perimeter of triangle LMO ~ 3 + 21 + 19.3 cm ~43.3 cm

Two of the angles are 35° and 82°.

..

the third angle = 180° — 35° — 82° __

pQo

=63

A

{angles in a triangle}

G0 b

Area of triangle = absin C

- 40 = labsin35°

.

a= i

Using

C

absin 35° = 80 80 bsin 35°

.

g

the sine rule:

!


0}

sinC c

sin 35°

__ 11.25in35° C= sin 63°

e~ T721l cm

So, the triangle has sides of length

~ 7.21 cm,

~ 11.2 cm,

In AADE,

DE =

In ACDG,

DG =+/42+32=125=5m.

{Pythagoras}

In AEGH,

EG =

{Pythagoras}

62 432 = /45 m.

and

~ 12.5 cm.

{Pythagoras}

/62+42=+52m.

By rearrangement of the cosine rule, cosf =

(V35)? VA +5% —e (v52)°

4542552 5m

T

10V45

18

"~ 10V45 6 = cos EDG

-

18

(10\/@)


0}

409

410

Chapter 9 (Non-right angled triangle trigonometry)

b For ABC ~ 69.5°,

Review set 9A

CAB =~ 180° —42° —69.5°

{angles in a triangle}

~ 68.5°

area of AABC ~ 1 x 7 x 5 x sin 68.5° ~ 16.3 cm?

For

ABC ~ 110.5°,

CAB ~ 180° — 42° — 110.5°

{angles in a triangle}

= 27.5°

area of AABC ~ 1 x 7 x 5 x sin27.5° ~ 8.09 cm? 14

Suppose dune buggy X travels to P, and dune buggy Y travels to Q.

NXY = 360° — 215°

{angles at a point}

= 145°

QYX = 180° — 145°

{co-interior angles}

= 35°

Using the cosine rule in AQXY,

QX? = 1007 4 500% — 2 x 100 x 500 x cos 35°

-

QX = /1002 4 5002 — 2 x 100 x 500 X cos 35°

- QX ~422.0 m

sinQXY _ sin35°

Using the sine rule in AQXY,

100

QX

100 sin 35° sin Q)A(Y = 422.0

Now,

-

QXY ~sin~!
180°

measures about 57.4°

and BAC measures 180° — 74° — 57.4° ~ 48.6°. b

Q

83°

PRQ = 180° — 46° —83° =51°

R

.

{angles in a triangle}

§

Using the sine rule, P

462-"10cm

L sin 51°

P

-

PQ=

1

and

sin 83° < 10>.' 0}

413

414 7

Chapter 9 (Non-right angled triangle trigonometry) Let the height of the tree be h m. sin 8°

180° — 90° — 18°=72°

- 50

_

— h

Using the sine rule,

Review set 9B

hm

sin 72° 50 x sin 8°

_

g

sinT2°

©

s h~T7.32

et

.

.

\4/

So, the tree is about 7.32 m high.

The unknown angle is .

180° — 68° — 71° =41°

i

Using the sine rule,

—

AB

==

sin 71°

50 m

1

20 sin41°

L

o

68°

150 m

sin41°

.

Ll

o

1

{angles in a triangle}

150 x sin 71° w AB= ————

.

.

AB~216.18 m

.

Also using the sine rule,

—

BC

sin 68°

-

= —

71°

150

¢

sin41°

BC=

B

150 >< sin 68° sin41°

.

BC~211.99

m

So, the perimeter of the triangle ~ 150 + 216.18 4 211.99 m

~ 578 m

Area of the triangle = $bcsin A 2L %X 150 x 216.18 x sin 68°

~ 15000 m?

ASP = 210° — 113° = 97° By the cosine rule:

AP? = 310% 4 430% — 2 x 310 x 430 x cos97° . AP = 1/310% + 4302 — 2 x 310 x 430 x cos 97°

{as AP > 0}

AP = 559.90 m Using the sine rule,

SO 430

AP

6inf ~ 430 X sin 97° 559.90

.

9

sin—? 0}

1275 m

Using the cosine rule in AACD:

= cos ADC

~

2

122

410

2 _

12.75

2x12x

=

1

10

(122

+10% —12.752

T i 12

=~ cos™! ADC

.

2

2x12x10

. ADC ~ 70.2° Using the sine rule in AABC:

Using the sine rule in AACD:

sinBAC _ sin105° 9 AC s

W

sin BAC ~ Sl -

BAC

in

sinCAD __ sin70.2° 10 T AC

105°

12.75

:

.

~

1

i

i28

X 2e mi0 sinCAD =~ 10

.

12.75

CAD =~ sin-1 (10 X sin 70.20)

~sin-! (9 X sin 1050)

12.75

12.75

- CAD ~47.5° Now BAD = BAC + CAD . BAD ~ 43.0° + 47.5°

- BAC ~ 43.0°

. BAD ~ 90.5° Also, BCD = 360° — 105° — BAD — ADC

{angles in a quadrilateral}

A~ 255° — 90.5° — 70.2° ~ 94.3° 11

Q

Using the sine rule, sin@Q _

sin47°

9.6

11m

.

12

a

9.6m

R

9.6 X sin47°

coosin@Q = ——u— 11

i

F

11

o

. Q=sin

—1(96x

(

sin47°)

=

Q397 By the cosine rule:

BD? = 1202 + 125 — 2 x 120 x 125 X cos 75° BD = /1202 + 1252 — 2 x 120 x 125 x cos 75°

{as BD >0} BD ~ 149.2 m

The area of the block = area of AABD + area of ABCD

~ 1 %120 x 125 x sin 75° + 1 x 149.2 x 90 x sin 30°

~ 10600 m*

416

Chapter 9 (Non-right angled triangle trigonometry)

b

13

Area~ 10600 m? ~ (10600 = 10000) ha ~ 1.06 ha

a

Review set 9B

{10000 m® =1 ha}

B

By the cosine rule,

62 =22 +8%—2x

1 x 8 x cos44°

. 36 =a? + 64 — 16z x cos44° Ly

2?1151z + 28 ~ 0

6m

TTR

A

C

8m

rA

1151+ /11512 — 4(1)(28) 11.51 +4.524 —2—

z~8.02

Frank needs additional information as there are two possible cases: and (1) when AB ~ 8.02m (2) when AB =~ 3.49 m

8.02m

44°

6

or

3.49

B

B

3.49m

8m

6m

44°

8m

Case (1)

b

The area of the plot is a maximum when

2

Case (2)

z ~ 8.02 m.

Volume = area x depth

=1x8xazxsindd®x0.1

{10 cm=0.1 m}

~ 4 x 8.02 x sin44° x 0.1

~2.23 m® 14

a

i

The sum of the interior angles of a regular hexagon is

M o omm A e

(6

B

-

2) x 180° = 720°.

AMB=

720°

=120°

By the cosine rule in AAMB:

AB? =122 4122 — 2 x 12 x 12 x cos 120°

il

.

AB=4/122+122 -2 x 12 x 12 x cos120°

.

AB

Area of the hexagon = 2 x area of AAMB

+ area of rectangle

= 16m mm?

Volume of nut = area of base x height = (area of hexagon — area of circular hole) x height

~ 2270 mm?

M

~2x L x12x12 xsin120° + 20.8 x 12

Area of circular hole in nut = 7 x (%)2

~ (374 — 16m) x 7 mm?

AB >0}

&~ 20.8 mm

~ 374 mm? b

{as

A an A

B

Chapter 9 (Non-right angled triangle trigonometry)

Let BAC=6,

Review set 9B

B

A= Lbesin 6

so

a

(%bcsin@)2

. A=

417

= %172122 sin?0 2.2

= b:

x (1 — cos®6)

p2c2

.

Using the cosine rule,

cosf = A2

: A27b2402

b2 +c?

—a?

2bc

14

4

(1

b2 + % —a?

'

2be

1__b2+62—(12

7b202

2bc + b2 + 2

2be

2bc

b2 +c? —a?

C

_b262

4

1

b2+2cZ—a2>

C

T4

b

(14 cos@)(1 — cosh)

b2c?

(1’*b2+2627a2)

C

— a?

1_1)2+62—¢z2

2bc

2bc — b2 — 2 + a?

2bc

(b + 2bc + ¢ — a®)(a® — (b* — 2bc + ¢?))

= 35((b+¢)* —a®)(a® — (b—¢)’) o N

o

N

|

/NN IS] S)

)

16(17 tceta)(b+c—a)(a—b+c)(a+

) () (e=b) () ) (gt ) (e ) (2222

s(s—a)(s—b)(s—c)

L

c)

A=+/s(s—a)(s—b)(s—c)

where

s=

{as

a+b+c

A >0}

2be

Chapter 10

POINTS IN SPACE

2

a

I AB=+/(6-0)2+(-4-0)2+(2-0) = /62 4 (—4)2 +22

=v36+16+4

= /56

= 2v/14 units

il

The midpoint is (0+6

0+—4

27

hichis

which

1s

2

(3,

7

0+2)

2 )

—2, 1)

, —2,1).

Chapter 10 (Points in space)

b

i

AB=4/(0—-4)2+(1-1)2+(-2-0)2

IR

— V65071

Exercise 10A

The midpoint is

(R

2+(z—4)?*=1 So, P lies on a sphere with centre 10

(2, 5, 4)

and radius 1 unit.

a

X.

A plane parallel_to the Y'-Z plane, passing b b (3.0.0

through

(3,

0, 0).

A llqe paralllel to the X-axis, passing through e 1

A

circle

in

plane, centre

(0,2, -1).

radius 2 units. f

the

X-Y

(0, 0, 0),

\Z 2

3

X,

A sphere, centre (0, 0, 0), radius 3 units.

A 4 by 1 rectangular plane 2 units above the X-Y plane (as shown).

All points on and within a 3 x5 x2 rectangular prism (as shown).

EXERCISE 10B 1

a

5

B b

Volume = length x width x height =3x3x3 = 27 units®

Volume = length x width x height =5x2x6 = 60 units®

>

Y

424

Chapter 10 (Points in space)

Exercise 10B

Volume = length x width x height =5x4x2 = 40 units® 2

a

Dis

(-7,0,3)

andEis

(-7,4,0).

b

Volume = area of end x length

= 1 x base x height x length =1x4x3x7 = 42 units®

¢ AB=+/(0-02+(4-0)2+(0—3)2 =/ 0121 (32 =V0+16+9

=25 =5

d

units

Surface area of prism

= area of base + area of 2 triangular faces + area of 2 rectangular faces =7x4 + 2 x area of AOAB + area of quadrilateral OADC + area of quadrilateral ABED

=28 +2x3x4x3 = 96 units® 3

a

+7x3

+ 7x5

To find the centre of the base, we locate the

midpoints of the diagonals.

The midpoint of [OB] is (% 0+62 M) 2 which is (3, 3, 0).

The midpoint of [AC] is (% 0+62 M) 2 which is (3, 3, 0).

the centre of the base is

the apex

(3, 3,9)

centre of the base.

b

(3, 3, 0).

lies directly above the

Volume = % (area of base x height)

=1x6x6x9

= 108 units®

Chapter 10 (Points in space)

¢

i Mis (% %

Exercise 10B

%) which is (6, 3, 0).

i MD=+/(3-6)2+(3-3)2+(9-0)? =/(-3)2+02+92 =V9+0+81 =90 = 3/10 units

i Area of triangle ABD = 1 x 6 x 3V/10 = 9v/10 units? Surface area of pyramid = area of base + area of 4 triangular faces

—6x6 + 4x9V10 =36 +36V10 = 36(1 4+ V/10) units?

& The midpoint of [OB] is (

L

which is

0+10

0+18

lies directly below the apex

)

0

N

the centre of the base is

(5, 9, 0)

)

0”0>

AL

1040

(5, 9, 0).

M

Il

which is (5, 9, 0).

The midpoint of [AC] is (

s

1800)

oL

O

which

(5, 9, 12).

M

Volume = 3 (area of base x height)

B(10,18,0)

= % x 18 x 10 x 12

= 720 units® Let the midpoint of [AB] be M.

M is (10;105 0;“,%)

which is (10, 9, 0). D

MD = /(5 —10)2 4+ (9 — 9)? + (12 — 0)?

AR

=25+ 0+ 144 =169 = 13 units Area of triangle ABD = % x 18 x 13 = 117 units?

=

B -~

18

425

426

Chapter 10 (Points in space)

Exercise 10B

Let the midpoint of [BC] be N.

D

which is (5, 18, 0).

M) 1BRIE Nis (1‘”0, 2 2 2

5)2+ ND =+/(5(9— 18)% + (12— 0)2

— VP

(R 12

=0+ 81+

144

=225

= 15 units

[1

B

N

-—

Area of triangle BCD = £ x 10 x 15

1)

c —>

= 75 units® Surface area of pyramid = area of base + area of 4 triangular faces =18 x 10 + 2 x area of AABD + 2 X area of ABCD

=180 + 2x 117 = 564 units? 5

a

Base radius of cone = distance from centre

+ 2x 75

(0, 0, 0) to point

(4, 5, 0)

=/A=0P?+(5-02+(0-0)2 = V42152402 =V16+25+0 =

b

v/41 units

Volume of cone = (area of base x height)

=1x7mx(V41)* x6 = 827 units® ¢

Let the slant height be s units.

52 =62 + (v/41)>

{Pythagoras}

52 =36+41

§ =77

s =717

6 units

{as s >0}

So, the slant height of the cone is v/77 units. d

S units

Surface area of cone = 7rs + 772

=7 x VAL x V7T + 7 x (V41)? ~ 305 units?

41 units

Chapter 10 (Points in space)

Exercise 10B

Radius of sphere

= distance from centre

(—4—2)2+

(2, 3, —1)

(-6

to point

—3)24 (10 —

— o oI

(—4, —6, 10)

—1)2

+ 121 =/36+81 =

/238 units

Volume of sphere =

Centre of sphere = midpoint of [PQ]

= (71+—5 147 2+—8) s D i) A =(-3,4,-3)

Radius of sphere = distance from centre

(—3, 4, —3)

to point P(—1, 1, 2)

=4/(-1-=-32+(1-4)24+(2--3)2 = /22 +(-3)2 + 52 =v4+9+25 /38 units

=

4.3 Volume of sphere = 77

=4 x7x (V38) ~ 981 units®

Surface area of sphere = 477>

=4 x 7 x (V38)? ~ 478 units?

Radius of cylinder = distance from centre of base

(1, —3, 0)

=(—2-1)2+(-2—-3)24+(0-0)?

— (BT ® =v9+1

=

V10 units

Volume of cylinder = 407 units® o

wrh = 40m

. mx (V10)? x h = 407 10h = 40 h=4

So, the height of the cylinder is 4 units.

to point

(—2, —2, 0)

427

428

Chapter 10 (Points in space) b

Exercise 10B

From a, the radius of the cylinder is v/10 units, and the height of the cylinder is 4 units. Since the base of the cylinder lies in the X-Y plane, a point P(z, y, z) lies on the curved surface of the cylinder if:

(-1 +(y+3)?=(/102=10 The point

(3, k, 2)

and

0< 2z 2

since the apex lies

Now, surface area of a square-based pyramid = area of 4 identical sides + area of square base = 210 units?

AB— T2 T G_0f

- VP =25+9 =34

area of square base = (v/34)? = 34 units? area of 4 identical sides + 34 units? = 210 units?

area of 4 identical sides = 176 units

2

area of triangular face = 44 units?

[AB] has midpoint M(

44 -1 0

WAL

043

OFR

2+2 a0

) which is M(%, &, 2).

VM=1/(3-0 - -1 2++( (2-3 2)?

(0, —1,2)

=@ @rr -

=4/E+(2-2)? {from above}

A(4,0,2)

3.2)

—< wies [ ol

AB=+/34

B(-1,3,2)

430

Chapter 10 (Points in space)

Exercise 10B

Area of triangle ABV = 44 units?

I XABx

VM =44

LxVBIx /U 172 +(2-2)2 =44 U 2

(9g_2=238 Z) \/fl

(

i

T +H2-2)"

2

3872

==

A2 =iaa

7455

2

7455

=

T

Vv

16.8

Since the base of the pyramid is in the plane place, is 16.8 — 2 = 14.8 units. ¢

a

Z = 2, the height of the pyramid, to 1 decimal

Volume of pyramid =

~ 10

z>2}

7455

Gl z~

{as

i

168 units

A has coordinates (10, 40) 2-dimensional plane.

Ais

plane.

(10, 40, 0)

B has coordinates

C has coordinates plane.

on the 3-dimensional

(110, 140)

(70, 20)

2-dimensional plane. Dis (70, 20, 0) plane.

il

e

on the

on the

50

(110, 140, 0) on the 3-dimensional

D has coordinates

.

i

2-dimensional plane.

Cis

B,

150

on the 3-dimensional

(50, 160)

2-dimensional plane. Bis (50, 160, 0) plane.

Ay (m

on the

on the

on the 3-dimensional

. A

D

. v

50

The apex is 15 m above the centre of the base. To find the centre of the base, we locate the midpoints of the diagonals.

The midpoint of [AC] is ( 1 ;110, 40 ;140, _O;U) which is (60, 90, 0).

The midpoint of [BD] is (50 - o 2z, %) the centre of the base is (60, 90, 0). the apex is (60, 90, 15).

which is (60, 90, 0).

= (m) 100

Chapter 10 (Points in space)

b

We need to find the side lengths of the base of the tent.

AD = /(70— 10)2 + (20 — 40)2 + (0 — 0)2 = /602 + (—20)2 + 02 = /4000 m CD = /(70— 110)2 + (20 — 140)? + (0 — 0)2 = /(~40)2 + (—120)2 + 02 — V16000 m

T(60,90,15)

Exercise 10B

431

B(50,160,0)

(o, 140 0)

/ A(10,40,0)

D(70,20,0)

Volume of air inside tent = % (area of base x height)

=1 % V16000 x V4000 x 15 = 40000 m* Let the apex

(60, 90, 15)

be T.

Let the midpoint of side [AD] be M.

M is (10270,

s

M)

which is (40, 30, 0).

MT = /(60 — 40)2 + (90 — 30)2 + (15 — 0)2

_ TR =/1225 =65 m

Area of triangle ADT = 1 x v/4000 x 65 /4000 m—>

4000 m? Let the midpoint of side [CD] be N.

N is (”0”0

140 s 120 0—;0) which is (90, 80, 0).

T

NT = /(60— 90)2 + (90— 80)2 + (15_ 0)2 + 152 =/(—30)2 + 102 = V1225

o1

=35m Area of triangle CDT = 1 x v/16000 x 35 3

D

"

r]l]

/16000 m—*>

16000 m?

Area of material needed for the tent = area of 4 triangular faces

=2 x area of AADT =2x8 \/400 %8540m

i

+ 2 X area of ACDT

+2x3 \/16000

C

432

11

Chapter 10 (Points in space)

Exercise 10C

Let the pyramid at the top of the structure have volume

Vi,

Z (m)4

and the rest of the structure have

(6,6,550)

volume Vpyse. The pyramid at the top has base side lengths

and height

(36,36,599) 1

A

(6,66, 550)

,

6,

(66,6,550)

66 — 6 = 60 m

599 — 550 = 49 m

Viep = 3 % 60 x 60 x 49

= 58800 m®

Y (m) (72,72,0)

X (m)

Now, the rest of the structure is part of a pyramid with apex

(36, 36, 2), z> 0.

§

Consider the cross-section of this pyramid through the apex and parallel to the Y-Z plane. Using similar triangles, z

Z—550

(36,36, 2)

72

60

0

—6k+19

26

. 400 — 240k + 36k* = 26(k? — 6k + 19) ©. 400 — 240k + 36k = 26k* — 156k + 494 o 10k? —84k—94=0 . 2(5k% — 42k —47) =0 . Bk?— 42k —47=0 (5k —47)(k+1)=0 .

k:%

or —1

The reasoning above is only true provided

When k=4,

20— 6k=20—6(4) = —1825 .

When k=—1, k= —1

k=%

20 — 6k > 0.

whichis

0

1is the only solution.

439

440 10

Chapter 10 (Points in space) a

Exercise 10C

To find the centre of the face ABCD,

we locate the

Ais

(0,6,5),

midpoints of the diagonals. (5,0,5),

Bis

(5,6,5),

Dis (0,0,5). The midpoint of [AC] is which is (3, 3, 5). The

is

of [BD]

midpoint

Cis

and

(% 2Lk 6y E) ’

60 seconds)

~ 0.487

i

78

~ 0.051

¢

P(between 20 and 59 seconds inclusive) = 387+819 ~ 0.731

3

a

Thesurvey lasted

2+7+114+84+7+4+34+0+1 = 43 days

b

i

12‘ frequency

P(0calls) = %

~ 0.0465

12

i P(>5 calls) & 2X3F0H1 ~ 0.186 NI S

2t

3

6

4

2 0

43

~ 0.465

01 23456738

number of calls per day

Chapter 11 (Probability)

Exercise 11A

459

Total frequency =37+ 81 +48+17+6+1 =190

a

P(4 days gap) ~ %

b

P(at least 4 days gap) ~ %

~ 0.0895

a

~ 0.126

.

13 30°C) ~ 2> ~0.366

Chapter 11 (Probability) ¢

Thereare

0.3+1.240.7 =2.2

in July which are > 40°C.

Exercise 11B

461

days in total during summer which are > 40°C, and 1.2 days 1.2

B

P(a summer day > 40°C is in July) ~ 55 0545

1

We extend the table to include totals for each row and column.

a

The total attendance for the match was 7510 people.

b

1

2439 out of the 7510 people at the match were children.

P(a child is selected) ~ 2439 0 0.325 7510

ii

4907 out of the 7510 people at the match were not season ticket holders. P(a non-season ticket holder is selected) ~ %

ili

1824 out of the 7510 people at the match were adult season ticket holders.

P(an adult season ticket holder is selected) ~ %

b

~ 0.653

i

~ 0.243

436 out of the 721 students surveyed play sport. 436

P(plays sport) ~ =1 & 0.605 ii

131 out of the 721 students surveyed play sport and are in the junior school. P(plays sport and is in junior school) ~ %

iii

~0.182

814176 = 257 students out of the 721 surveyed do not play sport and are in the middle or senior school. P(does not play sport and is in middle school or higher) ~ i—;: ~ 0.356

462 3

Chapter 11 (Probability)

Exercise 11C

We extend the table to include totals for each row and column.

| Peakseason | 225 [ 420 | 98 | 743 | a

1

743 out of the 1235 bookings made were in the peak season.

P(in the peak season) ~ % il

~ 0.602

148 out of the 1235 bookings made were for a single room in the off-peak season. P(a single room in the off-peak season) ~ 112;4385 ~0.120

iii

373+ 712 = 1085

bookings out of the 1235 were for a single or a double room.

P(a single or a double room) =~ % v

2254420+ 98+ 52 = 795 or were for a family room.

~ 0.879

bookings out of the 1235 were made during the peak season

P(during the peak season or a family room) ~ % b

~ 0.644

52 out of the 492 bookings made during the off-peak season were for a family room. P(booking made in off-peak season is for family room) ~ 4%22 ~ 0.106

¢

420+ 98 = 518 bookings out of 712+ 150 = 862 or family rooms.

were made in the peak season for double

P(booking made for a double or family room was in peak season) ~ Z—éi ~ 0.601

1

a {A,B,CD}

2

a U={1,234,5,6,729,10, 11,12, 13, 14, 15, 16}

b

b {1,2,3,4,5,6,7,8}

i A={4,8 12 16}




30

The modal class of the data is

x 100% =~ 36.7%

|

:‘

0

Exercise 12F

40 people waiting

20 - 29 people.

Tally

Hif HI H I H I il | Total b

164 frequency 14 12 10

8 6

L

4

B

2

B

0

0

10

20

¢

The modal class is

Alg

e

J 40

30

50

SR el

>

number of houses

30 - 39 houses.

100% = 67.5%

of the streets contain at least 20 houses.

533

534

Chapter 12 (Sampling and data)

1

H < 180

8

/

170 < H < 175

Exercise 12G

9

a

Height is a continuous variable as it is measured on a continuous scale.

b

Heights of a volleyball squad

SN

i; 4 frequency

175

180

185

190

195

200

205

height (cm)

¢

The modal class 185 < H < 190 cm occurs most frequently. heights in this interval than in any other interval.

d

The data is slightly positively skewed.

a

Travel time is a continuous variable, even though times have been rounded to the nearest minute.

V

b | Travel time (min)

W

N

=

0< 0< 0< 0