Focus PT3 2022 - Mathematics

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CC037232
PT3



FORM FOCUS
MATHEMATICS 1∙2∙3 PT3


KSSM
MATHEMATICS



REVISION FOCUS PT3 KSSM Form 1 • 2 • 3 – a complete
REVISION
REVISI
and precise series of reference books with special
ü Worked Examples ü Maths Info features to enhance students’ learning as a whole.
ü Tips ü Common Mistakes PT3
ü Concept Map This series covers the latest Kurikulum Standard
Sekolah Menengah (KSSM) and integrates
Pentaksiran Tingkatan 3 (PT3) requirements. FORM
REINFORCEMENT
REINFORCEMENT A great resource for every student indeed!
& ASSESSMENT MATHEMATICS 1∙2∙3
& ASSESSMENT
PT3 Model Paper
ü Formative Practices ü
ü PT3 Practices (QR Code) REVISION Penerbitan Pelangi Sdn Bhd. All rights Reserved
ü Complete Answers KSSM
REINFORCEMENT

EXTRA FEATURES
EXTRA FEATURES ASSESSMENT
ü HOTS Challenge ü PAK-21 EXTRA
ü Daily Application ü QR Codes
ü TIMSS Challenge FORM


TITLES IN THIS SERIES
• Bahasa Melayu • Geografi • Pendidikan Islam
• Tatabahasa • Sejarah • Reka Bentuk dan Teknologi 1•2•3
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CC037232
ISBN: 978-967-2720-76-8 Yong Kuan Yeoh (Textbook Writer)
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PELANGI Samantha Neo

CONTENTS











Form 1 4.5 Relationship between Ratios, Rates
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and Proportions with Percentages,
Fractions and Decimals 59
LEARNING AREA Number and Operations PT3 Practice 4 63

Chapter Chapter
1 Rational Numbers 1 5 Algebraic Expressions 65

1.1 Integers 2 5.1 Variables and Algebraic
1.2 Basic Arithmetic Operations Expressions 66
involving Integers 4 5.2 Algebraic Expressions involving
1.3 Positive and Negative Fractions 8 Basic Arithmetic Operations 71
1.4 Positive and Negative Decimals 12 PT3 Practice 5 76
1.5 Rational Numbers 14
PT3 Practice 1 16 Chapter
6 Linear Equations 78
Chapter
2 Factors and Multiples 18 6.1 Linear Equations in One Variable 79
6.2 Linear Equations in Two Variables 82
2.1 Factors, Prime Factors and Highest 6.3 Simultaneous Linear Equations
Common Factor (HCF) 19 in Two Variables 85
2.2 Multiples, Common Multiples and PT3 Practice 6 88
Lowest Common Multiple (LCM) 23
PT3 Practice 2 26 Chapter
7 Linear Inequalities 90
Chapter Squares, Square Roots, Cubes
3 and Cube Roots 28 7.1 Inequalities 91
7.2 Linear Inequalities in One
3.1 Squares and Square Roots 29 Variable 94
3.2 Cubes and Cube Roots 36 PT3 Practice 7 99
PT3 Practice 3 44
Measurement and
LEARNING AREA Geometry

LEARNING AREA Relationship and Algebra
Chapter Lines and Angles 100
Chapter 8
4 Ratios, Rates and Proportions 46 8.1 Lines and Angles 101
8.2 Angles related to Intersecting
4.1 Ratios 47 Lines 116
4.2 Rates 51 8.3 Angles related to Parallel Lines
4.3 Proportions 53 and Transversals 118
4.4 Ratios, Rates and Proportions 54
PT3 Practice 8 124





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0b Content Maths PT3.indd 5 12/22/21 8:23 AM

Chapter Measurement and
9 Basic Polygons 127 LEARNING AREA Geometry

9.1 Polygons 128 Chapter
9.2 Properties of Triangles and 13 The Pythagoras Theorem 195
the Interior and Exterior Angles
of Triangles 129 13.1 The Pythagoras Theorem 196
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9.3 Properties of Quadrilaterals 13.2 The Converse of Pythagoras
and the Interior and Exterior Theorem 199
Angles of Quadrilaterals 133 PT3 Practice 13 202
PT3 Practice 9 137

Chapter Form 2
10 Perimeter and Area 140
LEARNING AREA Number and Operations
10.1 Perimeter 141 Chapter
10.2 Area of Triangles, Parallelograms, 1 Patterns and Sequences 204
Kites and Trapeziums 146
10.3 Relationship between Perimeter 1.1 Patterns 205
and Area 154 1.2 Sequences 206
PT3 Practice 10 157 1.3 Patterns and Sequences 209
PT3 Practice 1 212


LEARNING AREA Discrete Mathematics LEARNING AREA Relationship and Algebra
Chapter Factorisation and Algebraic
Chapter 2 Fractions 215
11 Introduction to Set 161
2.1 Expansion 216
11.1 Set 162 2.2 Factorisation 220
11.2 Venn Diagrams, Universal 2.3 Algebraic Expressions and Laws
Sets, Complement of a Set of Basic Arithmetic Operations 224
and Subsets 166 PT3 Practice 2 226
PT3 Practice 11 173
Chapter
3 Algebraic Formulae 228

3.1 Algebraic Formulae 229
LEARNING AREA Statistics and Probability
PT3 Practice 3 234
Chapter
12 Data Handling 175 Measurement and
LEARNING AREA Geometry
12.1 Data Collection, Organization
and Representation Process, Chapter
and Interpretation of Data 4 Polygons 236
Representation 176 4.1 Regular Polygons 237
PT3 Practice 12 192
4.2 Interior Angles and Exterior
Angles of Polygons 239
PT3 Practice 4 245


vi






0b Content Maths PT3.indd 6 12/22/21 8:23 AM

Chapter Chapter
5 Circles 248 9 Speed and Acceleration 306

5.1 Properties of Circles 249 9.1 Speed 307
5.2 Symmetrical Properties 9.2 Acceleration 312
of Chords 251 PT3 Practice 9 314
5.3 Circumference and
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Area of a Circle 253 Chapter
PT3 Practice 5 259 10 Gradient of a Straight Line 316
10.1 Gradient 317
Chapter Three-Dimensional Geometrical PT3 Practice 10 323
6 Shapes 262

6.1 Geometric Properties Measurement and
of Three-Dimensional Shapes 263 LEARNING AREA Geometry
6.2 Nets of Three-
Dimensional Shapes 265 Chapter
6.3 Surface Area of Three- 11 Isometric Transformations 326
Dimensional Shapes 267
6.4 Volume of Three-Dimensional 11.1 Transformations 327
Shapes 274 11.2 Translation 328
PT3 Practice 6 280 11.3 Reflection 335
11.4 Rotation 340
11.5 Translation, Reflection
and Rotation as an Isometry 346
11.6 Rotational Symmetry 348
LEARNING AREA Relationship and Algebra PT3 Practice 11 350
Chapter
7 Coordinates 283
LEARNING AREA Statistics and Probability
7.1 Distance in the Cartesian
Coordinate System 284 Chapter
7.2 Midpoint in the Cartesian 12 Measures of Central Tendencies 352
Coordinate System 287
7.3 The Cartesian 12.1 Measures of Central Tendencies 353
Coordinate System 289 PT3 Practice 12 363
PT3 Practice 7 291
Chapter
13 Simple Probability 366

Chapter
8 Graphs of Functions 293 13.1 Experimental Probability 367
13.2 Probability Theory involving
8.1 Functions 294 Equally Likely Outcomes 369
8.2 Graphs of Functions 297 13.3 Probability of the Complement
PT3 Practice 8 304 of an Event 372
13.4 Simple Probability 374
PT3 Practice 13 376





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0b Content Maths PT3.indd 7 12/22/21 8:23 AM

Form 3 Chapter
6 Angles and Tangents of Circles 436

LEARNING AREA Number and Operations 6.1 Angle at the Circumference and
Central Angle Subtended by
Chapter an Arc 437
1 Indices 378 6.2 Cyclic Quadrilaterals 440

6.3 Tangents to Circles 443
1.1 Index Notation 379 6.4 Angles and Tangents of Circles 449
1.2 Laws of Indices 380 PT3 Practice 6 451
PT3 Practice 1 386

Chapter Chapter
2 Standard Form 387 7 Plans and Elevations 453

2.1 Significant Figures 388 7.1 Orthogonal Projections 454
2.2 Standard Form 391 7.2 Plans and Elevations 457
PT3 Practice 2 395 PT3 Practice 7 468

Consumer Mathematics: Chapter
Chapter Savings and Investments, 8 Loci in Two Dimensions 472
3 4.1 Scale Drawings Pelangi Sdn Bhd. All rights Reserved
397
Credit and Debt
8.1 Loci 473
3.1 Savings and Investments 398 8.2 Loci in Two Dimensions 474
3.2 Management of Credit and PT3 Practice 8 480
Debt 405
PT3 Practice 3 412


Measurement and LEARNING AREA Relationship and Algebra
LEARNING AREA Geometry
Chapter
Chapter 9 Straight Lines 483
4 Scale Drawings 414 9.1 Straight Lines 494
484
Penerbitan
PT3 Practice 9
415

Chapter PT3 Practice 4 423 Answers 497
5 5.1 Sine, Cosine and Tangent of Acute PT3 Model Paper & Answers
Trigonometric Ratios
425
https://plus.pelangibooks.com/
Resources/Focus2022/PT3/Mathematics/
Angles in Right-angled Triangle 426
PT3ModelPaper&Answers.pdf
434
PT3 Practice 5










viii






0b Content Maths PT3.indd 8 12/22/21 8:23 AM

Learning Area: Relationship and Algebra
Chapter Form 2
Mathematics PT3 Chapter 3 Algebraic Formulae
3 Algebraic Formulae










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KEYWORDS


• Value
• Constant
• Variable
• Subject of a formula
• Statement
• Algebraic formulae


























Concept
Map










Short-sightedness or myopia is a common eye disease which a person can see a clear vision
of near objects but blurry vision of distant objects. A short-sighted person has to use a
concave lens so that the light can be focused on the retina. Therefore, we can apply the
1 1 1
formula + = to find the focal length, f of a lens to correct short-sightedness, where
u v f
u and v represent the distance of the object and the image from the lens respectively.




228






F2 Chapter 3.indd 228 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae

3.1 Algebraic Formulae Number of rows, x 1 2 3 4 5 6
Number of squares, y 1 4 9

3.1.1 Form a formula based on a Solution:
situation
Number of 1 2 3 4 5 6
1. A formula is an equation that shows rows, x
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relationship between a few variables. Number of
2. For instance, A = πr + πrs is the formula squares, y 1 4 9 16 25 36
2
that shows the relationship between
variables A, r and s. The value of π is 1 × 1 3 × 3 5 × 5
always fixed and it is known as a constant.
2 × 2 4 × 4 6 × 6
Thus, y = x × x
MATHS INFO y = x 2

INFO
• A variable is a quantity which does not have a Try questions 1 – 8 in Formative Practice 3.1
fixed value. Variable is commonly represented
by using a letter.
• A constant is a quantity which has a fixed 3.1.2 Change the subject of formula Form
value. of an algebraic equation
Form
2
1 1. The subject of a formula is a variable that
is expressed in terms of other variables.

2. It must be a linear term with coeffcient
p
1. For example, the formula A = r h, A is
2
written in terms of h and r. Thus, A is the
2p subject of the formula.
Write a formula for area, A, of the rectangle
shown in the diagram above. 3. The variable of the subject of the formula is
usually written on the left of the equation
Solution: and the expression is on the right.
It is known that the area of a rectangle is
length × width.
MATHS INFO

INFO
Thus, A = 2p × p
A = 2p 2
In the formula, p = pq + r , p is not the subject of
2
the formula because the term p is on both sides
2 of the equation.
3

Express the variables in the bracket as the
subject of the formula.
One row Two rows Three rows
(a) a = b + c [b] (b) e = f – 3g [f ]
The diagram above shows a few arrangements i
of squares. Complete the following table and (c) h = 2j [i] (d) k = ml [m]
then write a formula that expresses y in terms

of x. (e) p = n [n] (f) L = j 3 [j]


229






F2 Chapter 3.indd 229 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae
Solution: (b) c = a + b 2
2
2
2
2
c – a = b 2
(a) a = b + c 2
 2
a – c = b + c – c Subtract both sides of the c – a = b
equation by c.
a – c = b Thus, b = c – a
 2
2
Thus, b = a – c
5
(b) e = f – 3g
e + 3g = f – 3g + 3g Add both sides of the 3n – np
equation by 3g.
e + 3g = f (a) Given the formula m = 4 , express n
Thus, f = e + 3g in terms of m and p.
2a – 3
i (b) Given the formula = 4, express a in
(c) h = a + b
2j terms of b.
i Multiply both sides of
h × 2j = × 2j Solution:
2j the equation by 2j.
2hj = i 3n – np
(a) m =
4
Thus, i = 2hj Bhd. All rights Reserved
4m = 3n – np
(d) k = ml 4m = n(3 – p)
k = ml Divide both sides of the 4m = n
l Form l equation by l. 3 – p
k = m Thus, n = 4m
Form
2 l 3 – p
Thus, m = k (b) 2a – 3 = 4
l a + b
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
(e) p = n 2a – 3 = 4(a + b)
2a – 3 = 4a + 4b

2
(p) =  n  Square both sides of –4b – 3 = 4a – 2a
2
the equation.
p = n –4b – 3 = 2a
2
Thus, n = p 2 –4b – 3 = a
2
(f) L = j 3 Thus, a = –4b – 3

3
3 L = j Take cube roots on both 2
 3

3 L = j sides of the equation.

3
Thus, j = L Alternative Method
4 (b) 2a – 3 = 4
a + b
Change the subject of the formula of the
following equations into the variable in the 2a – 3 = 4(a + b)
brackets. 2a – 3 = 4a + 4b
2a – 4a = 4b + 3

(a) p = r – s [r] –2a = 4b + 3
(b) c = a + b [b] a = 4b + 3
2
2
2
–2
Solution: Common (4b + 3)
mistakes

(a) p = r – s = – 2

p =  r – s  2 INFO –4b – 3
2
p = r – s Thus, a = 2
2
p + s = r
2
2
Thus, r = p + s Try questions 9 and 10 in Formative Practice 3.1
230
F2 Chapter 3.indd 230 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae

3.1.3 Determine the value of a Solution:
variable (a) P = 215 + 75t
= 215 + 75(3)
= 215 + 225
6 = 440
Given that r = 4s – 3t, find
(a) the value of r when s = 4 and t = 3, (b) 215 + 75t = 815
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(b) the value of t when r = 2 and s = 5. 75t = 815 – 215
75t = 600
Solution: t = 600
(a) Given that s = 4 and t = 3, t = 8 75
r = 4s – 3t Substitute the values of
= 4(4) – 3(3) variables into the formula. The number of days the customer stays in
= 16 – 9 the hotel is 8.
= 7
Try questions 11 – 19 in Formative Practice 3.1
(b) Given that r = 2 and s = 5,
r = 4s – 3t
3t = 4s – r Change the subject of the PISA
4s – r
t = formula into the variable The pacelength, p, is the distance between the Form
3 that need to be found. two consecutive footprints. A formula of n = 130
Form
t = 4(5) – 2 p
3 shows the relationship between n and p, where 2
20 – 2 n is the number of steps per minute and p is the
=
3 pacelength in metres.
18 (a) Rozaidi walks 52 steps per minute. Calculate
=
3 the pacelength of Rozaidi, in cm.
t = 6 (b) Halim walks with his pacelength is 0.5 metres.
Calculate the speed of Halim walked, in
kilometres per hour (km/h).
Alternative Method

3.1.4 Solve problems involving
r = 4s – 3t
2 = 4(5) – 3t formulae
2 = 20 – 3t
3t = 20 – 2 8 Daily Application
PAK-21
3t = 18 P AK - 21
t = 18 35 m
3
t = 6 5 m Park
k m
Garden
h m
7 Daily Application
The diagram above shows a piece of rectangular
The price, in RM, of a customer who stays land that is used to build a park and a garden.
in a hotel for t days follows the formula of (a) Hussin wants to fence the garden with wire.
P = 215 + 75t. Express the perimeter, p, of the garden in
(a) Find the value of P when t = 3. terms of h and k.
2
(b) Calculate the number of days the customer (b) If the area of the park is 265 m and k = 10,
stays in the hotel if he pays RM815. calculate the value of h.


231






F2 Chapter 3.indd 231 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae
Solution: (c) Perimeter, P
(a) Length of the garden = (35 – h) m
Width of the garden = (k – 5) m
a
Perimeter of the garden,
p = (35 – h) + (35 – h) + (k – 5) + (k – 5)
= 70 – 2h + 2k – 10
= 2k – 2h + 60 2. Write a formula to show the relationship
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(b) Area of the park = 35 × k – (35 – h)(k – 5) between the variables in each of the following
= 35k – (35 – h)(k – 5) statements.
(a) The sum of nine and square of a number,
When k = 10, x, is y.
35k – (35 – h)(k – 5) = 265 (b) Power, P, is the measure of work done, W,
35(10) – (35 – h)(10 – 5) = 265 divided by the time taken, t.
350 – 5(35 – h) = 265 (c) Einstein’s equation states that the energy,
–5(35 – h) = 265 – 350 E, is the product of mass, m, and the
square of speed of light, c.
–5(35 – h) = –85
–85
35 – h =
–5 3. A pen and an exercise book cost RM1.50 and
35 – h = 17 RM1 respectively. Remy bought x pens and y
h = 35 – 17 exercise books. The total amount paid is RMz.
h = 18 Write a formula for z.
Try questions 20 – 25 in Formative Practice 3.1 4. Yi Liang is m years old. His brother is 6 years
Form
Form
2 older than him. The total age of Yi Liang and
HOTS Challenge his brother is N years. Write a formula for N.
Praktis Formatif 1
Jerry bought p bottles of carbonated drinks with the 5. A shirt costs RM50. In a mega sale, customers
price of RMq per bottle. He sold all the carbonated get a discount of d% for the shirt. Write a
drinks with the price of RMx per bottle and earned formula for the new price, H.
a profit of RMy. Express p in terms of q, x and y.
Solution: 6. Write a formula for y based on the following
Purchasing price = p × RMq information in each tables.
Selling price = p × RMx (a) x 1 2 3 4 5
Profit = RMy
Profit = Selling price – Purchasing price y 1 8 27 64 125

Thus, y = px – pq (b) x 1 2 3 4 5
px – pq = y y 3 6 9 12 15
p(x – q) = y
p = y
x – q 7.


Formative Practice 3.1 1 triangle 2 triangles 3 triangles

1. Write a formula for each of the following based The toothpicks are arranged to form a few
on the given diagrams. triangles as shown in the diagram above.
(a) Volume, V (b) Area, A Analysing
(a) How many toothpicks are used to arrange
5 triangles?
r
t (b) Write a formula for the number of toothpicks,
q p, if n triangles are formed.
p s


232






F2 Chapter 3.indd 232 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae

8. p
2
15. Given the formula t = 3m – , find
4
(a) the value of t when m = 1 and p = –4,
(b) the value of m when p = 16 and t = 8.
h = 1 h = 2 h = 3 6m – 1
16. Given the formula = –3p, find
m + n
The diagram above shows a few patterns (a) the value of m when n = –2 and p = –1,
drawn in square grids. The height of each (b) the value of n when m = 4 and p = –2.
pattern is h units and the number of dots in each
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pattern is represents by d. Analysing 17. Given the formula pq + 5q = 2ps, find
4
(a) Complete the following table. (a) the value of q when p = 2 and s = –11,
h 1 2 3 4 5 (b) the value of p when q = –1 and s = 2.
d 22
18. Given the formula h = k(2m + k), find
(b) Write a formula for d in terms of h. 7
(a) the value of h when k = –7 and m = –2,
9. Express the variables in brackets as the subject (b) the value of m when h = –154 and k = –1.
of the formula.
(a) h = e + 4 [e] 19. Given the formula of the volume of a cylinder
2
(b) c + 2d = e [c] is V = πr h, where r and h are the radius and
(c) l = b – d [b] height of cylinder respectively. Calculate the
3
(d) –p + r = h [p] volume, in cm , for the cylinder with a radius
2
m of 5 cm and a height of 2 cm.
(e) = 4r [m] Form Form
n  Use π = 22  Analysing
V 7
(f) = R [I]
I 20. 2
(g) 2rt = s [r]
(h) 3v = nq [n] h
(i) y = 7(2 – w) [w] r
(j) v = u + 2as [u]
2
2
(k) j =  [I] Given the area of a circle and the area of the
3I

3
curved surface of the cylinder are πr and 2πrh
2
4k
L
  [L] respectively.
(a) Express the total surface area, A, of the
(l) T = k g cylinder in terms of π, r and h.
10. A stone is thrown vertically upwards with initial (b) Hence, find the value of h, in cm, when
velocity, u m/s. The final velocity, v m/s, of A = 528 cm , π = 22 and r = 2 cm.
2
the stone at t, in seconds, after throwing is 7
v = u + gt. Express t in terms of g, u and v. 21. Given the area of equilateral triangle with

length of 2a cm is  3 a cm . Find the length
2
2
Applying of side if the area of the equilateral triangle is

11. Given the formula a = b + c, find 4 3 cm .
2
2
(a) the value of a when b = 7 and c = –2,
(b) the value of c when a = 8 and b = –3. 22.
2
12. Given the formula q = n – t + 2m , find
6 h cm
(a) the value of q when m = 4, n = 2 and t = 1,
(b) the value of t when m = –1, n = 3 and q = 2. y cm
y cm
13. Given the formula x = b – 4ac, find
2
(a) the value of x when a = 6, b = –7 and c = 2, The diagram above shows a cuboid with
(b) the value of a when b = 5, c = –1 and x = 33. square base of y cm and height of h cm. Given
14. Given the formula y = 18a – 3ac , find the total surface area of the cuboid is A.
b 2 (a) Express h in terms of y and A.
(a) the value of y when a = 2, b = –1 and c = –2, (b) Hence, calculate the value of h when
(b) the value of b when a = 4, c = 3 and y = 36. y = 2 and A = 48.
233

F2 Chapter 3.indd 233 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae

23. 24. This year, Phang is k years old and his mother
C is three times of his age. The sum of their ages
2h cm five years ago is q years. Express k in terms
of q. Applying
B

A E 6 cm D 25. Chong has RMg. After spending RM(h – 3) a
day for 5 days, he still has RM10. Express h
in terms of g. Applying
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The diagram above shows two right-angled
triangles ACD and ABE. Given that BC = 2h cm,
DE = 6 cm, BC = 1 AE and 4DE = 3AB.
2
Express the perimeter, p, of right-angled
triangle ABE, in terms of h. Applying





PT3 Practice 3



Section A Form g = 3h – 2, then h = pk
6. Given 4 = 4p – 1, then p =
1. Given 1
Form
2 A g + 2 C 3(g + 2) A 4 – k C 4(16 – k)
4
4
3
g B 16 – k D k – 16
B + 2 D 3g + 2
3
7. Given R = 3  1  , express T in terms of R
b – 5 S – T
2. Given = a , express b in terms of a. and S.
2
2 9 9
A b = a + 5 C b = 2(a + 5) A T = R 2 – S C T = S + R 2
2
2 9 3
5 B T = S – D T = S –
B b = a + D b = 2a + 5 R 2 R 2
2
2
2
7(y + z)
3
3. Given 27 = (p – q) , express p in terms of q. 8. Given y = z , express z in terms of y.
A p = √27 + q C p = q + 3 A z = y – 7 C z = y
3
3
B p = 27 + q D p = q – 3 7y y – 1
3
B z = 7y D z = 7y
4. Given c – d(6 + c) = d, express c in terms of d. y – 7 y + 7
7d 7d p + r
A c = C c = 9. If p = 3, q = –2 and r = 5, therefore =
2 1 – d q
B c = 7d D c = 5d + 1 A –4 C 2
1 + d 2 B –2 D 4
5. Given k = l + 3 , express l in terms of k. 10. The distance, s, in m, travelled by an accelerating
2l rocket is given by s = ht + 1 at . Find the value
2
2k – 1 2
A l = C l = 2(3k – 1)
3 of s when h = 3 m/s, t = 100 s and a = 0.5 m/s .
2
3 A 300 C 1 400
B l = D l = 2k – 3
2k – 1 B 700 D 2 800
234






F2 Chapter 3.indd 234 12/22/21 8:44 AM

Mathematics PT3 Chapter 3 Algebraic Formulae
Section B (b) h
1. (a) A square has a side of 2q cm. Mark (✓) for the
correct formula for perimeter, P, of the square,
in cm. [1 mark] p
( ) P = 4q
2
( ) q = 4P Based on the diagram above, write a formula
2
( ) P = 8q of h in terms of p. Applying
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( ) q = 8P [3 marks]
(c) The diagram below shows a trapezium LMNP
(b) Complete the following steps to express m and a right-angled triangle PQR.
as the subject of the formula. [3 marks]
4 M (q + 1) cm N R
= 1
5m – c
(r + 2) cm
4 =
5 m = L (q + 7) cm P 2q cm Q


m = Given that the total area of both polygons is
3p cm . Express r in terms of p and q.
2
[4 marks]
2. (a) State the subject of the formula for each of Form Form
the following. i-THINK [3 marks] 2. (a) (i) Given 9b – 6 = 3, express b in terms
2b + c
1 (a + b)t = s of c. [2 marks] 2
2
3
Formulae v = u + at 2 √p + q = r (ii) Given k = (lm + n) × (l + m – 1) , find the
Subject (i) (ii) (iii) value of k when l = –4, m = 2 and n = 7.
[2 marks]
(b) Circle the possible values of the following (b) Nurul Huda is 10 years older than Najwa.
variable. [1 mark] After eight years, Aisyah’s age is twice
Najwa’s age. Analysing
Variable Possible value (i) If p represents Najwa’s present age and
J represents the total ages of Nurul Huda
r = Number of and Aisyah after eight years, express J in
students who terms of p. [3 marks]
attend extra –5 0 5 10.5
class (ii) Find Aisyah’s age after eight years if
J = 46. [1 mark]
(c) The formula of the area of a trapezium is
Section C 1
x z L = 2 (a + b)h. Find the value of a when
1. (a) Given – 2y = .
2
4 3 L = 128, b = 14 and h = 16. [2 marks]
(i) Express z in terms of x and y. [1 mark]
(ii) Calculate the value of z when x = – 8 and
y = – 2. [2 marks]















235






F2 Chapter 3.indd 235 12/22/21 8:44 AM

Learning Area : Measurement and Geometry
Chapter Form 3


5 Trigonometric Ratios










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KEYWORDS


• Sine
• Cosine
• Tangent
• Right-angled triangle
• Opposite side
• Adjacent side
• Hypotenuse
• Special angle
• Acute angle















Concept
Map


















Photo courtesy : Tourism Malaysia

Mount Kinabalu in Sabah is the highest mountain in Malaysia with a height of 4 095 m. Lake
Bera in Pahang with a length of 35 km and a width of 20 km is the largest natural lake in
Malaysia. Do you know how the height of the mountain and the distance across the lake are
determined?



425






F3 Chapter 5.indd 425 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

5.1 Sine, Cosine and Tangent of (b) PQ is the opposite side.
QR is the adjacent side.

Acute Angles in Right-angled
Triangles Try questions 1 and 2 in Formative Practice 5.1

5.1.1 Identify the opposite side and
adjacent side 5.1.2 Define trigonometric ratios
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1. In a right-angled triangle, the longest side 1. The ratio between two sides of a right-
which is opposite to the right angle is angled triangle is known as trigonometric
called hypotenuse. ratio. This ratio remains constant if the
size of the angle in the triangle changes
proportionally.
Opposite Hypotenuse
side 2. For an acute angle, A, in a right-angled
θ triangle ABC:
Adjacent side
A
2. The side which is opposite to the angle q Adjacent
is called opposite side and the side which side Hypotenuse
is next to the angle q is called adjacent
side. B Opposite C
side
TIPS
Ratio of the opposite side to the
For a right-angled triangle, hypotenuse always hypotenuse is named as sine.
remains the same whereas the opposite side
and the adjacent side change according to the sin A = opposite side
position of the given angle. hypotenuse

Ratio of the adjacent side to the
MATHS INFO hypotenuse is named as cosine.

INFO
adjacent side
cos A =
Besides alphabetical letters, the angle labelled in hypotenuse
a triangle can be indicated by using Greek letters
such as q (theta), a (alpha) and b (beta).
Ratio of the opposite side to the adjacent
side is named as tangent.
1 tan A = opposite side
adjacent side
Identify the opposite side and the adjacent side
with respect to the shaded angle for each of the
Form
Form
3 right-angled triangles below. MATHS INFO

(b)
(a)
INFO
C
Q
Abbreviation for sine, cosine and tangant are sin,
cos and tan respectively.
P
A B R
Solution:
(a) BC is the opposite side. Mnemonic for memorizing
trigonometric ratios
AB is the adjacent side. INFO

426






F3 Chapter 5.indd 426 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

2 5.1.4 Determine the values of
trigonometric ratios of acute
angles
y
z
1. The values of sine, cosine or tangent of an
θ x acute angle in a right-angled triangle can
be determined when the lengths of sides
State the following trigonometric ratios based
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on the length of sides of the right-angled of the triangle are given.
triangle above. 4
(a) sin q
(b) cos q Determine the value of trigonometric ratio
(c) tan q of the acute angle given in each of the right-
angled triangles below.
Solution: (a) sin x
y
(a)
z 4 cm
x 3 cm
(b) x
z
y 5 cm
(c) (b) tan y
x
Try question 3 in Formative Practice 5.1 y
6 cm 10 cm
INFO
5.1.3 The impact of changing the size 8 cm tan q = sin q
of the angles on the values of (c) cos z cos q
trigonometric ratios
13 cm 5 cm
When the size When the size z
Value of the angle of the angle 12 cm
increases decreases
sine increases decreases Solution:
cosine decreases increases 3 opposite side
(a) sin x = hypotenuse
tangent increases decreases 5

Opposite side
4 cm
3 3 cm
x Form Form
5 cm
Each of the following shows the sine, cosine Hypotenuse
and tangent of an angle. Arrange the values of
the sine, cosine and tangent in ascending order. 3
(a) sin 62°, sin 29°, sin 90°, sin 34° (b) tan y = 8 opposite side
(b) cos 44°, cos 76°, cos 6°, cos 15° 6 adjacent side
(c) tan 26°, tan 0°, tan 85°, tan 57° 4
=
Solution: 3
(a) sin 29°, sin 34°, sin 62°, sin 90° Adjacent side y 10 cm
(b) cos 76°, cos 44°, cos 15°, cos 6° 6 cm
(c) tan 0°, tan 26°, tan 57°, tan 85°
8 cm
Opposite side
Try questions 4 and 5 in Formative Practice 5.1

427






F3 Chapter 5.indd 427 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

12 adjacent side 6
(c) cos z =
13 hypotenuse
Determine the value of each of the following.
Give your answer correct to 3 decimal places.
Hypotenuse
13 cm (a) sin 41°
5 cm
z (b) cos 9.6°
12 cm (c) tan 73°
Adjacent side
Solution:
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Try question 6 in Formative Practice 5.1
(a) sin 41° = 0.656 Press sin 4 1 =
5 (b) cos 9.6° = 0.986 Press cos 9 · 6 =

In the diagram on the 15 cm R (c) tan 73° = 3.271 Press tan 7 3 =
3 P x
right, sin x = . Find the
5 Try question 8 in Formative Practice 5.1
value of
(a) cos x Q
(b) tan x 5.1.5 Determine the values of sine,
cosine and tangent of special
Solution: angles
PQ
(a) sin x =
PR 1. Angles such as 30°, 45° and 60° often seen
PQ = 3 in trigonometry. These angles are called
15 5 special angles.
3
PQ = × 15 2. The exact values of trigonometric ratios for
5 30° and 60° can be determined by using
= 9 cm a triangle of 30°-60°-90° whereas the exact
2
2
QR = 15 – 9 2 values of the trigonometric ratios for 45°
= 144 can be determined by using a triangle of
—
QR = √144 45°-45°-90°.
= 12 cm Triangle Triangle
QR 30°-60°-90° 45°-45°-90°
Therefore, cos x =
PR
12 A D
=
15 45°
2
4 30° 2 1
= 3
5
60° 45°
PQ E 1 F
(b) tan QR B D 1 C
Form x =
Form
3 = 9 q 30° 45° 60°
—
12 1 1 √3
3 sin q —
= 2 √2 2
4 —
√3 1 1
Try question 7 in Formative Practice 5.1 cos q —
2 √2 2
2. The values of trigonometric ratios can be 1 —
—
determined by using scientific calculator. tan q √3 1 √3
During calculation, calculator has to be in
degree mode. Try question 9 in Formative Practice 5.1

428






F3 Chapter 5.indd 428 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios
Solution:

The exact values are stated in (a) sin 56° = p opposite side
surd form for irrational numbers 25 hypotenuse
INFO p = 25 × sin 56°
= 20.73
HOTS Challenge (b) tan 60° = q
Praktis Formatif 1
8
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If q is the angle formed — q
between the hour hand and √3 = 8
—
the minute hand, what are the θ q = 8√3
exact values of sin q, cos q
and tan q at two o’clock? 8
cos 60° =
Solution: r
2 1 = 8
q = × 90° = 60° 2 r
3
sin q = sin 60° cos q = cos 60° r = 16
—
= √3 = 1
2 2 Alternative Method
tan q = tan 60°
—
= √3 Use Pythagoras theorem to find the value
of r.
—
r = 8 + (8√3 )
2
2
2
= 64 + 192
5.1.6 Perform calculations involving = 256
—
sine, cosine and tangent r = √256
= 16
1. We can use trigonometric ratios to find
the lengths of unknown sides of a triangle
if the measurements of one side and one 8
angle are given. S
x
7

Find the values of p, q and r of each of the
following right-angled triangles. P Q 4 cm R
(a) using calculator In the diagram above, PQR is a straight line.
2
Given sin x = , determine the length of PR.
5 Form Form
25 cm Solution:
p cm QR
sin x = 3
56° QS
4 2 2
that is, = Given sin x =
QS 5 5
(b) without using calculator
2QS = 5 × 4
QS = 10 cm
60°
8 cm r cm Therefore, PQ = QS = 10 cm
Hence PR = 10 + 4
= 14 cm
q cm
Try questions 10 – 13 in Formative Practice 5.1


429






F3 Chapter 5.indd 429 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

2. When the values of sine, cosine and 3. If the lengths of two sides of a right-angled
tangent of an angle are given, the size of triangle are given, then the angle of the
the angle can be determined by finding the triangle can be determined.
inverse values of the trigonometric ratios
using scientific calculator. 10

9 Find the angle q in the following diagram. Give
your answer correct to one decimal place.
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Find the size of angle q of each of the following.
Give your answer correct to 1 decimal place. θ
(a) sin q = 0.75 16 cm 10 cm
(b) cos q = 0.06
(c) tan q = 3.7
Solution:
Solution:
(a) sin q = 0.75 Press θ Opposite
side
−1
q = sin (0.75) SHIFT sin 0 . 7 5 = 16 cm 10 cm

q = 48.6° Hypotenuse
Display 48.59037789
(b) cos q = 0.06 Press sin q = 10
−1
q = cos (0.06) SHIFT cos 0 . 0 6 = 16


q = 86.6° Display 86.56018723 10
q = sin –1 1 2
16
(c) tan q = 3.7 Press = 38.7°
−1
q = tan (3.7) SHIFT tan 3 . 7 =


q = 74.9° Display 74.87599269 Try question 16 in Formative Practice 5.1
TIPS
5.1.7 Solving problems
You can check the answers by using
calculator.
(a) sin 48.6° = 0.7501... 11
(b) cos 86.6° = 0.0593…
(c) tan 74.9° = 3.706…

Wire Wire
MATHS INFO 16 m

INFO
58°
Inverse sine of x is written as sin x. 58° Wire
–1
Inverse cosine of x is written as cos x.
–1
Form
Form
3 Inverse tangent of x is written as tan x. 58°
–1
A 16-m tall telecommunication tower is built
Try questions 14 and 15 in Formative Practice 5.1
on a hill. To support the tower, three wires
are attached at the top of the tower at an angle
of elevation of 58° as shown in the diagram
Common mistake above. A utility worker brings a roll of wire
with the length of 60 m. Is the wire brought
INFO
enough?


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F3 Chapter 5.indd 430 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios
Solution: (b) In the right-angled triangle EAC,
Let x represents the length of a piece of wire tan ∠ACE = 8
needed. 11.31
16 8
sin 58° = ∠ACE = tan –1 1 2
x 11.31
16 x m 16 m = 35.3°
x =
sin 58° E
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58°
= 18.87
8 cm
Total length of three wires
= 3 × 18.87 11.31 cm
= 56.61 m , 60 m A C
Therefore, the length of wire brought is enough. Try question 19 in Formative Practice 5.1


Praktis Formatif 1

MATHS INFO HOTS Challenge
INFO
In the application of trigonometry involving right- According to the safety
angled triangle, angle of elevation and angle of requirement, a ladder is
depression are used in the measurements of height stable when the base of the
and distance which are difficult to determine. These ladder leans at an angle
angles exist when we look upward or downward 10 m in the range of 70° to 80°
at an object. with the horizontal ground.
A painter is standing on
• Angle of elevation • Angle of depression a ladder which is leaning
Angle of against the wall as shown in
depression the diagram. The length of
Angle of 1.5 m the ladder is 10 m and the
elevation distance from the base of the ladder to the wall

is 1.5 m at that moment. Assuming you are the
safety inspector, what advice will be given to the
Try questions 17 and 18 in Formative Practice 5.1 painter? Give justification to your answer.
Solution:
12 Let q represents the angle between
the base of the ladder and the
H The diagram on the left horizontal ground.
shows a cube with side 1.5 10 m
E G cos q =
F length of 8 cm. Calculate 10
1 2
D (a) the length of line AC, q = cos –1 1.5
(b) the angle between the 10 Form θ Form
A C = 81.4°
lines AC and EC. 1.5 m
B
The ladder is not safe to use as 81.4° lies outside 3
Solution: the range of 70° to 80°.
(a) In the right-angled triangle ABC, The painter is advised to move the base of the
ladder slightly away from the wall.
2
2
AC = AB + BC Pythagoras theorem
2
2
= 8 + 8
2
= 128 C
—
AC = √128 Ladder Safety using 4 : 1 Rule
= 11.31 cm 8 cm
INFO
A 8 cm B
431






F3 Chapter 5.indd 431 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios
Formative Practice 5.1 5. Mark ‘3’ for the correct statement.

(a) sin 25°, sin 10°, sin 70°, sin 32°, sin 54°
1. Identify the opposite side and the adjacent side
with respect to the shaded angle in each of the (i) sin 70° has the smallest value.
following right-angled triangles. (ii) sin 32° , sin 54°
(a) (b)
T (b) cos 0°, cos 15°, cos 30°, cos 45°, cos 60°
K
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(i) cos 0° has the largest value.
(ii) cos 30° . cos 15°
S
M (c) tan 0°, tan 40°, tan 90°, tan 20°, tan 72°
L R
(i) The values are arranged in
2. A ladder leaning against a vertical wall as shown descending order: tan 90°, tan 72°,
in the diagram below. Identify the opposite side tan 40°, tan 20°, tan 0°
and the adjacent side with respect to the given (ii) tan 90° has the smallest value.
angle q.
6. Determine the value of the trigonometric ratio
stated for each of the following right-angled
5 m triangles.
4 m
(a) tan x (b) cos y
θ 8 cm 15 cm y
3 m 12 cm 20 cm
x
17 cm
3. 16 cm
p q (c) sin z (d) sin q
x
r
Based on the diagram above, complete the 15 cm 9 cm
empty boxes. 21 m 29 m
z
(a) sin x = (b) cos x = 12 cm θ
20 m
(c) tan x = 7. In the diagram on the
7
right, cos x = 25 . Find
4. The following diagrams show the values of the value of x
sine, cosine and tangent for the corresponding 25 cm
angles in the table below. Complete the table. (a) sin x
(b) tan x
Sine 0.9063 0.2588 0 0.9848 0.6428
8. Determine the value of each of the following.
Form
Form
3 Cosine 0.4226 1 0.9659 0.1736 0.7660 Give your answer correct to 3 decimal places.
(a) sin 20°
Tangent 0.2679 5.6713 0.8391 0 2.1445 (b) cos 58°
(c) tan 31.5°
q 0° 15° 40° 65° 80°
9. Complete each of the following without using
sin q calculator.
cos q equal
to sin 30° as cos 30° as tan 45° as sin 45° as cos 60° as tan 60°
tan q (a) (b) (c) (d) (e) (f)




432






F3 Chapter 5.indd 432 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

10. Find the value of q of each of the following 16. Determine the angle q in each of the following
triangles. Give your answer correct to 2 decimal diagrams. Give your answer correct to 1
places. decimal place.
(a) (c) (a) (c)
7.1 cm 6.6 cm
18 cm q cm
22.6 cm
50° q cm 65 cm
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14° θ
(b) q cm
31°
θ
6.2 cm (b) 14 cm
θ
72 cm
11. For each of the following diagrams, find the 76.2 cm
values of x and y without using calculator.
(a) (b)

x cm 5 cm y cm 18 cm 17. A surveyor wants to know the distance
across a lake. The diagram below shows two
45° 30° measurements obtained by him. Show your
y cm x cm calculations to find the value of x, that is the
distance across the lake. Give your answer to
12. S the nearest metre. Applying

6 cm

θ
P Q 4 cm R x m
In the diagram above, PQR is a straight line.
Given sin q = 2 , find the length of PQ. 40°
3 576 m
13.
P
18. On a cliff of 120 m high, an observer saw a
buoy on the surface of the sea at an angle
30 cm
of depression of 15°. Calculate the horizontal
x distance of the buoy from the cliff. Give your
Q M R answer to the nearest metre. Applying
In the diagram above, M is the midpoint of QR. Form Form
Given tan x = 3 , find the length of MR. 19.
5 F 10 cm
14. Evaluate each of the following to the nearest E 3
degree. 3 cm
(a) tan (2.051) C
–1
(b) sin (0.3062) D B
–1
(c) cos (0.6528) A 4 cm
–1
15. Find the value of x of each of the following. The diagram above shows a right prism with a
Give your answer correct to 1 decimal place. uniform cross-section of a right-angled triangle
(a) tan x = 0.5 ABE. Calculate
(b) sin x = 0.24 (a) the length of AC,
(c) cos x = 0.08 (b) the angle between line AF and line AC.



433






F3 Chapter 5.indd 433 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios

PT3 Practice 5



Section A tan tan tan tan
1. R
x (b) Find two pairs with equivalent value.
[2 marks]
Q
kos 60°
y 1 1
P S 2 2
In the diagram above, PRS is a right-angled sin 60°
triangle. Q is the midpoint of side PR. Which of
the following sides is adjacent to the angle x? kos 45° tan 60°
A RS B QS C PR D QR 3
2. T 7 cm S R
x 2. (a) Based on the given diagram, write the correct
trigonometric ratio (sin, cos, or tan) in the
12 cm blanks. [2 marks]


P Q 4 cm 9 cm
The diagram above shows a square PQRT. TSR Diagram x 7 cm 15 cm
is a straight line. Determine cos x.
A 5 B 5 C 7 D 12 y
13 12 12 13
3. Penerbitan Pelangi Sdn Bhd. All rights Reserved 7
Trigonometric
9
P
ratio x = 4 y = 15
B
10 cm (b) Fill in the blanks with the correct numbers.
Q 20.3° = 20° + 0.3° [2 marks]
D = 20° + (0.3 × )9
2 cm
A C = 20° + 9
T 4 cm S R
In the diagram above, PQR and TSR are straight = 20°189
lines. Which of the following angles A, B, C and Section C
2 1. (a)
D, with the tangent value of ?
3 P
Section B Roof beam
1. (a) 1.5 m
Form
Form
3 z x
R
3 m
w Q The diagram above shows a truss supporting
x a roof. Determine the value of x. [2 marks]
y
(b) A ladder leans at 75° from the ground for
optimal safety. Kamal uses a ladder of
4.5 m long to clean the windows at the
The diagram above is drawn on square grid second floor of his house. What is the height,
with sides of 1 unit. Arrange the values of in m, that the ladder can safely reach? Give
tangent of the given angles in ascending your answer to 1 decimal place. [4 marks]
order. [2 marks] Analysing


434






F3 Chapter 5.indd 434 12/22/21 9:00 AM

Mathematics PT3 Chapter 5 Trigonometric Ratios
(c) (c)
3°
Vertical 250 m
distance
Aeroplane runway
10 m A pilot found that his plane was at an altitude
of 250 m above a runway with slope of an
Jib
60° angle of 3°. How far, in m, is the plane
30° needed to glide before it could land on the
runway? Applying [3 marks]
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A crane carries cargo from a hatch and 3. (a) The diagram below shows a cuboid. If the
—
swings it to the side before lowering it onto length of BD is 8√3 cm, find the length of
the dock. The length of the jib of the crane BH. [2 marks]
is 10 m. Determine the vertical distance, in
m, covered by the end of the jib when the H G
operator raises the jib from 30° to 60°. Give E F
your answer in surd form. [4 marks] 60°
Analysing
D C
2. (a) (i) S A B
y y (b)
15 cm The line
of sight
x
P Q R
Theodolite 24° 900 m
In the diagram above, PQR is a straight
line. Given sin x = 4 and tan y = 2.
5 1.5 m
Calculate the length, in cm, of PQR. A surveyor wants to determine the height of
[2 marks] a building. He used a theodolite to measure
(ii) C the angles. The theodolite was placed 1.5 m
above the horizontal ground and 900 m away
b 60° from the base of the building. The angle of
a elevation of the top of the building from the
theodolite is 24°. Determine the height, in m,
30°
A c B of the building. Give your answer to 1 decimal
place. Applying [4 marks]
The diagram above shows a triangle
ABC. Without using calculator, determine (c)
the values of a and c, if b = 12 cm.
[2 marks]
Wire A Wire B
(b) Form Form
y
48° 52°
P(a, b) Q(8 3 , c) 3
10 m 8.5 m
The diagram above shows a cellular telephone
6
3
tower which is supported by two wires. Wire A
60° 45° is attached to the tower at an angle of elevation
O T S R(d, 0) x of 48° whereas wire B is attached at an angle of
elevation of 52°. Both wires A and B are attached
The diagram above shows a trapezium at a distance of 10 m and 8.5 m away from
OPQR which is drawn on a Cartesian plane. the base of the tower respectively. Which wire
Find the values of a, b, c and d. [3 marks] is longer? Show your calculations. [4 marks]
Analysing


435






F3 Chapter 5.indd 435 12/22/21 9:00 AM

Mathematics
Mathematics PMR Answers
Mathematics PT3 Answers





6. (a) 491 (b) 530 7 1 5 1 3
Form 1 (c) 4 900 (d) 27 (d) (i) –1 8 , –1 2 , – , 1 4 , 2
8
(e) 2 804 (f) 120 3 1 5 1 7
Chapter 7. 15 m below sea level (ii) 2 , 1 4 , – , –1 2 , –1 8
8
1 Rational Numbers 8. Water level decreases by 6 m. 3
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9. (a) Leong will not receive a prize 4. (a) –1 8 (b) 3
Formative Practice 1.1 because he answered 13 questions (c) –1 1 (d) – 3
correctly, 2 questions incorrectly 18 4
6
1. Positive numbers: and 1 question not answered; The (e) – (f) 1 7
2 , 7.3, 5, +3 total marks that he obtained is 34. 7 9
5 (b) John will likely receive a prize. If (g) –1 1 (h) –4 1
Negative numbers: he answered 13 questions correctly 3 8
and 3 questions not answered, so
3
–10, –67, – , –8.9 the total marks obtained is 36 marks. (i) 4
4 15
If he answered 13 questions
(Zero is not a positive number and not correctly, 1 question incorrectly and 5. 2 250 m below sea level
a negative number) 2 questions not answered, maybe 7
2. (a) –7 levels (b) +500 m the total marks obtained is 35 6. Not enough. Still short of 10 kg of
(c) –RM450 (d) +RM0.50 marks. sugar.
(e) –150 m (f) –3 cm
3. (a) –100 Formative Practice 1.4
(b) +38 or 38
(c) –45 Formative Practice 1.3 1. (a)
4. Positive integer: 4; 1. (a)
Negative integer: –8, –3
5. (a) –0.4 –0.2 –0.1 0.2 0.3
– – 4 – – 2 – 1 – 3
–3 –2 1 2 4 7 7 7 7 (b)
(b)
(b) –5.2 –3.9 –1.3 2.6 3.9
–15 –9 –3 3 6 –1– 1 3 – – 2 3 0 – 1 3 2. (a) 2.9
6. (a) 8 (b) –15 (c) (b) –5.6
7. (a) Largest integer: 6; (c) –1.11
Smallest integer: –10 3. (a) (i) –6.7, –3.31, –1.4, 3.87, 4.5
(b) Largest integer: 14; – – 1 2 – – 1 3 1 – – 1 3 (ii) 4.5, 3.87, –1.4, –3.31, –6.7
6
Smallest integer: –16 (b) (i) –5.2, –3.0, –0.4, 0.9, 1.4
(ii) 1.4, 0.9, –0.4, –3.0, –5.2
8. (a) –6, –5, –4, –2, 1, 3 (d) (c) (i) –4.11, –3.22, –1.44, 1.55, 2.33
(b) –10, –8, –6, 3, 7, 9
(c) –16, –14, –3, 0, 11, 18 3 1 1 1 (ii) 2.33, 1.55, –1.44, –3.22, –4.11
(d) –19, –13, –4, –3, 4, 19 – – 8 – – 4 – 8 – 2 (d) (i) –5.44, –5.42, –2.9, 0.03, 0.3
9. (a) 5, 3, 0, –1, –4, –6 (ii) 0.3, 0.03, –2.9, –5.42, –5.44
(b) 7, 5, 3, –1, –2, –6 2. (a) 4 (b) –1 1 4. (a) –1.66 (b) –1.2
(c) –9.35
(d) 8.985
(c) 4, 2, 1, –1, –3, –7 5 2
(d) –1, –3, –5, –8, –10, –12 (e) 2 (f) –6.665
(c) –2 3 (d) –2 1 (g) –19.35 (h) –5.647
5 9 (i) –3.072
Formative Practice 1.2 1 3 1 1 5 5. –4°C
3. (a) (i) – , – , – , , 6. Danny gained a profit of RM185.
1. (a) 5 (b) 3 2 8 8 2 8
3
1
(c) –1 (d) –14 (ii) 5 , 1 , – , – , – 1
(e) –3 (f) 11 8 2 8 8 2 TIMSS Challenge
(g) –10 (h) –4 7 2 1 1 3
2. (a) –16 (b) –20 (b) (i) – 10 , – , , , 5 ‒; +; ‒
5 10 2
(c) 42 (d) 36 3 1 1 2 7
,
3. (a) –15 (b) –7 (ii) 5 , 2 10 , – , – 10 Formative Practice 1.5
5
(c) 7 (d) 12 1. All the numbers given are rational
5
4. (a) 0 (b) 1 (c) (i) –1 1 , –1 1 , – , 4 , 7 numbers.
(c) 12 (d) 27 3 6 6 9 9 –8 63 1 6
5
5. (a) 10 (b) 0 (ii) 7 9 , 4 9 , – , –1 1 6 , –1 1 3 –8 = 1 ; 3.15 = 20 ; 1 5 = 5
6
(c) 40 (d) 17 2. (a) –7.2 (b) 2.58
(e) 15 (f) –28 (c) –0.3125 (d) –3.905
(g) –6 (h) –4
497
F1 Answers.indd 497 12/22/21 9:05 AM

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