Preface
Study Smart Additional Mathematics – GEOMETRY & TRIGONOMETRY is designed to
provide students with structured practice in Additional Mathematics and to expose students to
varied question types to prepare them for major examinations.
What can you expect to find in this book?
Key Concepts
These are included to help you recap essential concepts that you must know for each topic.
These summarise what you learn in school and will save your precious time to get started on the
practice questions.
Worked Examples
These show detailed and specific steps to solving a problem of a particular question type.
Learning is also reinforced through similar question types following the examples.
Useful Notes
These are extra information that can help you better understand the concepts so that you can
work out the solutions faster.
Guided Exercise
These provide you with specific instructions to guide you through solving a problem of a
particular question type.
Practice Questions with Quick Answer Key
These consolidate what you have learnt in each topic and give a comprehensive overview of the
concepts covered. Quick answer key beside the questions also makes checking of answers much
easier for students.
Past O-Level Examination Questions Analysis
This analysis provides insight into the types of questions and trends in examinations.
Test Papers
There are 2 test papers to assess your overall understanding of the concepts taught and learned.
Full Worked Solutions
These are provided to help students better understand how each problem is being solved and also
serve as a tool for self-study and assessment.
The Editorial Team
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Contents
Chapter 1 Coordinate Geometry . ............................................................................ 1
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Chapter 2 Linear Law �������������������������������������������������������������������������������������������� 66
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Chapter 3 Circles ���������������������������������������������������������������������������������������������������� 88
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Chapter 4 Trigonometric Functions, Identities and Equations ������������������������ 113
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Chapter 5 Further Trigonometric Identities ������������������������������������������������������ 203
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Chapter 6 Proofs in Plane Geometry ������������������������������������������������������������������ 241
Key Concepts • Types of Questions • Worked Examples • Guided Exercise • Practice Questions with
Quick Answer Key • Analysis of Past O-Level Examination Question Types
Test Paper 1 �������������������������������������������������������������������������������������������������������������� 264
Test Paper 2 �������������������������������������������������������������������������������������������������������������� 265
Analysis of Past O-Level Examination Questions . ................................................ 266
Years 2008 to 2015
Worked Solutions ����������������������������������������������������������������������������������������������� S1–S76
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Chapter
1
Coordinate Geometry
• Condition for two lines to be parallel or perpendicular
• Midpoint of line segment
• Area of rectilinear figure
Key Concepts
Distance Between Two Points
1
Distance between two point A(x1, y1) and B(x2, y2) is given by the following.
AB = ​√(x2 – x1)2 + (y2 – y1)2 ​
_________________
________________
= ​√(x-step)2 + (y-step)
2 ​
B(x2, y2)
M(x1, y1)
y2 – y1
A(x1, y1)
θ
x2 – x1
Midpoint of Two Points
2
( )
x + x ______
y +y
Midpoint M of the line joining two points A(x1, y1) and B(x2, y2) is M = ​ ______
​ 1 2 2 ​, ​ 1 2 2 ​ ​.
Note that rectangles, rhombuses and squares are special parallelograms and their diagonals bisect
each other, i.e. the diagonals have a common midpoint.
Gradient or Slope of Line Passing Through Two Points
3
The gradient or slope of a line is a measure of its steepness.
4
Gradient m of a line passing through two points A(x1, y1) and B(x2, y2) is given by the following,
change in y-coordinates ____
rise ​ y-step m = ____________________
​ change in x-coordinates
​= ​ run ​ = _____
x-step ​
y2 – y1
= ​ ______
​
x – x 2
1
= tan θ
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Gradient
Size of θ
m>0
0º < θ < 90º
Rise from left to right
m<0
90º < θ < 180º
Falls from left to right
m=0
θ = 0º
Horizontal
m=∞
θ = 90º
Vertical
Nature of line
Gradient and Collinear Points
5
I f three points A, B and C are collinear, i.e. they lie on the same straight line, then gradient of AB = gradient of BC = gradient of AC and area of ∆ABC = 0.
Conversely, if gradient of AB = gradient of BC = gradient of AC, then A, B and C are collinear.
Gradient and Parallel Lines
6
If two lines are parallel, their gradients are equal.
Conversely, if two lines have the same gradient, then the lines are parallel.
Gradient and Perpendicular Lines
7If two non-vertical lines are perpendicular, the product of their gradients is equal to negative one
(–1). Conversely, if the product of the gradients of two lines is equal to negative one (–1), then
the lines are perpendicular.
8
A
B
θ
θ
C
If ∠ABC is equal to ∠ACB and BC is horizontal, then mAB = –mAC, where mAB is the gradient of
AB and mAC is the gradient of AC.
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Equation of a Non-Vertical Straight line
9
The equation of a line states the connection between the x and y values for every point on the line.
10 The forms of equation of a non-vertical straight line are summarised in the following table.
From
Equation
General
Ax + By + C = 0
Gradient-intercept
y = mx + c
y – y1 = m(x – x1)
Gradient-point
y2 – y1
y – y1 = ​ ______
(x – x1)
x – x ​ 2
1
y
​ a ​ + ​ __ ​ = 1
b
x
__
Intercept
Equation of a Vertical Straight Line
11 All vertical lines have equations of the form x = a, where a is a constant.
Coordinates of Midpoint
12The coordinates of the midpoint M(x, y) of a line segment AB can be found if the coordinates of
A and B are known.
B(x2, y2)
y2 – y
M(x, y)
x2 – x
y – y1
A(x1, y1)
x – x1
Based on congruent triangles,
( x – x ) ( x – x )
1
​ ​y2 – y​ ​= ​ ​ y – y ​ ​
x2 – x = x – x1 and y2 – y = y – y1
2x = x1 + x2 and 2y = y1 + y2
x +x
y +y
x = ______
​ 1 2 2 ​ and y = ______
​ 1 2 2 ​ 2
1
( )
x1 + x2 ______
y +y
The coordinates of the midpoint of A(x1, y1) and B(x2, y2) are ​ ​ ______
​, ​ 1 2 2 ​ ​.2 13Parallelograms (including rectangles, rhombuses and squares) have diagonals which bisect each
other, i.e. the diagonals have a common midpoint.
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Coordinates of the Vertices of a Parallelogram
14The coordinates of the vertex D(x4, y4) of a parallelogram can be found if the coordinates of the
other 3 vertices A(x1, y1), B(x2, y2) and C(x3, y3) are known.
B(x2, y2)
(y2 – y1)
A(x1, y1)
(x2 – x1)
D(x4, y4)
Based on congruent triangle,
x3 – x4
x2 – x1
​ ​y – y ​ ​= ​ ​y ​ ​
– y4 2
1
3
x2 – x1 = x3 – x4 and y2 – y1 = y3 – y4
) ( ( (y3 – y4)
C(x3, y3)
(x3 – x4)
)
x1 + x3 = x2 + x4 and y1 + y3 = y2 + y4
x4 = x1 + x3 – x2 and y4 = y1 + y2 – y2
 Coordinates of vertex D are (x4, y4), i.e.(x1 + x3 – x2, y1 + y3 – y2).
Ratio Theorem (Finding Coordinates of a Point that Divides a Line Segment in a Given
Ratio)
15 T
he coordinates of a point P(x, y), which divides a line segment AB such that AP and PB are in
the ratio m : n, can be found if the coordinates of A and B are known.
B(x2, y2)
n
y2 – y
P(x, y)
m
A(x1, y1)
x2 – x
y – y1
x – x1
Based on similar triangles,
x – x1
x2 – x
n​ ​y – y ​ ​ = m​ ​y – y​ ​
1
2
x_____
– x1 __
y – y1 __
m
_____
​ x – x ​ = ​ n ​ and ​ ​ = ​ m
y
–
y
n ​ 2
2
nx – nx1 = mx2 – mx and ny – ny1 = my2 – my
( ) ( )
mx + nx = nx1 + mx2 and my + ny = ny1 + my2
(m + n)x = nx1 + mx2 and (m + n)y = ny1 + my2
nx1 + mx2
ny1 + my2
________
x = ​ ________
​ and
y
= ​ ​
m+n
m + n nx1 + mx2 ________
ny1 + my2
Coordinates of P(x, y) = ​ ________
​ m
​, ​ m
​ ​
+ n + n x + x2 ______
y + y2
If P is the midpoint of AB, then m = n and coordinates of P(x, y) = ​ ______
​ 1 ​
, ​ 1 ​ ​.
2
2
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( )
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Coordinates of a Point On a Line Segment With Known Distance of the Line Segment And
Equation
16
he coordinates of A(x, y) on a line segment AB can be found if the following are known:
T
(a) Length D of the segment AB
(b) Equation of AB, y = mx + c
(c) Coordinates of B
A(x, y)
D
B(h, k)
Coordinates of A(x, y) can be stated
in terms of x as A(x, mx + c).
_________________
Applying distance formula, D = ​√(x – h)2 + (mx + c – k)2 ​
This equation can be solved for x, the only unknown.
Hence the coordinate of A(x, y) can be found.
Coordinates of Point(s) of Intersection
17 T
he coordinates of point(s) of intersection of straight lines and curves can be found by solving
simultaneous equations.
18 Axis intercepts are the values of x and y where a graph cuts the coordinate axes.
19A x-intercept is a point where a graph cuts the x-axis. The x-intercept is found by letting y = 0,
i.e. solving simultaneously the equations of the graph and the x-axis, y = 0.
20A y-intercept is a point where a graph cuts the y-axis. The y-intercept is found by letting x = 0,
i.e. solving simultaneously the equations of the graph and the y-axis, x = 0.
Area of a Polygon
21 T
he area of a polygon with vertices A(x1, y1), B(x2, y2), C(x3, y3), ......, K(xn, yn) taken in an
anticlockwise order is given by the following.
Area of polygon
x1 x 2
yn x 1
= __
​ 12 ​​ ​y ​ ​y ​
 ​y ​ ​y ​ ​
2
1
1
n
1
__
= ​ 2 ​|x1y2 + x2y3 + xny1 – x2y1 – x1yn|
| |
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© Singapore Asia Publishers Pte Ltd & Henry Loh
08 Analysis of 2008-2015 'O' Lev266 266
Analysis of Past O-Level Examination Questions
266
2/17/2016 7:17:33 PM
Q1
6
3
9
13
11
7
9
29
80
Grand Total
6
3
Calculus Total
Kinematics
Q6
Q8iii
100
21
3
4
Q8i,
10i,ii
Q4ii
Integration
Plane Area
2
Q8ii
Q13
Minima and Maxima
9
2
Q10iii
49
10
9
Q5ii,7
Q6
Q8i,10i
2
Q4i
19
14
14
12
30
Differentiation
Gradients, Tangents and
Normals
Rate of Change
Plane Geometry
Geometry and
Trigonometry Total
Q3i,9
Q3ii,7
3
Q8i
Further Trigonometry
8
Q1,8ii
Trigonometry
8
Q11
Q12
32
Circles
Linear Law
Coordinate Geometry
Algrebra Total
180
50
6
3
7
11
2
19
2
68
9
17
22
12
8
0
62
7
Q11
8
Binomial Expansions
8
0
Q4
Modulus Functions
Logarithm
11
5
9
7
27.8
37.8
34.4
%
100.0
2008
Total
Indices
Q2
Q5
Q2
Mark
N08/2
0
3
6
11
Qn
Surds
Partial Fractions
Q5i
Q10
Quadratic Functions
Polynomials
Q3,9
Mark
N08/1
Simultaneous Equations
Qn
Q12iii
Q2
Q12ii
Q5
Q12i
Q8i
Q8ii,9
Q10
80
20
3
5
2
6
4
33
3
12
9
9
27
6
5
4
6
6
Mark
N09/1
Q11
Q4
Q3
Q1
Q6
Q7
Qn
Q8
Q6
Q2ii
Q10
Q4
Q1,11
Q9
Q5
Q7
100
35
10
9
5
11
36
8
18
10
29
9
9
3
8
Mark
N09/2
Q2i
Q3
Qn
180
55
10
9
8
16
2
6
4
69
8
21
12
10
9
9
56
9
9
6
0
5
3
12
6
6
Total
30.6
38.3
31.1
%
100.0
2009
Q4ii
Q8i,
11ii,ii
Q2ii,6ii,
11
Q4i,8ii
Q6i
Q2i,10
Q12
Q7
Q5
Q1
80
31
11
9
2
7
2
26
9
10
7
23
6
4
13
Mark
N10/1
Q3,9
Qn
Analysis of Past O-Level Examination Questions
Q8
Q7ii
Q10iii
Q2
Q10ii
Q6
Q11ii,iii
Q1,11i
Q5i,7i,9
Q4
Q3,5ii
100
29
11
6
4
6
2
47
10
8
8
21
24
8
11
5
Mark
N10/2
Q10i
Qn
180
60
11
6
15
15
2
7
4
73
10
17
8
10
7
21
47
8
6
0
11
0
5
4
13
0
33.3
40.6
26.1
%
100.0
2010
Total
Chapter 1
Type C Question
Step
Worked Solutions for Guided Exercise
Type A Question 1
______________
Step 1 AB = ​ [7 – (–1)]2 + (1
– 2)2 ​
√_____
= ​√64 + 1 ​ __
= ​√65 ​ units
Type D1 Question
Step
Type A Question 2
______________
Step 1 AB = ​ (7 – 4)2 + [1 – (–2)]2 ​
√____
= ​√9 + 9 ​ __
= ​√18 ​ _
= 3​√2 ​ units
____________
BC = ​√(7
– 3)2 + (1 – 5)2 ​
______
= ​√__
16 + 16 ​
= ​√32 ​ _
= 4​√2 ​ units
______________
CA = ​√[5 – (–2)]2 + (3
– 4)2 ​
_____
= ​√49 + 1 ​
__
= ​√50 ​ _
= 5​√2 ​ units _
_
_
_
2 Perimeter = 3​√2 ​ + 4​√2 ​ + 5​√2 ​ = 12​√2 ​ units
| |
|
Type D2 Question
Step
6 – (–1)
1 mPQ = _______
​ 18 – 4 ​
7
= ___
​ 14
​ = __
​ 12 ​
Type B2 Question
09 Solutions.indd 1
| x1 x2 x3 x1
1 Area of ∆PQR = __
​ 12 ​​ ​y1​ ​ y2​ ​y3​ ​y1​ ​
6 6 ​ ​
2 ​ –2
= __
​ 12 ​​ ​–4
​ ​–1
​2 ​ ​–4
= __
​ 12 ​[(6)(–1) + (2)(2) + (–2)(–4)
– (–4)(2) – (–1)(–2) – (2)(6)]
= __
​ 12 ​(–6 + 4 + 8 + 8 – 2 – 12)
= 0 units2
2 Since area of ∆PQR = 0 units2,  P, Q and R
are collinear
Type B1 Question
_________________
1 AB = ​√___________
[–2 – (–1)]2 + [k – (–4)]2 ​
2
= ​√_________
1 + k + 8k + 16 ​
2
= ​√k______________
+ 8k + 17 ​ AC = ​√_____________
[5 – (–2)]2 + (8
– k)2 ​
= ​√__________
49 + 64 – 16k
+ k2 ​
2
= ​√k – 16k + __________
113 ​ _________
2 ​√k2 + 8k + 17 ​ = ​√k2 – 16k + 113 ​
3 k2 + 8k + 17 = k2 – 16k + 113
24k = 96
k=4
Study Smart Additional Mathematics – Geometry&Trigo
© Singapore Asia Publishers Pte Ltd & Henry Loh
Method 1
2 – (–1)
1 mPQ = _______
​ –2 – 2 ​
= –__
​ 34 ​
2 – (–4)
mPR = _______
​ –2 – 6 ​
6
= ___
​ –8
​ = –__
​ 34 ​
2 Since mPQ = mPR,  P, Q and R are collinear
Method 2
____________
Step 1 AB = ​ (5 – 2)2 + (1–
h)2 ​
√___________
= ​√_________
9 + 1 – 2h + h2 ​ = ​√h2 – 2h + 10 ​ _________
2
2 ​√h – 2h + 10 ​ =5
h2 – 2h + 10 = 25
h2 – 2h – 15 = 0
(h – 5)(h + 3) = 0
h – 5 = 0 or h + 3 = 0
h = 5 or –3
Step
4 – (–2)
______
1 Gradient = ​ 6 – 3 ​ _
6
= ​ 3 ​
=2
5 – (–1)
mQR = _______
​ 1 – 4 ​
6
= ___
​ –3
​ = –2
2 (mPQ)(mQR) = ​ __
​ 12 ​ ​–2 = –1
( )
3 S
ince (mPQ)(mQR) = –1,  PQ is perpendicular
to QR
S
Chapter 1
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