DUM10122 WORKBOOK

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.6 PARTIAL FRACTIONS

The Partial fraction for a linear denominator ax + b is given by


+

Example : Combined two algebraic fractions to become single compound fractions

Form the following algebraic fractions to a single b) 1 − 3
compound 3x +1 4x − 8

a) 2 + 3 Solutions:
3+x 4−x

Solutions:

3 2 x + 3 = 2(4 − x) + 3(3 + x)
+ 4−x (3 + x)(4 − x)

= 8 − 2x + 9 + 3x

(3 + x)(4 − x)

= x + 17 x)

(3 + x)(4 −

Example : Algebraic fraction as the sum of partial fractions

Find the sum of partial fraction for the following algebraic fractions.

a) x + 17

(3 + x)(4 − x)

Solutions:

(3 x + 17 x)  (3 A x ) + (4 B x)
+ −
+ x)(4 −

 x −17 = A(4 − x)+ B(3 + x)

For 3 + x = 0 For 4 − x = 0
x = −3 x=4

− 3 + 17 = A(4 − (− 3)) + B(3 + (− 3)) 4 + 17 = A(4 − 4) + B(3 + 4)

14 = 7A 21 = 7B
A=2 B=3

 (3 x + 17 x) = 3 2 x + 4 3 x
+ −
+ x)(4 −

51

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

b) −x−4

(3x + 1)(5x − 2)

Solutions:

c) 3x
x2 − x − 2

52

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

TUTORIAL 1.6
1. Form the following algebraic fractions to a single fraction.

a. 4 − 1 b. 3 − 3
x−3 x+2 1− x x +1

c. 3 + 1 d. 2 − 1
3 − 3x 10x + 2 x−2 x+1

53

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

2. Find the sum of partial fraction for the following algebraic fractions.

a. 6 b. 3x + 11

(x + 1)(x −1) (x − 3)(x + 2)

c. 2x + 3 d. 5x
x2 − 9
(x − 4)(5x + 2)

54

Unit 2

GEOMETRY

INTRODUCTION

OVERVIEW

The word geometry is Greek for geos (earth) and metron (measure). Geometry was extremely
important to ancient societies and was used for surveying, astronomy, navigation and building. Geometry
is the study of angles and triangles, perimeter, area and volume.

Geometry is the math related to proportions, or size, shape and position, so practical
applications of geometry come in measurement and spatial reasoning. Everything from wrapping a
gift to designing a backyard landscape is governed by geometry

People commonly use geometry area problems when working on their homes. Area problems
help them decide how much carpet or paint to buy and even which furniture will work in a given space.
Likewise, decorating a room or an outdoor area takes spatial reasoning and an arrangement of geometric
shapes

Volume problems are another daily-life use of geometry. People use volume equations to
determine how much water goes in a fish tank, how much sand is needed for a sandbox or even how

much soil is needed

CONTENT CONTENT

2.1. Pythagoras Theorem
2.2. Perimeter And Area Of Planes
2.3. Surface Area And Volume Of Solids
2.4. Circular Measures

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Use Pythagoras Theorem to find length of sides of right-angled
triangles.

2. Convert the angle in degree to radian and vice versa.

2.1 PYTHAGORAS THEOREM

Right-Angled Triangle

Pythagoras Theorem

+ =

Square of the longest side of a right-angled triangle (hypotenuse) is equal to
the sum of square of the other two sides.

(3,4,5) : 32 + 42 = 52 = √ +
(5,12,13) : 52 + 122 = 132 = √ −
(8,15,17) : 82 + 152 = 172 = √ −

Example

Question Solution

a Find length of c. 2 = 32 + 42
= √32+ 42
3 c = √25
b Find length of a. 4 = 5

a
13

5

c Find length of b.

8

17
b

d Find length of r.

P3R

r
Q

Question Solution
e Given the right-angled triangle has side 7 cm and

the hypotenuse 16 cm. Find length of the third
side.

7 cm 16 cm



f Given RS = 2PQ. Calculate the length of QS. 162 = 2 + 72
PQ 2 = 162 − 72
3 = √162−72
= √207
10
= 2(3) = 6

TS R
g Find the value of m.

3
13

4
m

TUTORIAL 2.1 C
1. Find AC.

A

41 m

32 m

12 m

2. Captain is on a geo-cache hunt. His GPS tells him that he is 40m away from the treasure. He
walks 24m due west. The GPS compass now tells him that the treasure is due south from
where he is standing. How far south does he need to go to find it?

Hint :
N

WE

S

3. Find the value of x.

17 cm

8 cm

x

4. Given KN = 13 cm, PN = 10 cm and PM = 8 cm. Find KL.
K

L MN

P T
5. PQR and URS are straight lines. Find RU. 8 cm

15 cm
U

R

5 cm Q

P S
12 cm

6. To get from point A to B you must avoid walking through a pond. To avoid the pond, you must
walk 34 meter to the south and 41 meter to the east. What is the nearest distance to point B
if it is possible to walk through the pond? (give your answer to the nearest meter).

7. A student was given three sticks to form a right angle triangle. The length of the two shorter
sides is 7cm and 24 cm respectively. Calculate the length of the third side.

8. Find the length of PC. C
D

AB
15 cm

S R
P 16 cm 12 cm
Q

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 1. RecogniSe the properties of common planes.
2. Calculate perimeters and areas of common planes.

2.2 PERIMETER AND AREA OF PLANES

Perimeter : The total length around the outline of a closed
figure.

Area : The measure of the size of a figure. Compute
as the product of two linear measures and is
express in square units.

Properties of Triangle

Triangle Shape Properties Perimeter Area
Scalene
Triangle h All three sides are
b NOT in the same
Isosceles length. No angles
Triangle h have the same
b
value.

Two sides are in the a+b+c 1  base  height
same length. The 2
angles opposite the
equal sides have

equal value.

Equilateral All three sides are
Triangle h in the same length.

All three angles=
have the same
value, which is

b 60 .

Right-Angled a c Have a right angle,
Triangle or 90

b

\

Properties of Parallelogram, Trapezium, Circle and Polygon

Name Shape Properties Perimeter Area

Parallelogram ah Two pairs of 2(a + b) bh
b parallel sides
and opposite
sides equal.

Trapezium a d Two sides are a+b+c+d 1 (a + b)h
ch parallel. 2

b

r A perfectly
round shape
Circle and have a 2r r 2

radius.

Polygon All sides are
straight lines
joined and
closed. First
polygon is

triangle.

Properties of Square and Rectangle

Name Shape Properties Perimeter Area
Square s2
s All four sides are 4s
Rectangle in the same length lw
w 2(l + w)
l and have four
right-angled.

Opposite sides
are in the same
length and have
four right-angled.

Example
1. Find the perimeter and area of the shape below :

Solution :
Perimeter =

Area =

35 cm
Calculate the perimeter and area of the
square shape clock.

Solution :
Perimeter =
0.6 m Area =

1.94 m
Calculate the perimeter and area of the
carpet.

2. Find the perimeter and area of the combined shape.

45 cm Solution :
26 cm Perimeter =

24 cm 25 cm

10 cm 7 cm Area =

3. Given that the area of the trapezium as shown in the figure above is 80 cm2. Find its perimeter.

Solution :

10 cm 8 cm

4. Find the area of shaded region. Solution :
QR

4 cm

PS
A circle is inside rectangle with the side
touches the rectangle. Given QR = 2QP.
Calculate area of shaded region.

Q 12 cm R Figure show a semicircle and right angle triangle
5 cm O PQR. Given PR is the diameter of the semicircle,
P PQ = 5cm and QR = 12 cm. Calculate the

a) Radius of semi-circle
b) Area of shaded region (use ℼ = 3.142)

TUTORIAL 2.2
1. Find the area of the parallelogram.

12 cm
18 cm

2. Figure shows a shape which is combining between a rectangle and a triangle. If the perimeter of
the shape is 28 cm, find p.

4 cm 5 cm

p cm 3 cm

3. Figure shows a rectangle, PQRS. T and U are midpoints of PQ and QR. Find the area of the shaded
region, in cm2.

U
QR

T 3 cm

P 12 cm S

4. Calculate the area of shaded region.
10 cm

| 3 cm

5 cm |2 cm

7 cm

5. A square kitchen has an area of 100 square feet. What is the kitchen’s perimeter?

6. Mr. Lim’s bathroom measures 6 feet by 10 feet. She wants to cover the floor with square tiles.
The sides of tiles are 6 inches. How many tiles will Mr. Lim need?

7. The triangle is an isosceles triangle with its base passing through the center of the circle. The
diameter of the circle is 14 cm. Calculate the area of shaded region.

8. A circle has radius 12 cm and the a) Value of θ in radian.
angle of minor sector, θ is 54o. By b) Area of shaded region.
using ℼ = 3.142, calculate

R

12 cm
54o O

P

9. Figure show a semicircle and right a) Radius of semi-circle
angle triangle PQR. Given PR is the b) Area of shaded region (use ℼ = 3.142)
diameter of the semicircle, PQ =
5cm and QR = 12 cm. Calculate the

Q 12 cm R
5 cm O
P

40 m 10. A farm has perimeter 360 m, what is the length of
x?
x 80 m
11. ADE is triangle and ABCD is rectangle in which AB
60 m = 7cm, BC = 15 cm and the height of triangle is 5
cm. Calculate the area of the whole diagram.
E
A 5 cm D
7 cm
B 15 cm C

12. ABC is right angle triangle and KLMN is a) The area of triangle ABC
trapezium. Given that the area of the triangle is
equal to the area of the trapezium. Calculate

AL

b) Length of MN

10 cm 5 cm
K M

O 3 cm
C 6 cm BN

13. LPRT is rectangle and HRT is a right-angled triangle. Given that LP = HL = 12.50 cm. Find the
length of HR and give your answer in 2 decimal places.

L 12.5 cm P

7.30 cm

TR

H

14. A circle inside a square. Each side is 10 cm. Calculate area of the circle.
|

-- -- 10 cm
|

15. PQRS is rectangle and KLS is right angle triangle. Find the length of LQ.

1cm
S KR

L

P 16 cm Q
16. Two quadrants PAD and RBC was cut from a
P 20 cm S
32 cm D A rectangular plane with PS = 20 cm and PQ = 32
cm. Given SB = BR. Calculate
Q -- a) Perimeter of the shaded region

B b) Area of the shaded region
-- 16 cm

CR

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 1. Recognise the properties of common solids.
2. Calculate surface areas and volumes of common solids.

3.3 SURFACE AREA AND VOLUME OF SOLIDS

Properties of Common Solids

A solid is a three-dimensional figure that has length, width and height.
Surface area of a solid is the total area of its external surface.
Volume is the quantity of three-dimensional space enclosed by a closed surface. Volume is often
quantified numerically using cubic metre.

Example

1. Find the volume of

8 cm 20 cm
12 cm
0.05 m
16 cm
4 cm
Solution : Solution :

2. Based on following figure, find a) surface area of shaded region
12 cm b) volume, in cm3, of the solid.

5 cm 4 cm
4 cm

3. A hemisphere with diameter 10 cm is removed from a cylinder with 16 cm height. Find the
volume of the remaining solid.

16 cm

TUTORIAL 2.3
1. Half of a cuboidal container is filled with water. Calculate the volume of the water.
8 cm

5 cm
2. As a gift, you fill the calendar with packets of chocolate candy. Each packet has a volume of 2 cubic

inches. Find the maximum number of packets you can fi t inside the calendar.

3. Find the volume of composite solids.

4. Diameter of cone is 10 cm and the volume of solid is 628 4 cm3. Find the height of cylinder, h cm
7

  = 22  .
 7

12 cm
h cm

5. Find the surface area of composite solids.

6. A solid sphere is placed
inside a cylinder.
Calculate the volume of
the empty space in the
cylinder.

7 cm 22 cm

3 cm 7. A composite solid is form
6 cm by combining a cylinder
and a hemisphere. Given
the radius of hemisphere
is 7cm. Calculate the
volme of the solids in
form of ℼ.

21 cm 8. A cone is taken out from
cylinder with height 21
7 cm cm and radius 7 cm. The

height of the cone is 1 of

3

the cylinder’s height. By

using = 22 calculate

7

a. The surface area of
the cylinder.

b. Total volume of the
remaining solid.

9. A cuboidal steel plate with length 7 cm and width have two cylindrical holes. The height of the holes is
equal to the cuboid and diameter is 2 cm. determine the volume of the plate. (Use = 22)

7

2 cm
3 cm

12 cm

7 cm

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Convert angles from degree to radian and vice versa.
2. Calculate the circular measurement of a circle.

2.4 CIRCULAR MEASURES

1 circle = 360 degrees
1 quadrant = 90 degrees
1 degree = 60 minutes
1 minute = 60 seconds

2.4.1 UNIT OF ANGLE

Angle in circle can be measured either in degree (°) or radian (rad).
1 radian is obtained when the value of arc length is equal to the length of radius in a
circle.

Arc length = L = 3.142 rad
Radius = r = 22
L = r when =1rad =
rad
1° = 7
180°
1 rad =


180
180



2.4.2 RELATION BETWEEN DEGREE AND RADIAN

Degree °  Radian rad
180

 180


Example Convert Radian to Degree
0.25 rad
Convert Degree to Radian
80° = 0.25  180

= 80  
180 = 0.25  180
3.142
= 80  3.142
180 = 14.32

= 1.396 rad 7 rad
6
170.4°

24°15’ 4  rad
3

1 = 60'
15' = 15 1 = 0.25

60

Radius, Diameter and Chord
Sector and Segment

Arc Length and Circumference

Circumference of a circle is
the distance around the circle.

2.4.3 LENGTH OF ARC

The distance along the curved line forming the arc. It can be measured in distance units, such as meter
and centimeter.

The arc length of a sector, s = r where r is the radius of a sector and  is the angle in radian

subtended at the center of the circle.
s
Arc length, s

s = r

where r radius

 angle in radian

s length of arc

.

Example

A AB

8 cm O

O rad Find the arc length of AB.

B Solution :
Find the arc length of AB.

Solution :

Length of Arc AB = 8(1.2)
= 9.6 cm

2.4.4 AREA OF SECTOR

A circular sector is the portion of a circle enclosed by two radius and an arc of the circle.

The area of a sector, A = 1 r 2 , where r is the radius of sector and  is the angle in radian
2

subtended at the centre of circle/sector.

Area of Sector, A

A =   r2
2

A = 1 r 2
2

Example Q
P

70
O

Calculate b)
a) area of minor sector POQ
b) area of major sector POQ.

Solution :

a)  = 70   = 1.22rad
180

Area of sec tor = 1 (7)2(1.22)
2

= 29.89cm2

2.4.5 AREA OF SEGMENT

The area of a segment is the area of a sector minus the triangular piece.
There is a lengthy reason, but the result is a slight modification of the sector formula.

Area of Segment

As = 1 r 2 − 1r 2 sin
2 2

= 1  r 2  ( − sin )
2

in radian

Other formula : Area of Segment, As = 1 r 2 − 1 ab sin
2 2

Example Solution :
A Area of segment APB

O 1.333 rad P = 1  r 2  ( − sin )
2
B
= 1  62  1.333 − sin1.333  180  
2   

= 1  36  (1.333 − sin76.38)

2
= 6.501 cm2

TUTORIAL 2.4 b) 8 rad to degree.
1. Convert 7

a) 66 to radian. d) 2 rad to degree.
f)  rad to degree.
c) 360o to radian.
2
e) 135o to radian.

2. A sector AOB with center O. The length of the arc AB is 7.5 cm and the perimeter of the sector
AOB is 25 cm. Find the value of θ, in radian.
A

θ B
O

3. Given that the length of the major arc a. Length of radius, OR
RS is 45 cm , find the
R

T 0.35 rad O b. Area of segment RTS
S

4. The length of the minor arc is 15 cm a. the value of θ, in radians. (Give your answer
and the angle of the major sector correct to four significant figures)

POR is 280o. Calculate

P

Oθ b. the length, in cm, of the radius of the circle

R

5. Given that OQ = 20 cm, PY = 8 cm, a. the value of θ, in radian
∠ XPY = 1.1 radians and the length
of arc PQ = 14cm, calculate

b. the area, in cm2, of the shaded region.

6. A circle has radius 12 cm and the a) Value of θ in radian.
angle of minor sector, θ is 54o. By b) Area of shaded region.
using ℼ = 3.142, calculate

R

12 cm

54o O

P

7. A circle with center O has radius 6cm. Given the angle AOB is 0.56 rad. Calculate the arc
length of AB.

A 0.56 rad B

6 cm
O

8. Given the area of sector OQR is 60cm2, ∠POR = 1.2 rad and PQ = 2cm. Calculate the length
of OP.

Q

P

1.2rad R
OS

9. OPS and OQR is two concentric sectors at O. Given OS = 6 cm, SR = 2 cm and ∠QOR = 0.5
Q

P

0.5 rad

O 6 cm S R
rad. Calculate the perimeter of shaded region.

Unit 3 DUM 10122: Engineering Mathematics 1
TRIGONOMETRY

INTRODUCTION OVERVIEW

Trigonometry is a branch of mathematics that deals with the relations between the sides
and angles of a triangle.

Nowadays, you also can find the trigonometric ratios of any angles by pressing
appropriate buttons on calculator. Until now we have defined the trigonometric function only
for acute angles. However, many application of trigonometry involves angles that are not
acute. Consequently it is necessary to extend the definition of the six trigonometry function
to general angles and you will learn in this topic. Furthermore in this chapter, we will derive
two new formulae, the sine rule and cosine rule to enable us to solve oblique triangle quickly.
We also can calculate the area of triangles of oblique triangles.

CONTENT CONTENT

3.1. Introduction to Trigonometry 1
3.2. Trigonometric Ratios
3.3. Graphing Trigonometric Functions
3.4. Trigonometric Equations
3.5. Solution of Triangle

3.5.1. Sine Rule
3.5.2. Cosine Rule
3.5.3. Area of Triangle

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Determine an angle in degree and radian.
2. Convert the angle in degree to radian and vice versa.

3.1 INTRODUCTION OF TRIGONOMETRY

ANGLE

The angle  can be measured in one of two units which are degree and radian. In trigonometry
radians are the most common. It can have a measure which positive (anticlockwise) or
negative (clockwise).

What is 'radian' ? One radian is the angle subtended at the center of a circle by an arc that is
equal in length to the radius of the circle.

RELATIONSHIP BETWEEN DEGREE AND RADIAN

360o = 2π radian

Convert degree to radian and vice versa


180

Degree °  180 Radian rad


2

DUM 10122: Engineering Mathematics 1

CONVERT DEGREE TO RADIAN AND RADIAN TO DEGREE.

Convert the following angles.

a) 1350 to radian b) 45° 15’ to radian
135° = 135o  
45°15’ =  45 + 15 0   
180  60  180

= 2.36 radian

= 45.25o  
180

= 0.79 radian

c) 4  radian to degree d) 2.4 radian to degree
9 2.4 radian = 2.4  180

4  = 4   180 
99 
= 137.5o
= 80o

3

DUM 10122: Engineering Mathematics 1

TUTORIAL 3.1 b) 1250
1. Express the following angles in radian.
a) 300

c) 500 17’ d) 250.20

2. Express the following angles in degree. b) 2 rad
a) 4 rad
3

c) 3.12 rad d) 8 rad

4

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Form the six trigonometric ratios of a given right-angled triangles.
2. Find the values of:

• trigonometric functions
• inverse trigonometric functions using calculator.
3. Solve right-angled triangles to calculate

3.2 TRIGONOMETRIC RATIOS

Hypotenuse c a Opposite Hypotenuse c ∡ B a Adjacent

∡A b
b Opposite

Adjacent

Hypotenuse side : is the longest side of a right angle triangles and
always the side opposite the right angle.

Opposite side : is opposite the reference angle.
Adjacent side : is next to reference angle.

PHYTAGORAS THEOREM:

https://www.mathsisfun.com/pythagoras.html

5

DUM 10122: Engineering Mathematics 1

Name the sides of each these right triangles as opposite, adjacent or
hypotenuse with the reference to the state angles.

a. Reference to ∡ A b. Reference to ∡ Q

e a
A b

fg Q
c

Hypotenuse : p Hypotenuse :
Opposite : Opposite :
Adjacent : Adjacent :
c. Reference to ∡ R d. Reference to ∡ y

m mp
R Y

d n

Hypotenuse : Hypotenuse :
Opposite : Opposite :
Adjacent : Adjacent :

6

DUM 10122: Engineering Mathematics 1

TRIGONOMETRIC FUNCTIONS a
=
c
A
=


=

b
The Figure shows a right angled triangle where the reference angle is A, the adjacent side is

b, the opposite side is a and the hypotenuse is c.

The six trigonometric functions are defined in Table below:

Function Symbol Definition of function

Sine of angle A Sin A opposite side
Sin A = hypotenuse =

Cosine of angle A Cos A adjacent side
Cos A = hypotenuse =

Tangent of angle A Tan A opposite side
Tan A = adjacent side = =

Cotangent of angle A 1 adjacent side
Secant of angle A Cot A = tan Cot A = opposite side =
Cosecant of angle A
1 hypotenuse
Sec A = cos A Sec A = adjacent side =

1 hypotenuse
Csc A = sinA Csc A = opposite side =

7

DUM 10122: Engineering Mathematics 1

Evaluate the trigonometric functions based on the figure below.

P

4

Q3 R

a) side PR b) sin P c) tan R d) cos R

e) sec R f) csc P g) cot P h) csc R

Use calculator to evaluate the trigonometric functions.

a) sin 120 b) cos 2430

c) cot 312.50 d) sec 1300

8

DUM 10122: Engineering Mathematics 1

Determine the value of angle in each of the following given functions.

a) Sin A = 0.7936 b) Cos A = 0.31236
A = Sin-1 0.7956 = 52.520

c) Tan A = 4.9781 d) Sec A = 3.6531
e) Cot A = 4.8673 1 = 3.6531



= 0.2737
= 74.12

f) Csc A = 2.039

9

DUM 10122: Engineering Mathematics 1
Determine the unknown angles or sides of the right angled triangles.

a) Determine ∡ B
3.9cm

4.9cm B

S p
q 72.30

16.6 cm

b) Determine:
i) ∡ S
ii) Side p
iii) Side q

10

DUM 10122: Engineering Mathematics 1

TUTORIAL 3.2

1. Find the trigonometric ratio of the following angles

a) sin 360 b) cot 1240

c) cos 530 d) sec 2560

e) tan 162.20 f) csc 3130

2. Determine the value of angle for each of following given functions

a) cos B = 0.5 b) sec x = 2

c) cos B = 0.37604 d) sec x = 4.0657

e) tan D = 0.6945 f) csc y = 2

g) sin E = 0.8304 h) cot θ = 0.1798

i) sin E = 0.5 j) cot θ = 1

11

3. Find the value of: DUM 10122: Engineering Mathematics 1

a) side AC A
c) Sin A 12
e) Tan A B 5C
g) Cos C
b) Sec C
d) Csc A
f) Cot A
h) Csc C

4. Use calculator to evaluate the trigonometric ratio below

Ө 77.200 125.300 202.300 313.700
sin Ө
cos Ө
tan Ө
csc Ө
sec Ө
cot Ө

12