Prime Optional Mathematics 8

Limits

Unit Test

Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

1.• Write down next two terms of the sequence 2.01, 2.001, 2.0001, 2.00001,

...,•...,•...,•...,•...,•...

2.• a.• Write down the limit value of the sequence by inding next 2 terms

of,•4.9,•4.99,•4.999,•4.9999,•...,•...,•...,•...,•...,•...

• b. What is limit? Write down a suitable example.

c. Evaluate: lim (3x•+•5)
x "1
3.• Divide a line segment of length 16cm making half continuously 5 times

for a part. Show it in diagram and write down the limit value of the

sequence•of•its•length.•

4.• Write down the sequence of area of triangle having 80cm2 for the given

diagrams•by•adding•two•more•diagram.•Also•write•down•its•limit•value.

Limits

46 PRIME Opt. Maths Book - VIII

Unit 3 Matrices

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 1 – 2 6 10

Weight –24–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to represents the numbers in matrix form.
• Students are able to know the concept of order of matrix and types of

matrices.
• Students are able to operate the matrix addition, subtraction, transpose

and multiplication by scalar.
• Students are able to know the properties of addition matrices.

Materials Required:
• Chart paper.
• List of price chart of goods of a market.
• List of properties of matrix addition
• List of types of matrices

PRIME Opt. Maths Book - VIII 47

3.1 Matrices:

The•cost•per•kg of vegetables in Kalimati vegetable market in a particular

day is as follows.

Potato Cauli•lower Cabbage

Shop•A• • 15 25 20

Shop•B• • 20 30 15

Shop•C• • 18 24 12

Above cost of the vegetables can be represented using square bracket as,

Matrices Here,• Rows represents the shops and column represents the type of
vegetables. This type of presentation of numbers is called a matrix.

The•rectangular array of the numbers in different
rows and columns which are taken into a square
bracket or round bracket is called a matrix.

• A matrix is denoted by capital letter A, B, C, ...
• The members used in a matrix are called elements which are

denoted by small letters a, b, c, d, e, f, ........................
• The horizontal arrangement of the number is a matrix are called

rows.
• The vertical arrangement of the number in a matrix are called

columns.
• The no. of rows (m) and no. of columns (n) can be written as

m × n which is called the order of the matrix. We read m × n as m
by n.
• An element of a matrix is represented by the no. of row and no. of
columns. It lies as lower subscript e.g. the element lies in 2nd row
and 3rd column is denoted by a23.

A=

48 PRIME Opt. Maths Book - VIII

Here, a11 = element in !irst row and !irst column.
a23 = element in second row and third column.

Order of the matrix A = 3 by 3 written as 3 × 3.

Types of matrices

1. Row matrix
The matrix having only one row is called row matrix.

Ex : A = [a11 a12 a13]1 × 3 B = [2 4]1 × 2

2. Column matrix:

The matrix having only one column is called column matrix.

Ex : A = TSSRSSSSSSaaa132111VWWXWWWWWW3×1 B = SRSSSSSSSSSST1352WWWWWWXVWWWWW4×1
3. Null matrix:
Matrices
The matrix having all the elements zero is called null matrix. It is

denoted•by•‘O’.

• Ex : O = <00 0 00F2×3
0

4. Rectangular matrix:
The matrix having unequal number of rows and columns is called

rectangular matrix.

Ex : A =

5. Square matrix:

The matrix having equal number of rows and columns is called square

matrix. B = <13 24F2×2 C = SSRSSSSSS174 2 369WWVWXWWWWW3×3
• Ex : A = [2]1×1• T 5
8

PRIME Opt. Maths Book - VIII 49

6. Diagonal matrix
The square matrix having the diagonal elements from left top to right

bottom•are non-zero and remaining elements zero is called diagonal

matrix.

• Ex: A = <0a 0bF2×2

7. Scalar matrix:

The diagonal matrix having all non-zero diagonal elements are equal

is•called•scalar•matrix.
Ex : A = SSSSSSSRS00a 00aXWWWWWWVWW3×3
• T 0
a
0

8. Identity matrix (Unit matrix):

The square matrix having diagonal elements from left top to right

bottom•all•one•and•remaining elements zero is called identity matrix.

Matrices It•is•denoted•by•‘I’. I = SRSSSSSSS100 0 100WWWWWWXWVW3×3
T 1
• Eg : I = <10 10F2×2 0
9. Equality of the matrices

Any two matrices having same order and same corresponding

elements•are•called•equal•matrices.

• Ex : A = <24 13F, B = <24 13F
Here, A = B

10. General element of a matrix:

Twhheereeleimise•nnot.•ooff•rtohwe smaantdrijxisisndoe•onfo•ctoeldumbynsa.ij as the general element

• Ex : If aijaaaa=221112212====i•–2222•j×××× 1 – 1 = 1
• Then, 1 – 2 = 0
2 – 1 = 3
2 – 2 = 2

\ 2 × 2 matrix•is,•A•=• <aa1211 aa1222F = <13 02F

50 PRIME Opt. Maths Book - VIII

Worked out Examples

1. The cost of fruits in a fruits shop in different days is given below represent

the informations in a matrix with appropirate meaning.

Apple Orange Banana

Day 1 150 80 75

Day 2 120 100 70

Solution :

The cost of fruits given in table of two days are represented by matix

as•follows where,

rows ® represents the days

columns ® represents the type of fruits

A=

2. If A = <–32 2 –51F i) Find the order of matrix A.
4 ii) Find the elements a12, a21, & a23.

Solution, Matrices

i) No. of rows = 2

No. of columns = 3

\ order of matrix A is 2 by 3 (2 × 3).

ii) aaa122312 = element•in•1st row 2nd column•=•2
= element•in•2nd row 1st column•=•–2
• = element•in•2nd row 3rd column•=•–1
•

3. If A = B, where A = =x + 2 13G and B = <6 3 13F , !ind the value of x & y.
y – 2y

Solution : A = =x + 2 13G, B = <6 3 13F
y – 2y

Here, matrix A & B are equal.

By equating the corresponding elements

x+2=3 and y = 6 – 2y

or, x = 3 – 2 y + 2y = 6

or, x = 1 3y = 6

\x=1 y= 6
3
\ y=2 \ x = 1, y = 2

PRIME Opt. Maths Book - VIII 51

Exercise 3.1

1.• Write down the following informations in matrix form with appropriate

meaning.•Also•write•down•the•order•of•the•matrices.

• i) The number student in a class are given below.

1st column• 2nd column

• 1st row 4 3

2nd row 5 2

3rd row 6 4

4th row 3 5

ii) The cost of clothes in three shops are given below.

Shirt Pants Vest

1st shop• 500• 700• • 300

2nd shop• 450• 600• • 350

3rd shop• 400• 550• • 250

Matrices iii) The production of crops in Jhapa district in different years in

metric•tonne•is•given below.

Rice Wheat Maize others

2072 BS 300 150 100 150

2073 BS 350 170 120 130

2074 BS 475 180 200 300

2.• Write down the types of the matrices from the followings. Also write

di)ownRSSSSSSSS13t2he154oWWWWVWWWW rder of the matrices. RSSSSSSSS–141 132WWWWVWWWW
ii) <00 00F
iii) 2
5
TX SSSSSSSSRT100 0 100WVWWWWWWW X
iv) <20 02F
v) RSSSSSSSS00a 0 00c WWWWWWWWV vi) 0
b 1
0 0
TX TX
viii) SSSSSRSSS324WWWVWWWWW
vii) <–22 3 –12F TX ix) 63 –2 1@
2

52 PRIME Opt. Maths Book - VIII

3.• Answer the following questions from the matrix SSSSSSSSR132 3 –151WVWWWWWWW
i) Element in irst row and second column. 0
2
ii) aETW3lh1ereim+teeaele2dn3mot+waeainjn3wt3tahh2ee2r.oerid=er3o&f TX
iii)
iv) j = 2.
v) matrix.

4.• If•the•general•element•of•a matrix is aij = 3i – 2j, what will be the matrices
of•following•order.

• i) 2 × 2 matrix ii) 2 × 3 matrix iii) 3 × 3 matrix

iv) 3 × 2 matrix v) 1 × 3 matrix

5.• Find•the•value•of•‘x’•and•‘y’•from•the•following•equal•matrices.
• i) A = <53 x6–1F & B = <2y3+ 1 76F

ii) A = ,B= Matrices

iii) P = =2x3+ y 2 7G , B = <73 2 7 3F
5 y 5 2y –

iv) M = , N = SSSSRSSSS124 3 –772WWWWWWWWV
1
5
TX
A = <2x5+ y 17F= <57 3x – 2yF
v) 1

Answer

1.• Show•to•your•teacher. 2. Show to your teacher.

3.• i)• 3 ii) 0 iii) 9 iv) 2 v) 3 × 3
SSSSSRSSS174 30–3WWWWWWWWV
4 i) <14 –21F ii) <14 –1 0–3F iii) –1 v) 61 –1 –3@
2 2
5
ii)• x•=•1,•y•=•3•T Xiii)• x•=•2,•y•=•3
5.• i)• x•=•8,•y•=•2•

• iv) x = 4, y = 3 v) x = 3, y = 1

PRIME Opt. Maths Book - VIII 53

3.2 Operation•on•matrices

The• simpli•ication• of• two or more matrices in a single matrix by using
any kind of mathematical operations indicates the operation on matrices.
Which•are addition, subtraction, multiplication with scalar, transpose etc.
Here•we•are•discussing•some•of•them•in•grade•VIII.

3.2.1.•Addition of matrices
Marks obtained by three students Sita, Pranav and Pranisha in two monthly

test examinations in optional maths are as follows of two months.

Ashadh 1st 2nd Shrawan 1st 2nd

Sita 60 75 Sita 70 75

Pranav 85 90 Pranav 95 80

Pranisha 95 85 Pranisha 80 95

Total marks obtained by them in two tests in two months as,

1st 2nd

Matrices Sita 60•+•70•=•130 75•+•75•=•150

Pranav 85•+•95•=•180 90•+•80•+•170

Pranisha 95•+•80•=•175 85•+•95•=•180

This•information•can•be•expressed•in•matrix•form•as,
SSRSSSSSS689055 789550WWWWWVWWW + SSSSSSRSS789005 789505WWWWWVWWW = = RSSSSSSSS111873050 111785000WWWVWWWWW
T XT X TX

The•sum•of•any two matrices having same order
is•called•a new single matrix obtained by adding
the•corresponding•elements•of•the•matrices.•The•
single•matrix•so•formed has the order same as the
given matrices.

54 PRIME Opt. Maths Book - VIII

eg. If A = <–21 53F, B = <–34 –21F Matrices
Then,
A + B = <–21 53F + <–34 –21F
==2 + 3 3 + 2G
–1–4 5 – 1
= <–55 54F

3.2.•2.•Difference•of•the•matrices.

The•difference of any two matrices having same
order is called a new single matrix obtained by
subtracting• the• corresponding• elements• of• the•
matrices.• The• single• matrix• so• formed has the
order same as the given matrices.

eg. If P = <52 3 12F, Q = <–21 1 53F
–3 1

Then,

P–Q = <52 3 12F – <–21 1 53F
–3 1

=

= <33 2 ––41F
–4

PRIME Opt. Maths Book - VIII 55

3.2.3.•Transpose of the matrix.

Let us taking an example where cost of apple of a shop of three days

are given.

1st shop• • 2nd shop

• Sunday 150 140

Monday 160 145

Tuesday 170 150

It can be written in such a way by changing the information of rows

and•column•as,

• Sunday Monday Tuesday

1st Shop• 150 160 170

2nd Shop• 140• • 145 150

Matrices ASu=chSSSSSSSSR111e675x000am111p445l500eWVWWWWWWWs can be written in matrix form as,
TX

After changing row and columns =

It is called transpose of the matrix A and denoted by AT.

The•new•matrix•obtained•by interchanging the rows
and• columns• of• a matrix is called transpose of the
matrix.•
• Transpose of A is denoted by AT or•A`•or• A .

Note : Order of the transpose matrix is different than matrix A for the
rectangular matrix but same for the square matrix.

eg. If A = <32 4 95F then.
6

AT = SSSSRSSSS534 629WWWWWWVWW
TX

56 PRIME Opt. Maths Book - VIII

3.2.4.•Multiplication•of•a•matrix•with•a•scalar.

Let•us•consider•the•cost•of•apple•given above becomes double in the next
week which can be written as

(old•cost)••••••(new•cost)

The•new•matrix•formed by multiplying each element of
a given matrix by a given scalar quantity is called the
multiplication•of•a•matrix•with•the•scalar.

i.e.•If•A•=•<ac dbF; then,•KA•=• <kkac kkdbF

Eg. If A = <13 ––21F, !ind 3A. 3A = 3<13 ––12F = <39 ––36F Matrices
Solution : A = <13 ––12F

Worked out Examples

1. If A = <32 –11F and B = <43 –12F !ind 3A – 2B.
Solution :

A = <32 –11F B = <34 –12F
Then,

3A – 2B = 3<32 –11F – 2 <34 –12F

= <69 3–3F– <68 –24F

= =69 – 6 –33+–42G
– 8

= <10 –75F

PRIME Opt. Maths Book - VIII 57

Matrices 2. If A = <–11 32F , B = <32 –12F , prove that (A + B)T = BT + AT.
Solution :
A + B = <–11 32F + <32 –12F = <14 04F

\ L.H.S. = (A + B)T = <14 04FT = <04 14F
\ R.H.S. = BT + AT = <32 –12FT + <–11 32FT = <–32 12F + <12 –31F = <04 14F
\ L.H.S. = R.H.S. proved.
3. If A + B = <43 14F and A – B = <––52 32F, !ind the matrices A and B.

Solution :
A + B = <34 14F ................•(i)
• A – B = <––52 32F ................•(ii)
• Adding (i) and (ii), we get,

A + B = <34 14F
A – B = <––52 32F

2A•=• <–22 64F
\ A = <–11 32F
Putting the value of ‘A’ in equation (i)
B = <34 14F – <–11 32F = <34 –11F

\ A = <–11 32F & B = <34 –11F

58 PRIME Opt. Maths Book - VIII

Exercise 3.2

1.• Add the following matrices.
i) A = <–11 32F and•B•=• <34 –11F ii) M = <–31 –42F, N = <26 17F

iii) P = <–42 –31F, Q = <–24 –13F iv) A = <–32 75F, B = <5–2 ––33F

v) M = <34 2 15F, N = <–32 –1 02F
1 1

2.• Subtract•the•matrices•given below.
i) A = <–11 32F and•B•=• <34 –11F ii) M = <–31 –42F, N = <62 17F

iii) P = <–42 –31F, Q = <–24 –13F iv) A = <–32 57F, B = <5–2 ––33F

v) M = <34 2 15F, N = <–32 –1 02F Matrices
1 1

3.• If•A•=•<13 –22F, B = <24 –11F, ind•the•following•operations.

• i) 2A + B ii) 3A – B
iii) A + 3B iv) 3A – 2B
v) 4A – 3B

4.• i)• If•M•=•<12 –32F and•N•=•<36 –96F prove that 3M – N is a null matrix.

ii) If A = <–31 –22F and B = <–52 –45F, prove that 2A – B is an identity

matrix.

• iii) If A = <2x 13F, B = =4–+1x y 2 2G , C = <18 15F and A + B = C, ind the
–
value of ‘x’ and ‘y’ .

iv) If A = <13 –22F , ind•AT.

PRIME Opt. Maths Book - VIII 59

Matrices v) If A = <13 –21F and•B•=• <13 –43F, ind•(A•+•B)T.
5. PRIME more creative questions.

i) If A = <13 –22F and•B•=• <–14 15F , prove that (A + B)T = BT + AT.
ii) If A = <–31 24F and•(A•+•B)T = <62 –11F, ind•the•matrix•B.
• iii) If A = <24 13F and•A•+•B•=• <64 53F, ind•B•and•AT.
iv) If A + B = <64 53F, A – B = <20 –11F, ind•the•matrices•A•&•B.
• v) If P + Q = <–43 –71F & P – Q = <52 ––33F, ind•the•matrices•P•and•Q.
6. Project work
Collect the price of potato and tomato from local shop of two days in
the•month•of•Baishakh•and•Jestha.•Present the price of such month in
matrix•form.•Also• ind•the•difference of the cost using matrix.

60 PRIME Opt. Maths Book - VIII

Answer

1.• i)• <34 14F ii) <55 55F iii) <00 00F
iii) <–84 –62F
iv) <13 24F v) <17 1 53F iii) <175 –11F
2 iii) x = 2, y = 2

2.• i)• <––25 32F ii) <–17 –39F

iv) <–57 180F v) <15 3 –51F
0

3.• i)• <140 –33F ii) <15 –77F

iv) <–11 –88F v) <–02 –1111F

4.• i)• • ii) Matrices
iv) <12 –32F v) <–41 34F

5.• i)• <30 –03F ii)

iii) B = <22 22F , AT = <34 12F

iv) A = <24 13F, B = <22 22F

v) P = <13 –22F, Q = <–14 15F

PRIME Opt. Maths Book - VIII 61

Matrix

Unit Test - 1
Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

1.• What•do•you•mean•by•order•of•a•matrix?

2.• a.• If•A•=• <–21 34F and•B•=• <13 –12F, ind•the•matrix•3A•–•2B

• b. If <2x3–y y 54F = <64 25zF, ind•the•value•of•x•and•y.

• c. If A = SSSSRSSSS–203 3 –144WWWWWWWWV , ind the value of a23 – a32. Also write down the
–2
–1
order oTf the matrXix A.
Matrices
3.• a.• If•aij = 3i – 2j, ind the matrix of order 2 × 2. Also ind its transpose.
b. If A = <12 –32F and•B•=• <32 –11F , prove that (A + B)T = BT + AT.

4.• If•A•+•B•=• <53 –21F and•A•–•B•=• <––11 –43F ind•the•matrices•A•and•B.

62 PRIME Opt. Maths Book - VIII

Unit 4 Co-ordinate Geometry

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 1 – 2 6 10

Weight –24–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to know concept of co-ordinate system.
• Students can plot the points in graph.
• Students can !nd distance between any two points.
• Students can use section formula and mid-point formula to !nd the section

point.
• Students can !nd the centroid of a triangle.

Materials Required:
• Graph board
• Graph paper
• Geo-board
• Sheet of paper
• Scissors
• Chart of sign used in quadrents

PRIME Opt. Maths Book - VIII 63

4.1 Coordinate•system Y

As• shown• in• •ig.• let• a person starts B r P(x,•y)
moving from ‘O’ to reach the point P. He O
can•reach P in a straight way OP = r units q y
making•an•angle•q with the horizontal x AX
line•(OX). The position of the person at
P is denote as (r, q). On the•other•hand,•
the•person•can•move •irst OA = x units
along•OX direction and then AP = y units
along•OY direction where OX^OY. Then
the•position•of•P•is•expressed•as•(x,•y).

Co-ordinate Geometry In•particular,
OA = 4 equal parts = 4 units = run
AP•=•OB•=•3•equal•parts•=•3•units•=•rise

Y

4

B3 P(4,•3)

2

1

X’ -4 -3 -2 -1 O-11 2 3 4 A X

-2

-3

-4

Y’
Then•the•point•P•is•written•as•(4,•3).

The•horizontal•distance•travelled is called run and the
vertical distance travelled is called rise for a point.
Run and Rise may be taken from origin in standard
position.

64 PRIME Opt. Maths Book - VIII

Rectangular Coordinate axes:
The• horizontal• line• XOX‘ and
vertical line YOY‘ are intersected Y
at• point• ‘O’• perpendicularly 4
which are called x- axis and 3

y-axis respectively and the 2

intersecting• point• ‘O’• is• called• 1

the•origin•point. X’ -4 -3 -2 -1 0-11 2 3 4 X
Here, -2
-3
• In the line OX, only +ve

numbers are counted

in horizontal axis.
• In the line OX‘, only –ve
-4

numbers are counted Y’

in horizontal axis.

• In the line OY, only +ve numbers are counted in vertical axis. Co-ordinate Geometry
• In the line OY‘, only –ve numbers are counted in vertical axis.
• The plane is divided by the lines XOX‘ & YOY‘ into four equal parts

as below.

Plane YOX " 1st quadrant.
Plane YOX‘ " 2nd quadrant.
Plane Y‘OX‘ " 3rd quadrant.
Plane XOY‘ " 4th quadrant.

A point•in•the•quadrant. Y
Let• us• consider• a point
P(3,•4)•in• irst•quadrant• 4 P(3,•4)

where PM^OX. 3
Here,
OM = 3 units 2

= X-coordinate X‘ 1
= abscissa
PM = 4 units -4 -3 -2 -1 O -11 2 3 4 X
= Y-coordinate M
= ordinate -2
The two values abscissa
and ordinate can be written -3

-4

Y‘

PRIME Opt. Maths Book - VIII 65

as (3, 4) which is called co-ordinate. It is written in the form of
P(x, y) in general in the co-ordiante system.

Sign•in•quadrants; Y

(–,•+) (+,•+)
X’ O
X
(–,•–)
(+,•–)

Co-ordinate Geometry Y’

The coordinate of the points according to sign in different quadrants

can•be•written•as•for•a•point•P(3,•4).

• (+, +) ® 1st quadrant•=•P(3,•4)
• (–, +) ® 2nd quadrant•=•P‘(–3,•4)
• (–, –) ® 3rd quadrant•=•P‘‘(–3,•–4)
• (+, –) ® 4th quadrant•=•P‘‘‘(3,•–4)
Y

P‘(–3,•4) P(3,•4)

X’ O X

P‘‘(–3,•–4) P‘‘‘(3,•–4)

Y’
66 PRIME Opt. Maths Book - VIII

Distance•between•any•two•points.

Distance•between•the•points•A(x1, y1) and•B(x2, y2).

Y B (x2, y2)

d

(x1, y1) C
A

OM NX

DLerta,w•‘d, ’A•bMe•^thOeX•d, BisNta^nOceX•baentdwAeCe^n•BthNe.•points•A(x1, y1) and•B(x2, y2)

Then,•• OBAOCCMN ==•==MB•xxN21N,, B–A=NMCO•N=•N=•y=•y–21BONM–=AxM2 Co-ordinate Geometry
AB = d
–=xy12 – y1

In right angled DABC,
• h2 = p2 + b2

or, AB2 = AC2 + BC2
or, d2 = (x2 – x1)2 + (y2 – y1)2

\ d=
It is the distance formula,

Some important informations:
• Origin point is O(0, 0)
• A point on x – axis is (x, 0)
• A point on y – axis is (0, y)
• To prove equilateral triangle. All sides should be equal.
• To prove right angled triangle.
h2 = p2 + b2
• To prove isosceles triangle.
Any two sides should be equal.

PRIME Opt. Maths Book - VIII 67

Co-ordinate Geometry • To prove equilateral triangle.
All sides should be equal.

• To prove rhombus.
All 4 sides should be equal.

• To prove square.
All 4 sides should be equal and
h2 = p2 + b2 [diagonal is taken as hypotenuse] [or diagonals are equal]

• To prove parallelogram.
Opposite sides should be equal. (Mid points of diagonal must be
same.)

• To prove rectangle.
Opposite sides should be equal and h2 = p2 + b2 for a diagonal h.
(or diagonals are equal)

Worked out Examples

1.• Plot•the points A(1, 3) and B(4, 7) in graph paper and ind the distance
between the points A and B by drawing perpendicular lines to the axes
using•run•&•rise.
Solution: The given points A(1, 3) and B(4, 7) in graph as.
Y

B

AC

X’ O M NX

Y’
Here, AC = MN = ON – OM = 4 – 1 = 3 = run

BC = BN – CN = BN – AM = 7 – 3 = 4 = rise
\ AB =

68 PRIME Opt. Maths Book - VIII

= 32 + 42
= 25
= 5 units

2. Find•the•distance•between the points (2, –3) and (5, 1). Co-ordinate Geometry
Solution:
The given points are:
AUB((s52in,, g1–3)d)=is=t(ax(n2x,c1y,e2y)f1o)rmula,

d=

d(AB) =

= 32 + 42
= 5 units.
3. Prove that the line joining the points A(3, 4), B(7, 7) and C(11, 10) are
collinear points.
Solution :
The given points are A(3, 4), B(7, 7) & C(11, 10)
Using distance formula,

d=

d(AB) = = 5 units

• d(BC) = = 5 units

• d(AC) = = 10•units

• Here,

AC = AB + BC

or, 10 = 5 + 5

\ 10 = 10

• Hence, They are collinear points.

4. Prove that the points A(–2, 3), B(–2, –4), A(–2,•3) D(5,•3)
C(5,•–4)
C(5,•–4)•&•D(5,•3)•are•the•vertices of a square.

Solution :

The given points are

A(–2, 3), B(–2, –4), C(5, –4), D(5, 3)

Using distance formula

d = B(–2,•–4)

PRIME Opt. Maths Book - VIII 69

d(AB) = = 7units

• d(BC) = = 7units

• d(CD) = = 7units

• d(AD) = = 7units

• d(AC) = = 7 2 units

• Here, AB = BC = CD = DA

Also, AB2 + BC2 = AC2

or, 72 + 72 = (7 2 )2

or, 98 = 98

Hence, They are the vertices of a square.

5.• Find•the co-ordiante of a point on Y
O
x-axis which is 5 units distance

from a point (5, 4). (5,•4)

Co-ordinate Geometry Solution : Let, the point on x-axis

be•A(x,•0) d = 5units
(x,•0)
• The given point is B(5, 4)

Using distance formula, X

or d2 = =(x(2x–•–x•51))22 ++((y02•––•4y)12)2
(5)2

or, 25 = x2 – 10x•+•25•+•16

• or, x2 – 10x•+•16•=•0

• or, x2 – (8•+•2)x•+•16•=•0

• or, x2 – 8x•–•2x•+•16•=•0

• or, x(x – 8) – 2(x – 8) = 0

or, (x – 8) (x – 2) = 0

Either, or

x–8=0 x–2=0

\x=8 \x=2

\ The•required•point•is•(8,•0)•or•(2,•0)

70 PRIME Opt. Maths Book - VIII

Exercise 4.1

1.• Plot•the•following•points•in•graph•and•join•to•each•other.

• i) (1, 2), (3, 6) and (7, 0) ii) (1, 1), (3, –3) and (–2, –2)

iii) (–1, 5), (–7, –1) and (–3, 1) iv) (2, –3), (1, 7) and (–2, 1)

v) (–1, –1), (3, –5), (–2, –6) and (–5, 2)

2.• Plot•the•following points in the same graph and join each points with
a straight•line•one•after•another•continuously.
(5, 1), (6, 2), (4, 5), (–4, 5), (–5, 2), (–7, 4), (–6, 0), (–7, –3), (–5, –2),
(–4,•4),•(4,•–4),•(6,•–1),•(5,•1)

3.• Find•run•and•rise•of•the•following pairs of points. Also •ind the distance

between•the•points•using•run•and•rise.

• i) (3, 2) and (6, 6) ii) (1, –2) and (–5, 6) Co-ordinate Geometry

iii) (7, –9) and (3, –6) iv) (10, –1) and (6, 1)

v) (–5, –1) and (1, 3)

4.• Find•the•distance•between•the•following•pairs•of•points.

• i) A(2, 1) and B(5, 5) ii) M(3, –2) and N(9, 6)

iii) P(3, –1) and Q(–1, –4) iv) X(–6, –7) and Y(–3, –1)

v) A(a, b) and B(b, a)

5.• Find•the•distance•between•the•following•points.

• i) M(a, 0) and N(0, b)

ii) P(p + q, p –q) and Q(p – q, p + q)

iii) A(–1, 1) and B( 3 , 3 )
iv) (–3, –3) and (–3 3 , 3 3 )
v) (3, 3 ) and•(0,•2 3 )

6.• Solve that following problems.

i) Prove that the points (3, –4), (5, 0) and (–4, –3) are equidistance

from the origin.

ii) Prove that AB = BC where the points are A(–1, 5), B(5, –3) and

C(11,•–11).

• iii) P(–1, 3), Q(7, –3) and R(4, 1) are the three points. Prove that QR
1
= 2 PQ.

PRIME Opt. Maths Book - VIII 71

Co-ordinate Geometry iv) A(3, 4), B(7, 7) and C(11, 10) are the collinear points, prove that,
B is•the•mid-point•of•AC.

v) Prove that P(–2, 0) is equidistance from the points A(2, 3) and
B(–5,•4).

7.• Solve the following problems.
i) Prove that the vertices A(4, 8), B(3, –1) and C(–5, –7) are of a
scalene•triangle.

• ii) Prove that the points P(3, 5), Q(–2, 0) and R(7, 0) are the vertices
of•an•isosceles•triangle.

• iii) Prove that the points (–3, –3), (–3 3 , 3 3 ) and (3, 3) are the
vertices of an equilateral triangle.

iv) Prove that the points (–4, 2), (1, 2) and (1, –1) are the vertices of
a right•angled•triangle.

• v) Prove that the points (3, 2), (0, 5) and (–3, 2) are the vertices of a
right•angled•isosceles•triangle.

8.• Answer the following questions.
i) Prove that the points (1, –2), (3, 6), (5, 10) and (3, 2) are the
vertices of a parallelogram.
ii) Prove that the points (–2, 3), (–2, –4), (7, –4) and (7, 3) are the
vertices of a rectangle.
iii) Prove that the points (3, 2), (–5, 2), (–5, –6) and (3, –6) are the
vertices of a square.
iv) Prove that the points (–4, 3), (5, 4), (4, –5) and (–5, –6) are the
vertices of a rhombus
v) Prove that the points (7, 9) (3, –7) and (–3, 3) are the vertices of
a isosceles•right•angled•triangle.

9.• Solve the following problems.
i) If distance between the points A(2, –3) and B(x, 1) is 5 units, ind
the•value•of•‘x’.

• ii) If distance between the points P(3, y) and Q(9, –2) is 10 units,
ind•the•value•of•‘y’.

iii) Find the co-ordiante of a point on M(x, 0) which is 13 units distant
from the point N(–3, 12).

iv) Find the co-ordiante of a point on P(0, y) which is 17 units distant
from the point Q(15, –5).

72 PRIME Opt. Maths Book - VIII

v) The distance between the points C(a, 2a) and D(4, 1) is 13
units,• ind•the•value•of•‘a’.

PRIME more creative questions
10.• i)• Find•the•co-ordiante of a point on x-axis. Which is 5 units distance

from the point (1, 3).
ii) Find the co-ordinate of a point on y-axis which is 10 units distant

from the point (8, 4).
iii) Find the co-ordinate of a point on x-axis which is equidistance

from the points (5, 4) and (–2, 3).
iv) Find the co-ordinate of a point on y-axis which is equidistance

from the points (8, –3) and (–6, –5).
v) If a point P(x, y) is equidistance from the points A(1, 2) and

B(3,•–2),•prove that x – 8y – 8 =0.

11.• i)• Prove that the points (3, 3), (–3, –3) and (–3 3 , 3 3 ) are the Co-ordinate Geometry
ii) vertices of an equilateral triangle.
Prove that the points (3, 4), (7, 7) and (11, 10) are the collinear
• iii) points.
iv) Find the co-ordiante of a point (2a, a) which is 10 units distance
from the point (0, –5)
• v) Find the value of ‘a’ where distance between A(2a, 3) and B(8, a)

is• 17 units.
If A(3, 4), B(3, –3) and C(6, –3) are the points, prove that
AB^BC.

PRIME Opt. Maths Book - VIII 73

Answer

1.• Show•to•your•subject•teacher.

2.• Show•to•your•subject•teacher.

3.• i)• 3,•4,•5,•units• ii)• –6,•8,•10•units

• iii) –4, 3, 5 units iv) –4, 2, 2 5 units

• v) 6, 4, 2 13 units

4.• i)• 5•units• ii)• 10•units
• iii) 5 units iv) 3 5 units

• v) 2 (a2 + b2) units•

5 i) a2 + b2 units• ii)• 2q 5 units
iv) 6 5 units
• iii) 2 5 units•

Co-ordinate Geometry • v) 2 3 units

9.• i)• –1•or•5• ii)• 6•or•10
iv) 3 or – 13
• iii) 2 or – 8
2 ii)• (0,•–2)•or•(0,•10)
v) 2 or 5 iv) (0, 3)

10.• i)• (–3,•0)•or•(5,•0)•
• iii) 2, 0

11.• i)• (6,•3)•or•(–10,•–5)• iii)• (6,•3)•or•(–10,•–5)

• iv) 2 or 28
5

74 PRIME Opt. Maths Book - VIII

4.2•Section•Point

Let•us•consider•a point P cuts a line segment AB at a point where the point

divides the line segment in two sectors AP and BP. Hence the point P is

called• the• section• point• of• a line segment AB as shown in diagram. The

two sections AP and PB may be equal or

may be in a certain ratio. Here, we discuss B
the• process of •inding the coordinates of P
the• point• which divides a line segment in A
two parts (may be equal or unequal) in the

given ratio.

The•point•which cuts a line segment in equal sections is Co-ordinate Geometry
called•mid-point.
The•point•which cuts a line segment in unequal sections
either•externally or internally is called section point.

Centroid of a triangle

The•intersecting•point•of•medians•of•a triangle is called
the•centroid•of•the•triangle.•It•cuts•the•medians•in•the•
ratio 2:1 from the vertex.

Here,• A
In•DABC,•
P, Q and R are the mid-point of sides of DABC•where
AP, BQ and CR are called medians.

The•intersecting•point•of•the•medians•AP, BQ and R Q
G
CR•is•‘G’•which•is•called•the•centroid•of•DABC. B C
P
Here,• AG : GP = 2 : 1

BG : GQ = 2 : 1

CG : GR = 2 : 1

PRIME Opt. Maths Book - VIII 75

Co-ordiante of section point of a line segment.

i. The point cuts internally to a line segment joining the points

LAe(tx,1a, yp1o)inatnCd(xB,(yx)2c, uyt2s).the Y

line•joining•the•points• B(x2, y2)
iAn(txe1r,nya1)llya•nind••tBh(ex•2r,ayt2io) •m1:m2. C(x, y)m2 S

Draw the perpendiculars, A m1 Q
(x1, y1)
AM^OX, BN^OX, O M P NX

CP^OX, AQ^CP,

CS^BN,

Co-ordinate Geometry • TCOABCASQhQCSNe==:===nCPB,CMxBNONP2P,=M=–B–=mNQO=SON•1PN=x:Pm1•==–y,–2A2CBO,OMPONPM•P–==–•=•=ACyx•M1x2Px,,––•==CxxPyy.1•2.=––•yyy1..
•
•

In DACQ and DCBS

i) \ Q = \ S ® Both being 90°
ii) \ A = \ C ® Being•corresponding•angles.
• iii) \ C = \ B ® Being•remaining•angles
\ DACQ ~ DCBS ® By AAA axiom.

\ AQ = CQ = AC
CS BS CB
y – y1
or, x – x1 = y2 – y = m1
x2 – x m2

Taking,

or, mxx22–x–x–x1m=2xmm1 =12 m1x2 – m1x
or,

or, x(m1 + m2) = m2x1 + m1x2

\ x= m1 x2 + m2 x1
m1 + m2

76 PRIME Opt. Maths Book - VIII

Again taking,

y – y1 = m1
y2 – y m2

or, ym(2my1–+mm2y2)1 = mm21yy12 – mm11yy2
or, = +

\ y= m1 y2 + m2 y1
m1 + m2
\ Co-ordiante of section point is
tyh)e=c(o-mo1rmxd12in++ammte22oxf1 i,ntmer1mnya12 l++dmmiv22isyi1o)n
C(x, point.
It is

ii) Mid-point of a line segment

If point ‘C’ be the mid-point of the line segment AB in the above

pdioaignrta•omf•,t•hthee•lnin•me•s1e=gmm2e.nTth•AeBs,e•wcthieornep, oint so formed is called the mid-
+ m1 y2 + m2 y1 Co-ordinate Geometry
(x, y) = ( m1 x2 + m2 x1 , m1 + m2 )
m1 m2
If m1 ===m(( 2xm1fo12+(r2xxmm12 +i1d, x-yp2)1o2+,inymt22)(2ym1 +2
or, (x, y) y2) )
\
(x, y)

It is the co-ordinate of the mid-point of a line segment.

iii. If the section point cuts a line segment externally.

Let, a point C(x, y) cuts the line Y
jmBo(i1xn:m2i,n2yg. 2•)theex•tpeorinnatlsly• Ai(nx1t,hye1)raatnido
) C(x,•y)
Draw the perpendiculars,
m1 B(x , y 2

2 m2

S

AM^OX, BN^OX and CQ^OX

AR^CQ and BS^CQ A R
(x1, y1) N QX
Then, OM
• ABONRM===Myx21Q,,OA=QMO•=•=Q•x•y–,•1CO, QOM•N==••=yx•x–2 x1

PRIME Opt. Maths Book - VIII 77

BS ==:=BCNCCQQQ=––=mSROQ1QQ:m==–2CCOQQN––=BAxNM–==xy2y––yy21
CR
CS
AC

Now, In DARC and DBSC,

• \R = \S ® Both being 90°
\A = \B ® Corresponding angles
\C = \C ® Common angles
® By AAA axiom.
\ DARC @ DBSC••

\ AR = CR = AC
BS CS BC
x –x1 y –y1 m1
or, x – x2 = y – y2 = m2

Co-ordinate Geometry Taking,

1st and•last•ratios,

• x –x1 = m1
x – x2 m2

or, mm12xx2––mm22xx11==mx(1xm–1 –mm1x22)
or,

\ x= m1 x2 – m2 x1
m1 – m2

Again, taking,

2nd and•last•ratios
y –y1
• y – y2 = m1
m2

or, mm12yy2––mm22yy21==my(1ym–1 –mm1y22)
or,

\ y= m1 x2 – m2 x1
m1 – m2

\ co-ordinate of section point externally is,

(x, y) = ( m1 x2 – m2 x1 , m1 y2 – m2 y1 )
m1 – m2 m1 – m2

78 PRIME Opt. Maths Book - VIII

iv. Co-ordinates of centroid of a triangle.

Let ‘G’ (x, y) be the centroid oafmaetdriiaanngalendha‘Gv’incugtvsetrhteicmeseAdi(axn1, AyP1),i•nB(thx2e,
rya2)tiaon2d:1C.(x3, y3) where AP is

A(x1, y1)

G(x,•y)

B(x2, y2) P C(x3, y3)

Now,

using mid-point formula for BC, y2 + y3
+ 2
the co-ordiantes of P = ( x2 2 x3 , ) Co-ordinate Geometry

Again, using section formula for AP [G cuts AP in the ratio 2:1.]

(x, y) = ( m1 x2 + m2 x1 , m1 y2 + m2 y1 )
m1 + m2 m1 + m2

G(x, y) =

\ G(x, y) = ( x1 + x2 + x3 , y1 + y2 + y3 )
3 3
It is the co-ordinates of centroid of DABC.

PRIME Opt. Maths Book - VIII 79

Worked out Examples

1.• Find•the co-ordinate of a point which divides the line joining the points

(3,•1)•and•(–2,•6)•in•the•ratio•2:•3•internally.
23

•A P(x,•y) B

Solution:

The given points are,

ABLe((–3t,2,A1, 6)p)=o=i(nx(t1x,P2y,c1y)u2t)s AB in the ratio 2:3 = m1 : m2.

By using section formula, m1 y2 + m2 y1
+ m1 + m2
x = m1 x2 + m2 x1 , y =
m1 m2

Co-ordinate Geometry = 2 (–2) + 3 (3) = 2 (6) + 3 (1)
2+3 2+3

= –4 + 9 = 12 + 3
5 5
=1 =3

\ The required point is (1, 3).

2.• Find•the co-ordinate of mid-point of a line joining the points P(3, 4) and
Q(5,•–•6).

P(3,•4) M(x,•y) Q(5,•–6)

Solution:

The given points are,

PLQe((3t5,,,A4–)p6=o)i(=nxt(1x,My2(1,)xy,2y)) cuts the line PQ at mid-point.

Using mid-point formula, y1 + y2
x1 + x2 2
x = 2 , y =

= 3+5 = 4 + (–6)
2 2
8 –2
= 2 = 2

= 4 = –1

\ Mid-point of PQ is M(4, –1).

80 PRIME Opt. Maths Book - VIII

3. Find•the co-ordinate of centroid of a triangle having vertices A(1, – 2),

B(3,•5)•and•C(5,•–3). B(3,•5)
Solution:

The given vertices of DABC•are,

LCBAe(((531t,,,,G–5–(32)x))=,==y()x((x2xb31,e,y,yt2yh3)1)e) centroid of DABC, RP
G

A(1,•–2) Q C(5,•–3)

Using centroid formula, y1 + y2 + y3
x1 + x2 + x3 3
x = 3 y =

= 1+3+5 = –2 + 5 – 3
3 3
9 0
= 3 = 3

=3 =0 Co-ordinate Geometry

\ The required point is (3, 0).

4. If•(2,•3)•is•the mid-point of line joining the points (a, –2) and (3, 0) ind
the value of ‘a’ and ‘b’.

A(a,•–2) M(2,•3) B(3,•b)

Solution:

Let, M(2, 3) = (x, y) be the mid-point of line joining the points,

NAB((o3aw,, ,–b2))==(x(x2 1, y, y2)1)

By using mid-point formula, y1 + y2
x1 + x2 2
x= 2 , y=

or, 2= a+3 3= –2 + b
2 2
or, a + 3 = 4 –2 + b = 6

\ a=1 b = 8.

PRIME Opt. Maths Book - VIII 81

5.• Prove that the vertices A(–1, 3), B(0, –2), D(4,•0) C(5,•–5)
B(0,•–2)
C(5,• –5)• and• D(4,• 0)• are the vertices a
S(a,•b)
parallelogram. R(7,•5)

Solution:

The given vertices are A(–1, 3) B(0, – 2),

C(5,•–5)•and•D(4,•0). A(–1,•3)
• Now,

Using mid-point formula for diagonal AC.
x1 + x2 y1 + y2
x = 2 , y = 2

= –1 + 5 = 3–5
2 2

= 4 = –2
2 2
= 2 = –1

Co-ordinate Geometry \ Mid-point of diagonal AC is (2, –1)

Again,

For diagonal BD, y1 + y2
x1 + x2 2
x = 2 , y =

= 0+4 = –2 + 0
2 2
= 2 = –1

\ Mid-point of diagonal BD is also (2, –1).

Hence, ABCD is a parallelogram.

6.• The• three vertices of a parallelogram P(3,•1)
are P(3, 1), Q(5, 7) and R(7, 5). Find the

co-ordinate of fourth vertex.

Solution:

The three vertices of a parallelogram

are P(3,1), Q(5, 7) & R(7, 5). Q(5,•7)
Let, fourth vertex be S(a, b).

Now,

82 PRIME Opt. Maths Book - VIII

Using mid-point formula for diagonal PR,
x1 + x2 y1 + y2
x = 2 y = 2

= 3+7 = 1+5
2 2

= 10 = 6
2 2
=5 =3

Again,

The mid-point (5, 3) is also the mid-point of diagonal QS in a

parallelogram,

• So, x1 + x2 y1 + y2
2 2
x= y =

or, 5= 5+a 3= 7+b
2 2
or, 5 + a = 10 7+b=6
Co-ordinate Geometry
or, a = 5 b = –1

\ Fourth vertex is (5, –1).

Exercise 4.2

1.• Find•the•Co-ordinate of a point which cuts the line joining the following
points•in•the•ratio•given below.
i) A(1, 1) and B(4, 7); ratio 2:1.
ii) P(3, – 1) and Q(3, 8); ratio 1:2.
iii) M( –1, 5) and N(4, 5); ratio 2:3
iv) A(2, –3) and B(2, 3); ratio 3:4.
v) P(–3, –2) and Q(–3, 5); ratio 3:4

2.• Find• the• co-ordinate of mid-point of a line joining the points given
below.
i) A(3, 2) and B(5, 0)
ii) P(5, 1) and Q(1, 7)
iii) M(–2, 0) and N(–4, 6)
iv) A(–5, –1) and B(1, –7)
v) A(a, 3b) and B(3a, 7b).

PRIME Opt. Maths Book - VIII 83

Co-ordinate Geometry 3.• Find• the• co-ordinate of centroid of a triangle having vertices given
below.
i) A(1, 2), B(3, 6) and C(5, 1).
ii) P(3, 4) , Q(1, 1) and R(2, 1).
iii) K(2, –1) , L(4, –5) and M(3, 3).
iv) A(–3, –2), B(–4, 5) and C(–5, –3).
v) P(–1, 7) , Q(–5, –4) and R(–3, 3).

4.• Prove that the following vertices are of a parallelogram.
i) P(1, 2), Q(3, 6), R(7, 8) and S(5, 4)
ii) A(3, 4), B(0,1), C(5, 6) and D(8, 9).
iii) K(3, 1), L(1, –4), M(7, –5) and N(9, 0).
iv) A(–3, 2), B(–5, 8), C(–7, 6) and D(–5, 0).
v) P(–1, –3) , Q(–7, –5), R(–11, –1) and S(–5, 3)

5. PRIME more creative questions:
i) If (3, 2) is the mid-point of line joining the points A(–2, b) and
B(a,•5),• ind•the•value•of•a•and•b.

• ii) If A(m, n), B(2, 5) and C(6, 7), ind the value of m and n where
AB•=•BC.

• iii) If (1, 3) is the centroid of a triangle having vertices A(–2, 1),
B(p,•5)•and•C(3,q), ind•the•value•of•p•and•q.

• iv) The three vertices of a parallelogram are A(3, 1), B(5, 6) and
C(7,•3).•Find•the•co-ordinate•of•fourth•vertex D.

v) The three vertices of a parallelogram are P(2, 3), Q(1, –7) and
R(4,•–11),• ind•the•co-ordinate•of•fourth•vertex S.

6. Project work
Find the co-ordinate of the corners of your optional mathematics book
by putting in a sheet of graph paper. Verify it is a rectangle by calculate
distance•of•edges.

84 PRIME Opt. Maths Book - VIII

1.• i.• (3,•5)• Answer iii.• (1,•5)
• iv. (2, 3) iii.• (–3,•3)
2.• i.• (4,•1)• ii.• (3,•2)• iii.• (3,•–1)
• iv. (–2, –4) v. (–3, 1) iii.• p•=•2,•q•=•3
3.• i.• (3,•3)• ii.• (3,•4)•
• iv. (–4, 0) v. (2a, 5a)
5.• i.• a•=•8,•b•=•–1• ii.• (2,•2)•
• iv. (5, –2) v. (–3, 2)
ii.• m•=•–2,•n•=•–2•
v. (5, –1).

Co-ordinate Geometry

PRIME Opt. Maths Book - VIII 85

Co-ordinate Geometry Time•:•30•minutes
Unit•Test•-•1

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

1.• Write down the co-ordinate of a point on x-axis and y-axis.

2.• a.• Find•the•distance•between•the•points•A(3,•–•1)•and•B(7,•2).
• b. Find the co-ordinate of mid-point of a line segment joining the
points•A•(3,•–1)•and•B(–5,•7).
• c. Find the section point which cuts a line segment joining the points
A(3,•–2)•and•B(3,•–5)••in•the•ratio•1:2.

Co-ordinate Geometry 3.• a.• Prove that the vertices A(3, 5), B(–1, 2) and C(6, 1) are the vertices
• b. of•an•isosceles•triangle.
Prove that the points (–3, 2), (5, 4), (7, – 2)and (1, –4) are the
vertices of a parallelogram.

4.• Find•the•co-ordinate of a point on y-axis which is 10 units distant from
the•point•(6,•–5).

86 PRIME Opt. Maths Book - VIII

Unit 5 Trigonometry

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 1 2 – 4 11 20

Weight 128–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can solve the right angled triangle.
• Students are able to know the types of measurement of angles.
• Students can !nd the trigonometric ratios.
• Students can prove the trigonometric identities.
• Students can !nd the value of the ratio of the standard angles.
• Students can !nd height and distance.

Materials Required:
• Chart paper.
• Graph paper.
• Chart of list of formulae used in trigonometry.
• Chart of values of standard angles.
• Model of right angled triangle.

PRIME Opt. Maths Book - VIII 87

Trigonometry 5. Trigonometry

The• word trigonometry is de•ined using three words; tri ® three, gones
® angles, metron ® measurement. Tri + gones + metron = trigonometry
(Measurement•of•three•angles•of•a triangle.) It is said that the origin of the
trigonometry•is•taken•from•the•ancient•Hindu•Civilization•as,

• Tri – lq
gono – sf0] f

• metry – dfk0f

The•measurement•of•three•angles•of•
a triangle•is•called•trigonometry.

It• is• useful• to •ind the angles of a right angled triangle and to •ind the
length• of• • the sides of it. It is used by engineers to •ind the heights and
distance•during•constraction•of•structures•which is useful to estimate the
constructing• materials• and• cost. It is also useful for various purposes in
physics, mathematics, statistics as well as other scienti•ic purposes.

5.1 Measurement•of•angles

Here,• we discuss the important three ways of measuring an angle viz.
Sexagesimal measurement (British System), Centesimal measurement
(French System) and Circular measurement (Radian measurement).

i. Sexagesimal measurement:
In this system, the right angle is divided into 90 equal parts, called
degrees,•one•degree•is•divided into 60 equal parts, called minute and
one•minute•is•divided•into•again•60•equal•parts,•called•Second.

• Thus,

One•right•angle•=•90°
\ 1°•=•60’
\ 1’•=•60’’

88 PRIME Opt. Maths Book - VIII

ii. Centesimal measurement:
In this system, one right angle is divided into 100 equal parts, called
grade, one grade is divided into 100 equal parts, called minute and
one•minute•is•divided•into•100•equal•parts,•called•Seconds.

• Thus,

One•right•angle•=•100g
\ 1g = 100’
\ 1’•=•100’’

iii. Circular measurement:

As shown i$AnBig=uOreA,•=O•Ar = radius = r
Length of

• Then the angle \AOB is ixed for all circles irrespective of the length
of•the•radius•which•is•called•one•radian•angle.

• We denote it by 1c Trigonometry

i.e. \AOB = 1c B

radian angle (1c)

OA

Here, Radius OA = arc length $
AB
\ \ AOB = radian angle (1c).

1c angle : The central angle formed by an arc of
a circle equal to the radius of the circle is called
one•radian•angle.

Also we know C = 2pr (perimeter or circumference i.e. in circumference,
there•are•2p number•of•arcs•times•radius.)
\ Central angle formed by circumference is 2pc.
But, central angle by circumference in degree is 360° or 400g.

PRIME Opt. Maths Book - VIII 89

\ 2pc = 360°•=•400g

pc = 180°•=•200g 200 g
r
And, 1c = 180° =
r

Relationship table between the measurement of angles.

2rt•angle•=• \ 180°•=•200g = pc
rc
1 rt•angle•=•90°•=•100g = 2

1°•=• a 10 g = ` r jc
9 180
k

1g = a 9 kc = ` r jc
10 200

1c = ` 180 jc = ` 200 jg
r r

Trigonometry Worked out Examples

1. Convert into Seconds of 12°15’36’’.
Solution
12°15‘36‘‘ = (12 × 60 × 60 + 15 × 60 + 36)‘‘
= (43200 + 900 + 36)‘‘
= (44136)‘‘

2. Convert into grades of 25 g35’’45’’

Solution

25g35‘‘45‘‘ = (25 + 35 + 45 )g
100 100 ×100
= (25 + 0.35 + 0.0045)g

= (25.3545)g

3. Convert into degrees of ` r jc .
Solution 125

` r jc = [\ 1c = 180g ]
125 r

= a 180 kc
125

= a 36 kc
25

= 1.44°

90 PRIME Opt. Maths Book - VIII

4. Find the ratio of 40 g and a 3r c . A
Solution 10 70°
k B 76°

1st angle••=•40g c

2nd angle•= a 3r k C
10

= a 3r × 200 g [\ 1c = 200g ]
10 r r
k

= 60g

Ratio = 1st angle
2nd angle
40 g
= 60 g

= 2
3

= 2:3

5. If any two angles of a triangle are 70° and 76°, !ind the third angle in

grades. A Trigonometry
Solution :

Let, ABC be a triangle, 54°
Where,

\ A = 70° ` 3r c C
• \ B = 76° 4
• \C = ? j

We have, B

\ A + \ B + \ C = 180°• [\Sum•of•internal•Angles•of•a D in indegre]

or, 70° + 76° + \ C = 180°

• or, \ C = 180°•–•146°

• \ \ C = 54° 10
9
• \ Third•angle• =•(54•×• )g
= 60g

6. One angle of a triangle is 54° and the Second A
c 2x°
Saonlguletioisna:38r , !ind the third angle in grades. 3x° 5x° C
k
91
Let, ABC be a triangle where,

\ A = 54°• B

• = (54 × 10 )g = 60g
9

PRIME Opt. Maths Book - VIII

\ B = a 3r c
8
k

= a 3r × 200 g
8 r
k

= 75g

\C = ?
We have,

\ A + \ B + \ C = 200g [\Sum•of•internal•Angles•of•a•D in grades]
• or, 60g + 75g + C = 200g

or, C = 200g – 135g
\ C = 65g
\ The third angle is 65g.

Trigonometry 7. Find the angles of a triangle in grades which are in the ratio 2:3:5.
Solution :
Let, ABC be a triangle and the angles are in the ratio 2:3:5 in grades.
Where,
\ A = 2xg
\ B = 3xg
\ C = 5xg
We have,
\ A + \ B + \ C = 200g
or, 2x + 3x + 5x = 200g
or, x = 20g

Then, the angles are,

\ A = 2 × 20g = 40g
\ B = 3 × 20g = 60g
\ C = 5 × 20•=•100g

92 PRIME Opt. Maths Book - VIII

Exercise 5.1

1.•• Express•the•following•angles•in•sexagesimal Seconds.
i) 25° 26‘ 27‘‘ ii) 36° 25‘‘ iii) 45‘ 36‘‘
iv) 75° 50‘ 55‘‘ v) 42° 45‘

2.• Express•the•following•angles•in•centesimal•Seconds.
• i) 42g 35‘‘ 42‘‘ ii) 75g 65‘ 72‘‘ iii) 36g 38‘‘
iv) 55g 36‘ v) 25‘ 32‘‘

3.• Express•the•following•angles•in•degrees. iii) 50° 24‘‘
• i) 22° 36‘ 42‘‘ ii) 72° 42‘ 54‘‘
iv) 55° 45‘ vi) 65g 75‘ 42‘‘
v) 80g c

vii) 42g 36‘ 45‘‘ viii) a 2r k ix) ` r jc
c 5 3
x) a 7r
45 k

4.• Convert the following angles in grades. Trigonometry
i) 85g 52‘ 45‘‘ ii) 27g 52‘ 46‘‘ iii) 44g 45‘‘
iv) 47g 27’•35‘‘ vi) 66° 42‘
v) 72° c

vii) 56° 24‘ 36‘‘ viii) a 3r k ix) ` r jc
c 10 40
x) a 2r
25 k

5.• Convert the following angles into radian.

i) 72° ii) 240° iii) 144°
vi) 350g
iv) 125g v) 400g

6.• Answer the following problems:

i) Find the ratio of the angles 48° and 80g.
25arnkgcleasnd9•208g °•ainn•ddeag13re0reksc.
• ii) Find the sum of the angles a in degree
iii)
Find the difference of the
measurement.

• iv) Find the angles of a triangle in degrees which are in the ratio
1:2:3.

• v) Find any two angles in degrees whose sum is 50° and difference
is•14°.

PRIME Opt. Maths Book - VIII 93

7.• Find•the•angles•of•the•triangle•from•the•followings.

• i) Find the angles of a triangle which are in the ratio 3:4:5 in degree

measurement.

• ii) If two angles of a triangle are in the ratio 3:5 and the third angle

is•60°,••ind•the•angles•in•degrees.
OIisfn•t32ewaoonfag•anle•grlioegfshato•tafrnaiagtnlregi,al••eninigsdle•53thaoreef•t•ah4ir0ri°dg•haantndagnl6eg8•lien° •ardenesdgprteehceetsisv. eeclyo,n•dinadntghlee
• iii)
• iv)

third•angle•in•grades.

• v) Find the angles of the right angled isosceles triangle in grades.

8.• Solve the problems given below.

i) One angle of a triangle is 60g, Second angle is 36°. Find the third

angle•in•degrees.

• ii) Two angles of a triangle are in the ratio 3:5 and the third angle is

40g, •ind•the•angles•in•degrees. jc
7r
Trigonometry • iii) One angle of a triangle is ` 20 , Second angle is 81°, •ind the
third•angle•in•grades.
TOishn• aee3s5arunmkgcl,eo•ifonafdna•ytthrteiwa•tnohgailrnedgi•aslens29golefo•afinat•rdriiaegnghrgteleaensis.gl1e0a0n°danthdetSheecdoinffderaenngclee
• iv)
• v)

is•20°,••ind•the•angles•of•the•triangle.

9. PRIME more creative question:
i) Convert 70° 45‘ 30‘‘ into•centesimal•measurement.
• ii) Convert 81g 15‘ 75‘‘ into•sexagesimal measurement.
iii) Express the angle 82° 24‘ 42‘‘ into circular measurement. (p =

3.1416)

• iv) The angles of a triangle are a 3x kc , a 2x g and ` rx jc , •ind the
angle•in•degrees. 5 3 75
k

• v) The •irst angle of a triangle in grades, Second angle in degrees

and•the•third angle in radians are in the ratio 280 : 288 : p, •ind

the•angles•in•degrees.

94 PRIME Opt. Maths Book - VIII

1.• i)• 91587”• Answer iii)• 2736”

• iv) 273055” ii)• 129625”•
v) 153900”
2.• i)• 423542”• ii)• 756572”• iii)• 360038”
v) 2532”
• iv) 553600” ii)• 72.715°•
v) 72°
3.• i)• 22.61167°• viii) 72° iii)• 50.0067°
vi) 59.17878°
• iv) 55.0125° ix) 60°

vii) 38.12805°

x) 28°

4.• i)• 85.5245g ii) 27.5246g iii) 44.0045g
v) 80g vi) 74.1111g
iv) 47.2735g viii) 60g ix) 5g

vii) 62.6778g

x) 16g c c c

5.• i)• a 2r k ii) a 4r k iii) a 4r k
5 3 5

iv) a 5r c v) 2pc vi)• ` 7r jc Trigonometry
8 4
k

6.• i)• 2:3• ii)• 100°• iii)• 27°
v) 32° and 18°
• iv) 30°, 60°, 90° ii)• 45°,•75°,•60°•
v) 100g, 50g, 50g
7.• i)• 45°,•60°,•75°• ii)• 54°,•90°,•36°• iii)• 54°,•60°,•66°
v) 40°, 60°, 80°
• iv) 80g ii)• 73°2’30”•
v) 63°, 72°, 45°
8.• i)• 90°• iii)• 40g

iv) 52°

9.• i)• 78g62’03”• iii)• 1.40345c

iv) 30°, 30°, 120°

PRIME Opt. Maths Book - VIII 95