Prime Optional Mathematics 7

Exercise : 4.5

1. Find the followings.

i. Express all the trigonometric ratios in terms of CosA.

ii. Express all the trigonometric ratios in terms of SinA.
4
iii. If SinA = 5 , �ind all the trigonometric ratios.

iv. If CosA = 5 , �ind all the trigonometric ratios.
13
v. If 3TanA = 4, �ind all the trigonometric ratios.

2. Answer the followings.
3
i. If SinA = 5 , prove that Cosec²A – Cot²A = 1

ii. If Cosθ = m , prove that Sec²θ – Tan²θ = 1
n
iii. If 4CotA = 3, �ind the value of 3SinA + CosA.

iv. If 3 TanA = 1, �ind the value of Sin²A + Cos²A.
3
v. If Cosecθ = 2 , �ind the value of Tanθ + Secθ.

3. Prime more creative questions.
i. SinA = CosA, �ind the value of CosecA.
ii. If xCosθ = y, �ind the value of Cos²θ – Cosec²θ.

iii. If 3TanA =Co4s,e�cinAd=th0e, �vinadlutehoefv3CalCousoeAsoA+f–3SeSSciinnAAA.
iv. If SecA –

v. Find Sinα and Tanθ from the given diagram.
A

13cm θ 12cm
P 3cm

C αB

96 PRIME Opt. Maths Book - VII

Answer

1. Show to your teacher.

2. iii. 3 iv. 1 v. 5

3. i. 2 ii. 1 iii. 1
iv. 3
1
2 v. 4 , 5
5 12

PRIME Opt. Maths Book - VII 97

4.6 Trigonometric Ratios of some standard angles

The trigonometric ratios of some standard angles 0°, 30°, 45°, 60° &
90° are going to be discussed and geometrical interpretations are also
discussing in this chapter. The values of the trigonometric ratios are
calculated as follows.

i) Trigonometric ratios of 0°. A
B
In the right angled DABC,
2a
\ B = 90° aC

\ C = q (reference angle)

If A approaches to B, the value of q will be zero. q

As q → 0, AC → BC and AB → 0 C

Then,

\ Sinq = AB ( Sin0° = 0 =0
AC ( Cos0° BC =1
( Tan0° =0
\ Cosq = BC ( Cot0° = BC =∞
AC ( Sec0° BC =1
( Cosec0° =∞
\ Tanq = AB = 0
BC BC

\ Cotq = BC = BC
AB 0

\ Secq = AC = BC
BC BC

\ Cosecq = AC = BC
AB 0

ii) Trigonometric ratios of 30° & 60°. A
2a 30°
Let, ABC is an equilateral triangle
60°
Where, aD

AD⊥BC

AB = BC = AC = 2a (Say) B
BD = DC = a

\ A = \ B = \ C = 60°

Then,

AD = AB2 – BD2

= ^2ah2 – ^ah2

= 3 a2

=a 3

98 PRIME Opt. Maths Book - VII

Trigonometric ratios of 30˚
Then, In right angled DADB,
\ BAD = 90° – 60° = 30°
AB = 2a = h
BD = a = p

AD = a 3 = b

Sin30° = BD = a = 1
AB 2a 2

Cos30° = AD = a3 = 3
AB 2a 2

Tan30° = BD = a = 1
AD a3 3

Also,

Cosec30° = 2

Sec30° = 2
3

Cot30° = 3

Trigonometric ratios of 60˚

Again, In right angle DABD,

\ ABD = 60°

AD = a 3 = p

BD = a = b

AB = 2a = h

Sin60° = AD = a3 = 3
AB 2a 2

Cos60° = BD = a = 1
AB 2a 2

Tan60° = AD = a3 = 3
BD a

Also,

Cosec60° = 2
3

Sec60° = 2

Cot60° = 1
3

PRIME Opt. Maths Book - VII 99

iii) Trigonometric ratios of 45°. A

Let DABC is an isosceles right angled triangle

where, B 45° C
\ B = 90°
AB = BC = a (say)
\ \ A = \ C = 45°

Then
\ AC = AB2 + BC2

= a2 + a2

=a 2

Also, Taking reference angle A.
p
Sin45° = h = BC = a = 1
AC a2 2

Cos45° = b = AB = a = 1
h AC a2 2

Tan45° = p = BC = a =1
b AB a

Cosec45° = AC = a2 = 2
BC a

Sec45° = AC = a2 = 2
AB a

Cot45° = AB = a =1
BC a

iv) Trigonometric ratios of 90°. A

Let,

ABC is a right angled triangle.

Where, \ B = 90°

\ C = Reference angle B C

When C approaches to B, the reference angle will be 90°.

As C apporaches to B,

BC → 0 and AB → AC

Now,

SinC = AB & Sin90° = AB =1
AC AB

CosC = BC & Cos90° = 0 =0
AC AC

100 PRIME Opt. Maths Book - VII

TanC = AB & Tan90° = AB =∞
BC 0

Also,

Cosec90° = 1

Sec90° = ∞

Cot90° = 0

Table for the values with respect to angles.

Write down = 0, 1, 2, 3, 4

Dividing by 4 = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Taking square root = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Result = 0, 1 , 1 , 3 , 1
2 2 2

Tabulation the above values respectively.

Angles 0° 30° 45° 60° 90°
Ratios 1
3 0
Sin 0 1 1 2 ∞
22 1
1
Cos 1 3 1 2 101
22
3
Tan 0 1 1
3 2
3
Cosec ∞ 2 2

PRIME Opt. Maths Book - VII

Sec 1 2 2 2 ∞
3

Cot ∞ 3 1 1 0
3

To remember

Sin0° = Cos90° = Tan0° = Cot90° = 0
Sin90° = Cos0°
= Tan45° = Cosec90°
Sin30° = Cos60°
Sin60° = Cos30° = Sec0° = Cot45° = 1
Tan30° = Cot60°
= 1
2

= 3
2

= 1
3

Tan60° = Cot30° = 3

Cosec30° = Sec60° = 2

Cosec60° = Sec30° = 2
3

Worked out Examples

1. Find the value of Sin30° + Cos60° + 1 Tan45°,
2

Solution:

Sin30° + Cos60° + 1 Tan45°
2
1 1 1
= 2 + 2 + 2 ×1

= 1+1+1
2

= 3
2

= 1 1
2

102 PRIME Opt. Maths Book - VII

2. Find the value of Sin²60° + Cos²30° + Tan²45°.

Solution:

Sin²60° + Cos²30° + Tan²45°

=c 3 2 + c 3 2 + ^1h2
2 2
m m

= 3 + 3 +1
4 4

= 3+3+4
4

= 10
4

= 5
2

= 2 1
2

3. If A = 0°, B = 30°, C = 60° and D = 90° find the value of SinA + CosB –
SinC + CosD.

Solution:
Here, A = 0°, B = 30°, C = 60° and D = 90°
then,
= SinA + CosB – SinC + CosD
= Sin0° + Cos30° – Sin60° + Cos90°

33
=0+ 2 – 2 +0
=0

4. Prove that: 1– tan 30° = 2– 3
1 + Cot 60°

1– 1
L.H.S.= 3

1+ 1
3

3 –1
3

= 3+1
3

PRIME Opt. Maths Book - VII 103

= 3 –1 × 3-1
3+1 3 –1

= ^ 3 - 1h2
^ 3h2 - ^1h2

= ^ 3h2 –2. 3 .1 + ^1h2
3–1

= 3–2 3 + 1
2

= 4–2 3
2

= 2 ^2– 3h
2

= 2– 3 R.H.S proved.

Exercise : 4.6

1. Find the value of the followings:
i. Sin0° + Cos0° + Cos60° + Sin30°
ii. Tan45° + Cos0° + Sin90°
iii. Sin60° + Cot45° – Cos30°
iv. Tan60° +Tan45° – Cot30°
v. 2Sin30° + 2Cos30° – Tan60°

2. Find the value of the followings:

i. 2Sin²45° + 4Cos²45° – 3Tan²45°
4 3
ii. 3 Tan²60° + 2 Sec²30° – 4Sin²90°

iii. Cos²0° + Cos²30° – Sin²45°

iv. Sin60°.Cos30° – Cos60°.Sin30°

v. Cos30°.Cos60° – Sin30°.Sin60°

3. Prove that the followings :
i. Tan²30°.Tan²60°.Sin²30°.Sec²60° = 1
ii. 2Sin²45° + 4Sin²60° – Tan²45° = 3

104 PRIME Opt. Maths Book - VII

iii. Cot30° + 1 = 2+ 3
Tan60°–1

iv. 1–Sin60° = 7–4 3
1 + Cos30°

v. Tan45°–Tan30° = 2– 3
1 + Tan45°.Tan30°

4. Find the value of the followings where A = 0°, B = 30°, C = 60°,
D = 90°
i. Sin²B + Cos²D
ii. 3(TanB + CotC) – 2
iii. Sin²A + Sin²B + Cos²C
iv. Cos(A + B) + Sin(C – A)
v. Tan²B + Cot²C – Cot²D

5. Prime more creative questions:
1 – Tan30° 1 – Cos30°
i. Prove that : 1 + Cot60° = Sin30°

ii. Prove that : 3Tan²30° + 4Sin²60° – 2Cos²45° = 3

iii. Simplify : Sin²30° + Cos²60° – Sin²45° + 2Cot²45°.

iv. In right angled isosceles ∆ABC, prove that Cos45° = 1
2

v. In an equilateral ∆ABC, prove that Sin60° = 3
2

Answer

1. i. 2 ii. 3 iii. 1 iv. 1 v. 1
v. 0
2. i. 0 ii. 2 iii. 1 1 iv. 1
2 2

3. Show to your teacher.

4. i. 1 ii. 0 iii. 1 iv. 3 v. 2
4 2 3

5. iii. 1 1
2

PRIME Opt. Maths Book - VII 105

4.7 Solution of right angled triangle

Calculation of all three angles and all three sides of a right angled triangle

can be done by using trigonometry and using their values for standard

angles.

Such type of problems are given to be discussed in this chapter.

In right angled ∆ABC, A
∡B = 90° ph
∡C = q
∡A = 90° – q

also, B bqC

h² = p² + b²
p
Sinq = h = AB
AC

Cosq = b = BC
h AC

Tanq= p = AB
b BC

Also, If q = 30° and p = 20cm

Then,
p

Sin30° = h

or, 1 = 20
2 h

or h = 2 × 20

\ h = 40cm

Worked out Examples

1. If ∡B = 90°, ∡C =30° and BC = 3 , find the other parts of ∆ABC.

Solution: A
In∆ABC,

∡B = 90°

BC = 3 C 30° B
∡C = 30°
then, 3
∡A = 90°– ∡C

106 PRIME Opt. Maths Book - VII

= 90°–30°

= 60°

again,

BC = 3 cm = b

we have,

Cos30° = b
h

3 = BC
or, 2 AC

33
or, 2 = AC

or, AC 2× 3
=
` AC
Again, 3
= 2 cm.
d
d = h2 –b2
= AC2 –BC2
d = ^2h2 –^ 3h2

d = 4–3
d =1
` AB = 1 cm.

2. In right angled ∆PQR, ∡Q = 90°, ∡R = 60° and PR = 20cm, find the

length of QR. P
Solution:

In rt. ∡ed ∆PQR,
∡Q = 90°
∡R = 60°
PR = 20 cm. = h 60°
QR

QR = ? =b

We know,

Cos60° = b
h

or, 1 QR
2 = PR

PRIME Opt. Maths Book - VII 107

or, 1 QR
2 = 20

20 10
or, QR = 2

` QR = 10cm

3. In the given diagram AB is a house and AC is a ladder where ∡B = 90°,
∡C = 30° and AC = 20m, find the height of the house.

Solution: A
In rt. ∡ed ∆ABC, 20m

∡B = 90° B 30° C
∡C = 30°
AC = 20m

AB = ?

We have,
p

Sin30° = h

or, 1 = AB
2 AC

or, 1 = AB
2 20

or, 2AB = 20

20 10
or, AB = 2

` AB = 10m.
\ Height of the house is 10m.

4. A boy is flying a kite where the kite is at a height of 50m from the
ground and the string makes and angle of 30° with the ground
find the length of the string.

Solution : A

Let,

AC be the length of the string 50m

AB be the height from the ground. B 30° C
Given

∠C = 30°

∠B = 90°

108 PRIME Opt. Maths Book - VII

AB = 50m

AC = ?

Now, In right angle DABC,
p

Sin30° = h

or, 1 = AB
2 AC

or, 1 = 50
2 AC

or, AC = 2 × 50

\ AC = 100m

\ Length of the string is 100m

Exercise : 4.7

1. Find the remaining parts of triangles given below.
i) ∡B = 90°, ∡C = 30°, AB = 4 cm, in ∆ABC.
30° ii) ∡A = 90°, ∡B = 45°, AC = 6 cm in ∆ABC.

iii) ∡P = 90°, ∡R = 60°, QR = 10 cm, in ∆PQR.
iv) ∡Q = 90°, ∡P = 45°, PQ = 8 cm, in ∆PQR.
P

8cm 45°

QR

v) ∠X = 90°, ∠Y = 30°, XZ = 20cm, In DXYZ.

XZ

30°

Y

PRIME Opt. Maths Book - VII 109

2. Answer the followings from a right angled triangle.
i) Find the height AB from the given diagram.

A

B 45° C
20 m

ii) Find the length of AC from the given diagram.

A

10 m

C 30° B

iii) Find the length of side PQ from the given triangle.

R

60°

20m

Q 60° P

iv) A ladder of length 20 3 m is taken against the wall as shown
in diagram. Find the height of the wall.

D

3m
20

F 60° E

v) A boy is flying a kite as shown in diagram where the kite is at a

height of 60m from the ground and the string makes 30° with

the ground. Find the length of the string.

A

60m

B 30° C

110 PRIME Opt. Maths Book - VII

3. Prime more creative questions.
i. A ladder of length 20 m is taken against a wall where angle

made by ladder with the ground is 30°. Find the height of the

wall.

ii. An electric pole of height 10m is tied with a rope where the

rope makes an angle of 45° with the ground. Find the distance

between the rope and foot of the pole on the ground.
iii. A boy is flying a kite where 200m long string makes an angle

of 30° with the ground. Find the height of the kite from the

ground.

iv. Find the distance on the ground in the given diagram where

height of the pole is 20 3 m.

A

C 60° B
Distance

v. A tree is broken by the wind and the top 8m

touches the ground making an angle 8m
30°
of 30° as shown in diagram. Length of
the broken part is 8m, find the original
height of the tree.

Answer v. 120m
v. 12m
1. i. ∠A = 60°, AC = 8cm, BC = 4 3 cm
ii. ∠C = 45°, AB = 6cm, BC = 6 2 cm 111
iii. ∠Q = 30°, PQ = 5cm, PQ = 5 3 cm
iv. ∠R = 45°, QR = 8cm, PR = 8 2 cm
v. ∠Z = 60°, YZ = 40cm, XY = 20 3 cm

2. i. 20m ii. 20m iii. 20 3 m iv. 30m
3. i. 10m ii. 10m iii. 100m iv. 20m

PRIME Opt. Maths Book - VII

Trigonometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Write down the relation of 1g in degree measurement. A qC
2. a. Find the trigonometry ratio of Sinθ and

Cotθ from the given right angled ∆ABC.
b. Prove that : Tanθ.Cosθ.Cosecθ = 1.

c. Find the value of Sin²60° + Cos²30° – B
2Sin²30°.

3. a. One angle of a triangle is 40g and the other angler is 60°, Find the

third angle is degree.

b. Prove that : 1 – Sin4 A = 1 + 2Tan²A
Cos4 A

4. If SinA = 5 , find the value of Sec²A– Tan²A.
13

Unit Test - 2
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. Convert 45° into grades.

2. a. If CosA = 3 , find the value of SinA.
5

b. Prove that : 1 – CosA = SinA
SinA 1 + CosA

c. Find the value of : Tan²30° + Cot²60° + Cosec²60°.
3. a. One angle of a right angled triangle is 36°, find the third angle in

grades.

b. Prove that : Tan²A – Sin²A = Tan²A.Sin²A.

3
4. Prove geometrically that Sin60° = 2 by using an equilateral

triangle.

112 PRIME Opt. Maths Book - VII

5 Vector

Specification Grid Table TM Periods

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

No. of Questions 1 – 1 – 25 4

Weight 1–4–

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

Objectives : Materials Required:
At the end of the lesson • Chart paper.
• Students can identify the • Graph paper.
• Chart of list of types and
vector and scalar quantity.
• Students can find magnitude, application of vectors.
• List of the examples of
direction and unit vectors.
• Students can operate vector and scalar.
• Geo-board.
vector addition and scalar • Graph board.
multiplication.
• Students can use the vector in 113
our daily life.
• Students can find the position
vector in different situation.

PRIME Opt. Maths Book - VII

Introduction

Some of the quantities can be measured but some can not be measured
out of different quantities in our surroundings. The quantities which can
be measured are called Physical quantities. The Physical quantities are
classified into two groups scalar and vector.

The Physical quantities which can be measured and has magnitude as
well as direction is called vector and that which has only magnitude is
taken as scalar quantities.

The Physical quantities which has only magnitude
is called scalar.
Examples: Distance, time, mass, area, speed,
density etc.

The Physical quantities which has magnitude as well
as direction is called vector.
Examples: Displacement, velocity, acceleration, work,
power etc.

Scalar quantities is represented by using a line segment only where
vector is represented by using a directed line segment.

BB
AA

Scalar – AB Vector – AB

• Vector is represented by AB, a , PQ, b etc.
• A line segment with arrow is called directed line segment.
• Directed line segment is used to denote the vector quantity.

114 PRIME Opt. Maths Book - VII

5.1 Vector quantities using co-ordinate

Let us consider paoniynttws roepspoeincttsivAe(lyx1t,oyd1)ea�innde Ba(vxe2c, tyo2r) which are taken as
initial and �inal AB .

Y

(x2, y2)
B

(x1, y1) C
A

OM NX

Here,
A(x1, y1) and B(x2, y2) are any two points where,

AC = MN
= ON – OM
= x2 –x1

BC = BN – CN

= BN –AM

= y2 –y1

Then,
Vector AB is de�ined by using two component for the directed line

segment AB . where,
x - component = AC = x2 – x1

y- component = BC = y2 – y1

` AB = e x - component o = d x2 –x1 n
y - component y2 –y1

Component of vector in graph:

The x- component and y- component can be expressed taking horizontal
and vertical displacement of the given vector respectively. The sign of the
components can be taken as the quadrents rule to take the components
in graph.

PRIME Opt. Maths Book - VII 115

P
B

M
AD

Q

NC

Here,

The directed line segment which represent the vector with their

components can be expressed as follows.
3 –3
AB = ` 2 j, CD = ` 4 j

PQ = ` 3 j , MN = ` –2 j
–4 –4

Note:
Horizontal displacement $ Right side = +ve

Left side = –ve

Vertical displacement $ Upward = +ve
Downward = –ve

Types of Vector:

Column Vector:

The Vector quantity where the components are taken in a column is

called column Vector.
-
i.e. AB = c x - component m
y component

Row Vector :
The Vector quantity where the components are taken in a row is called
row vector.

116 PRIME Opt. Maths Book - VII

i.e. AB = (x- component y- component)

Null Vector :

The Vector having both the components zero is called null Vector.

i.e. a = a 0 k
0

Equal vectors:
Any two vectors are said to be equal if both the components of the
vectors are same. i.e. The vectors having same direction as well as equal
magnitude are called equal vectors.

Examples:
i. 4 units

AB
4 units

PQ

Here, AB = PQ

ii. AB = ` 3 j and PQ = ` 3 j
4 4

Here, AB = PQ

Worked out Examples

3 –2
1. Show the vector a =d n and b =d n with directed line segment
in graph paper. –4 5

Solution:

The given vectors are:

3 –2
a = d–4 n and b = d 5n

b
a

PRIME Opt. Maths Book - VII 117

2. If A(1, 2) and C(–3, 5) are any two points, find the vector AB .

Solution:

The given points are:

A(1, 2) = (x1 , y1)
B(7, 5) = (x2 , y2)
Then,

AB = dx2 – x1 n
y2 – y1

7–1
= d5 – 2n

6
= d3n

6
` AB = d3n

3. If P(3, 1), Q(5. 4), R(4, 3) & S(6, 6) are the four points, prove that

PQ = RS .

Solution:

The points for PQ are,

P(3, 1) = (x1 , y1)
Q(5, 4) = (x2 , y2)
Then,

PQ = dx2 – x1 n
y2 – y1

5–3
= d4 – 1n

2
= dn
3

Again, the points for RS . are,

R(4, 3) = (x1 , y1)
S(6, 6) = (x2 , y2)
d x2 – x1 n
Then, RS = y2 – y1

6–4
= d6 – 3n

118 PRIME Opt. Maths Book - VII

2
= d3n

` PQ = RS proved.

Exercise : 5.1

1. i. What is vector quantity? Write down four examples of it.
ii. What is scales quantity? Write down four examples of it.
iii. Write down the difference between the following straight lines.
BB

AA
iv. What do you mean by column vector?

v. What do you mean by null vector?

2. Find the vectors represented by the following directed line segments.
BM

i. AB A C v. RS
QN
D
ii. CD P
RS

iii. PQ iv. MN

3. Show the following vectors in graph paper using directed line

segment.

2 –3 2
i. AB = d4n ii. CD = d 4 n iii. PQ = d–5 n

–4 3
iv. RS = d n v. MN = d n
–3 0

PRIME Opt. Maths Book - VII 119

4. Find vectors AB and CD from the followings.

i. A(1, 5) and B(4, 10) ii. A(–3, 2) and B(–5, 6)

iii. C(4, 2) and D(6, 0) iv. C(1, 4) and D(4, 2)

v. A(–2, –1) and B(–4, –5)

5. Prime more creative questions:
i. If A(1, 3), B(4, 4), C(3, 2) and D(6, 3), find the vectors AB and
CD .

ii. Write down the relation of vectors AB and CD of Q. No. 1

iii. If P(1, 5), Q(3, 8), R(2, 6) and S(4, 9) are any four points, prove

that PQ = RS .

iv. Find AB where A(1, –3) and B(4, 1) are the any two points.
Also find the vector BA .

v. Find OA from the adjoining diagram.

Y`

A(3, 4)

X` O X

Y`

Answer

1. Show to your teacher.

3 –4 –3 5 5
2. i. dn ii. dn iii. dn iv. dn v. dn
4 –4 2 –3 0

3. Show to your teacher.

3 –2 23 –2
4. i. dn ii. dn iii. dn iv. dn v. dn
5 4 –2 –2 –4

33 3 –3 3
5. i. d1 n, d1 n ii. AB = CD iv. d4n, d–4n v. d4n

120 PRIME Opt. Maths Book - VII

5.2 Vector operations

Magnitude of a vector:

The length of the directed line segment used in the vector is called

magnitude of the vector. It is also called the modulus of the vector.
-
For the vector AB = c x - component m
y component

Magnitude of AB ,

AB = ^x - componenth2 + ^y - componenth2
x
For the vector a = ` y j

a = x2 + y2 3
4
for the vector of a = ` j

Magnitude of a is

|a | = x2 + y2

= 32 + 42

= 25

= 5 units.

The modulus of a vector which represents the length
of the directed line segment is called magnitude of the
vector.

i.e. |a | = x2 + y2

Unit vector:

The vector having magnitude one is called unit vector.

i.e. If AB = 1 unit, then the vector AB is called unit vector.

The vector having magnitude one is called unit vector.
1
Unit vector of a = a (a ).

PRIME Opt. Maths Book - VII 121

Examples:

If a = ` 3 j, find the unit vector of a .
4

Solution:

a = ` 3 j
4

|a | = x2 + y2

= ^3h2 + ^4h2

= 25

= 5 units.

Then, unit vector of a

= 1 ^ah
a

= 1 ` 3 j
5 4

= KKKKLKKKJ 3 OOOOOOPON
5
4
5

Addition of vectors:

The addition of any two vectors having same direction is the combination

of the vectors which is obtained by adding their corresponding

components.

i.e. a = a x1 k and b = a x2 k
y1 y2
Then,
+
a+b =a x1 k+a x2 k = a x1 + x2 k
y1 y2 y1 y2

Examples:

If a = ` 1 j and b = ` 3 j
2 4

Then,

a+b = ` 1 j + ` 3 j = e1 + 3o = ` 4 j
2 4 2 + 4 6

`a + b = ` 4 j
6

122 PRIME Opt. Maths Book - VII

Note : Subtraction of any two vectors also can be done as the rule of

addition.

i.e. a = d x1 n and b = d x2 n
y1 y2

∴ a –b = d x1 n – d x2 n
y1 y2

= d x1 – x2 n
y1 – y2

Worked out Examples

1. SIfoalu=tieon37: o, Find the magnitude of a .

a = e 7 o
3

We have,

Magnitude of a is

|a | = x2 + y2

= ^ 7h2 + ^3h2

= 7+9

= 16
= 4 units

2. If A(7, 1) and B(–1, 7), �ind the magnitude of AB .

Solution:

The two points are:

A(7, 1) = (x1 , y1)
B(–1, 7) = (x2 , y2)
Then,

AB = d x2 – x1 n
y2 – y1

= d–71––17n

= d–68n

PRIME Opt. Maths Book - VII 123

Magnitude of AB is,
; AB ; = x2 + y2

= (–8)2 + 62

= 64 + 36

= 100
= 10 units.

3. If magnitude of a = ` 4 j is 5 units find the value of ‘m’.
m

Solution:

a = ` 4 j
m

|a | = 5 units

We have,

|a | = x2 + y2

or, 5 = 42 + m2

squaring on both sides,

or, (5)2 = ^ 16 + m2h2

or, 25 = 6 + m2

or, 25 – 16 = m2

or, 9 = m2

\ m=3

4. Find the unit vector of a = ` 4 j .
3
4
Solution : a = ` 3 j

Then,

|a | = x2 + y2 = 42 + 32 = 5 units.

Also,

unit vector of a is,
a = 1 ^ah

a

= 1 ` 4 j
5 3

= c 4/5 m
3/5

124 PRIME Opt. Maths Book - VII

5. If a = ` 3 j and b = a 5 k , find the value of a +b. Also find the
2 4
magnitude of a + b .

Solution :

Here, a = ` 3 j
2

b = a 5 k
4

Then

a + b = ` 3 j+ a 5 k
2 4

= a 3 + 5 k
2 + 4

= a 8 k
6

Again, magnitude a + b is,
| a + b | = x2 + y2

= 82 + 62
= 100
= 10 units

Exercise : 5.2

1. Find the magnitude of the vectors form the followings:

3 6 5
i. a = d n , ii. b =d n iii. p =d n
4 8 12

q =e 3o 2
iv. 1 v. q=d n
5

2. Find the unit vector of the followings.

1 4 8
i. a =d n ii. b =d n iii. c =d n
3 3 6

12 q =e 5o
iv. p =d n v. 2
5

PRIME Opt. Maths Book - VII 125

3. Add the following vectors:

13 41
i. a = dn and b = dn ii. c = d n and d = d n
2 4 3 –1

21 –2 1
iii. p = d3n and q = d2n iv. a = d 3n and b = d–2 n

31
b = dn c =d n
v. 4 and –3

4. Subtract the following vectos.

31 –2 1
i. b = dn c =d n ii. a =d n and b =d n
4 and –3 3 –2

21 41 n
iii. p = dn and q = dn iv. c = d n and d = d
3 2 3 –1

13
v. a = d2n and b = d4n

5. Prime more creative questions:

i. Find the magnitude of AB where A(1, 3) and B(4, 7) are any

two points.

ii. If a = 1 and b = 5 n , find the magnitude of a +b.
dn d
2 6

iii. If magnitude of a = x is 5 units, find the value of x.
d4n

1
iv. Prove that a = dn is the unit vector of a .
0

v. Prove that a = KKKKKJLKK34 5 OOPOOOOON is the unit vector of a .
5

126 PRIME Opt. Maths Book - VII

Answer

1. i. 5 units ii. 10 units iii. 13 units

iv. 2JKLKKKKKKKJKKLKKKK15u123n1212i3t3sOOOOOONPOOOOOOOONP v. 3 units OOOOOONOP
i. KKLKKKKJKJLKKKKKKK43255533OOOOOOONP OOOPOONOO KJLKKKKKK34
2. ii. iii. 5
iv. v. 5

45 3 –1 4
3. i. dn ii. dn iii. dn iv. dn v. dn
6 2 5 1 1

–2 3 1 –3 2
4. i. dn ii. dn iii. 1 iv. dn v. dn
–2 4 5 7

5. i. 5 units ii. 10 units iii. 3

PRIME Opt. Maths Book - VII 127

Vector

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. If vector a = 2 n , find the vector 2a .
d3

2. a. Define vector quantity with examples.

b. If a = –1 and b = 4 n , find the vector a + b .
d3 n d1

c. If A (3, –2) and B (5, 2) are any two points, find the vector AB .

3. a. If magnitude of a vector a = m is 10 units, find the value of m.
d6n

b. Prove that AB = CD where the four points are A(3, –1), B (1, 5),

C (1, –3) and D (–1, 3).

4. If a = 2 and b = –2 n , find the magnitude of vector 3a – b .
dn d
1 –3

128 PRIME Opt. Maths Book - VII

6 Transformation

Specification Grid Table TM Periods

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

No. of Questions – 1 – 1 2 7 10

Weight –2–5

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

Objectives : Materials Required:
At the end of the lesson • Chart paper.
• Students can define the • Graph board.
• Sample of
transformation.
• Students can identify the transformations.
• List of the formula of
isometric transformations.
• Students can find the image transformations.
• Geo-board
of a point under isometric • Chart paper.
transformations geometrically. • Shape of objects
• Students can find the image
using co-ordinate. 129
• Students can plot the object
and image in graph.

PRIME Opt. Maths Book - VII

Introduction of transformation

Let us see,
i)

Here, the image of a girl is seen into a mirror, where
• A girl in front of mirror is the object and
• The girl into the mirror is called the image.
ii)

1st 2nd

4th 3rd
Here,
• 1st star is taken as the object.
• Second star is taken as the image.
• Others also can be drawn as other images.

130 PRIME Opt. Maths Book - VII

Above examples show that one object can be transferred from one place
to another place with same shape & size or different size which is called
transformation.

The process of changing the position or
size of an object under any geometrical
conditions is called transformation.

By using transformation patterns of objects can be drawn which is useful
to fill the pictures in clothes walls and in any kind of objects to make

them very attractive.

Here in grade VII we discuss only two types of transformations
i) Reflection iii)Translation

6.1 Reflection A P

The image of a line segment AB is A‘B‘ which is Q A‘
m B‘
formed at equal distance from the mirror line
m as the distance of object where AP = PA‘ or
BQ = QB‘. The image A‘B‘ so formed is laterally B

inverted as the image formed in the looking

glass but size is same as the object.

Here, AB is called the object
m is called reflection axis (mirror line)
A‘B‘ is called the image.

The transformation of an object from one place to
another place by a mirror line is called reflection.
• Object and image so formed are always

congruent in reflection.

PRIME Opt. Maths Book - VII 131

Reflection using co-ordinate
i) Reflection about x-axis (y = 0).

Y

4

3 M(2, 3)

2

1

X‘ P X
-4 -3 -2 -1 O 1 2 3 4
-1

-2

-3 M’(2, –3)

-4

Y‘
Draw MP⊥OX and produce MP to M‘ such that MP = PM‘

Here, M(2, 3) is an object.
x-axis (XX‘) is the reflection axis.
M‘(2, –3) is the image.

It shows that: P(x, y) → P‘(x, –y)

ii) Reflection about y-axis (x = 0)

Y

P‘(–3, 4) M4 P(3, 4)

3

2

1

X’ -4 -3 -2 -1 O 1 2 3 4 X
-1

-2

-3

-4

Y’

132 PRIME Opt. Maths Book - VII

Draw PM⊥OY & produce PM to MP‘ to make PM = MP‘.
Here, P(3, 4) is an object.
Y -axis (YY‘) is the reflection axis.
P‘(–3, 4) is the image.
It shows that : P(x, y) → P‘(–x, y)

Worked out Examples

1. Draw the image of ∆ABC by under the given mirror line.

A

B
C

Solution: AM

B

CP A`

C`

N
B`

Here.
AM, BN and CP are the perpendicular drawn on the mirror line.
Also, AM = MA`
BN = NB`
CP = PC`
Then, the ∆A`B`C` is the image of ∆ABC after reflection.

2. Find the image of a point P (3, 2) under reflection about y-axis.
Solution:
Under reflection about y -axis,
P (x, y) → P` (–x, y)
P (3, 2) → P` (–3, 2)

PRIME Opt. Maths Book - VII 133

3. Find the image of ∆ABC having vertices A(–1, 3), B( 2, 5) and C(4, 0)
under reflection on x-axis. Also plot the object and image in graph.

Solution:
under reflection about x - axis,
P (x, y) → P` (x, –y)
A (–1, 3) → A` (–1, –3)
B (2, 5) → B` (2, –5)
C (4, 0) → C` (4, 0) Y

B(2, 5)

A(–1, 3)

X’ C(4, 0) X

O C`(4, 0)

A`(–1, –3)

B`(2, –5)

Y’

134 PRIME Opt. Maths Book - VII

Exercise : 6.1

1. Draw the image of given triangles under reflection on mirror line M.

i. A ii. R M

B C S
M Q

T

iii. iv. L

R P J
M Q
K

I
M

v. O

P

S Q
R M

2. Find the image of following points under the given reflection axis.
i. A (2, 3) ; under x - axis
ii. P (–3, 5) ; under y - axis
iii. B (–4, –5) ; under y = 0
iv. Q (–5, 2) ; under y - axis
v. P (0, 5) ; under x = 0

PRIME Opt. Maths Book - VII 135

3. Find the image of the triangles under the given reflection axis. Also
plot the object and image in graph.

i. Having vertices A (2, 3), B (4, 6) and C (7, 1) under x- axis.

ii. Having vertices P (4, 2), Q (1, 5) and R (3, –3) under y- axis.
iii. Having vertices A (–3, 0), B (–6, 4) and C (0, 3) under reflection

about y = 0.

iv. Having vertices P (–6, 1), Q (–3, 5) and R (–5, –4) under x=0.

v. Having vertices K(–3, 0), L(0, 5), L(0, 5) & M(3, 2) under x-axis.

Answer

1. Show to your teacher.

2. i. A`(2, –3) ii. P`(3, 5) iii. Q`(–4, 5)
iv. Q`(5, 2) v. P`(0, 5)

3. i. A`(2, –3), B`(4, –6) and C`(7, –1); graph
ii. P`(–4, 2), Q`(–1, 5) and R`(–3, –3); graph
iii. A`(–3, 0), B`(–6, –4) and C`(0, –3); graph
iv. P`(6, 1), Q`(3, 5) and R`(5, –4); graph
v. K`(–3, 0), L`(0, –5) and M`(3, –2); graph

136 PRIME Opt. Maths Book - VII

6.2 Translation

The transformation of an object under the magnitude and direction of a
vector from one place to another place is called translation. The object
and image are always congruent in translation.

P’

P

Q’ B R’

QA

R

Here, vector AB is the magnitude & direction of translation.
DPQR is an object.
PP‘ = QQ‘ = RR‘ = AB
Then,
PP‘ || QQ‘ || RR‘ || AB are drawn
where, after translation.
DP‘Q‘R‘ is the image of DPQR

The transformation of an object from one place
to another place according to the magnitude and
direction of the given vector is called translation.

i) Translation using co-ordinate:

Y

A’(7, 5)

A(4, 3)

X’ B’(6, 1) C’(9, 1)
PRIME Opt. Maths Book - VII X
O
B(3, -1) C(6, -1)

Y’

137

Here, A(4, 3) is translated to A’(7, 5)
B(3, –1) is translated to B’(6, 1)
C(6, –1) is translated to C’(9, 1)

i.e. All the points are translated with constant number 3 for
x-component and 2 for y - component.

i.e. A (4, 3) → A‘(4 + 3, 3 + 2) = A‘(7, 5)
3

i.e. Translation vector is T = <2F

• a
i.e. under translation T = <F
b

P(x, y) → P‘(x + a, y + b)

iii) Translation using vector:

Let us consider a translation vector is AB where A(1, 2) and B(3, 5)

are any two points.

Then, according to the concept of column vector AB

AB = dx2 – x1 n = 3–1 = 2
y2 – y1 d5 – 2n d3 n

2
\ Translation vector T = AB = dn
3

Worked out Examples

1. Draw the image of ∆PQR under translation on given vector.

P

Q v
R

Solution:

138 PRIME Opt. Maths Book - VII

P P`

Q v
Q`

R R`

Here,PP`, QQ` and RR` are drawn parallel to the vector line V where,

PP` = QQ` = RR` = V
\ ∆PQR is the image of ∆PQR.

2. Find the image of a point P (2, 3) under a vector T =:32 D
Solution:
under translation about T = :32 D
then,

P (x, y) → P` (x + a, y + b)

→ P` (x + 2, y + 3)

P (2, –3) → P` (2 + 2, –3 + 3)

→ P` (4, 0)

3. Find the image of ∆ABC having vertices A (1, 2), B (3, 5) and C (6, 1)

2
under in the translation about T = <F . Also plot the object and
4
image in graph.

Solution:

2
Under translation about T = <4F
P (x, y) → P` (x + a, y + b)

→ P` (x + 2, y + 4)

A (1, 2) → A` (1 + 2, 2 + 4) = A` (3, 6)

B (3, 5) → B` (3 + 2, 5 + 4) = B` (5, 9)

C (6, 1) → C` (6 + 2, 1 + 4) = C` (8, 5)

PRIME Opt. Maths Book - VII 139

Y

B`

A`
B C`

A

C

X’ O X

Y’

Exercise : 6.2

1. Draw the image of the triangles given below under translation on

given vector ‘v’.

i. A ii. P

V Q R
B V

C

PS

iii. D iv. Q R
A V
V

B
C

140 PRIME Opt. Maths Book - VII

v. R

SP
V

QT

2. Find the image of the points given below under the given translation

vector ‘T’.
i. A (3, –2) ; T = 812 B

ii. P (–3, 2) ; T = 832 B

iii. M (3, –4) ; T = 8 –1 B
4

iv. B (–1, –2) ; T = :26 D

v. N (2, 3) ; T = : –2 D
–3

3. Find the image of the triangles given below under the translation
‘T’. Also plot the object and image in graph.
i. Having vertices A (1, 2), B (3, –2) and C (4, 5) under T =:32 D

ii. Having vertices P (0, 1), Q (3, 5) and R (1, –4) under T = 834 B

iii. Having vertices A (–1, 2), B (3, 6) and C (4, –3) under T = :–32 D

iv. Having vertices A (5, 2), B (3, 5) and C (7, –4) under T = 8––24 B

v. Having vertices P(–3, 1), Q(0, 5) and R(3, –4) under T = :53 D

PRIME Opt. Maths Book - VII 141

Answer

1. Show to your teacher.

2. i. A`(4, 0) ii. P`(0, 4) iii. Q`(2, 0)
iv. Q`(1, 4) v. P`(0, 0)

3. i. A`(3, 5), B`(5, 1) and C`(6, 8); graph
ii. P`(3, 5), Q`(6, 9) and R`(4, 0); graph
iii. A`(–3, 6), B`(1, 9) and C`(2, 0); graph
iv. P`(1, 0), Q`(–1, 3) and R`(3, –6); graph
v. K`(0, 6), L`(3, 10) and M`(6, 1); graph

Transformation

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 transformation of an object ?
2. a. Find the image of an object A (2, –3) under reflection about x-axis.

b. Find the image of an object P(0, 6) under transformation about

2
of T= <F .
3

c. Find the image of a point A (a, –b) under reflection about y-axis.

3. a. Find the image of a triangle having vertices A(3, 2), B(6, 5) and
C(8, 0) under reflection about x = 0. Also plot the object and

image in graph.

a
b. If a translation T = <F given image of an object A(2, 3) to A`(4, 6),
b

find the translation vector T.
4. Find the image of ∆PQR having vertices P(–1, 2), Q(2, 4) and

R(4, 0) under a translation vectorPQ . Also plot the object and

image in graph.

142 PRIME Opt. Maths Book - VII

7 Statistics

Specification Grid Table TM Periods

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

No. of Questions – 1 – 1 27 6

Weight –2–5

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

Objectives : Materials Required:
At the end of the lesson • Chart paper.
• Students can find the values • Sample of data

of central tendency mean, collection.
median & mode. • Sample of tabulation of
• Students can find the partition
values like quartiles and data.
deciles. • List of formula used in
• Students can find the
measures of dispersion range statistics.
and quartile deviation. • Graph paper.
• Students can tabulate the
collected data. 143
• Students can present the dat
in diagrams.

PRIME Opt. Maths Book - VII

Introduction of statistics

Data Collection

The set of numerals collected for any fact for the purpose of investigation
or analysis during study is called statistical data. It can be collected by the
individuals directly for personal investigation by travelling to the related
area, through interviews and questionnaires etc, is called primary data. It
can be collected by taking the informations from related offices like Village
Municipality, Municipality, sub metropolitan city, metropolitan city, district
office, province office, central office etc is called secondary data.

In statistics, the collected raw data should be tabulated, represented
diagrammatically and processed for different analysis. Among them
here we discuss about central values mean, median and mode as well as
simple measure of dispersion range.

Tabulation of data

The collected data which is written roughly is called raw data. The
categorical presentation of data in a table using frequency is called
the frequency distribution table. There are three types of frequency
distribution table viz. individual, discrete and continuous frequency
distribution tables.

i) Discrete frequency distribution (ungrouped) :

The raw data collected for the investigation is tabulated individually

with respect to their repeated numbers with tally marks in it. The number

of repeatation of any variable is called the frequency of the variable.

Example : The number of family members collected in a village is as

3, 2, 4, 4, 3, 2, 3, 3, 4, 4, 4, 5, 5, 6, 5, 6, 4, 3, 2, 5, 6, 4, 3, 4, 4, 3, 4, 5, 2, 5

No. of members Tally Marks Frequency

2 |||| 4

3 |||| || 7

4 |||| |||| 10

5 |||| | 6

6 ||| 3

Total 30

144 PRIME Opt. Maths Book - VII

7.1 Measure of Central Tendency

The calculation of a central numerical value of the given statistical data
which suppose to represent the entire data is called the measure of
central tendency. They are mean, median and mode.

Arithmetic Mean

The average value of the statistical data which is the ratio of sum of

observations to the total number of observations is called the arithmetic

mean or simply mean. Sum of the observation
Total no. of observation
i.e. Arithmetic Mean =

The average value of the observations of the collected

data is called arithmetic mean.
Sum of the observation
i.e. Mean (x) = Total no. of observation

i) For individual observations: Mean (x) = Rx
n

Where, x = observation (value of variable)

∑x = sum of the observations

n = no. of observations

ii) For discrete observation
Rfx
Mean (x) = N

Where,

x = observations (value of variable)

f = No of observations (No. of repeatation of particular variable)

fx = product of observations and respective no. of observations

R fx = sum of (f × x)

N = R f = sum of number of observations

PRIME Opt. Maths Book - VII 145