Vedanta Excel in Opt. Mathematics Book 7 Final (2078)

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7Book

Author
Piyush Raj Gosain

Hukum Pd. Dahal Editors P. L. Shah
Tara Bdr. Magar

vedanta

Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

vEMexdcAaenlTtaiHn AEMddAiTtiIoCnSal

Author 7Book
Piyush Raj Gosain

All rights reserved. No part of this publication may
be reproduced, copied or transmitted in any way,
without the prior written permission of the publisher.

¤ Vedanta Publication (P) Ltd.

First Edition: B.S. 2077 (2020 A. D.)
Second Edition: B. S. 2078 (2021 A. D.)

Layout and Design
Pradeep Kandel

Printed in Nepal

Published by:

Vedanta Publication (P) Ltd.
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Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
[email protected]
www.vedantapublication.com.np

Preface

The series of Vedanta Excel in Opt. Mathematics for class 9 and 10 is completely based on the

contemporary pedagogical teaching learning activities and methodologies. It is an innovative

and unique series in the sense that the contents of each textbooks of the series are written and

designed to ful ill the need of integrated teaching learning approaches.

Vedanta Excel in Opt. Mathematics has incorporated applied constructivism the latest trend of

learner centered teaching padagogy. Every lesson of the series is written and designed in such

a manner that makes the classes automatically constructive and the learner actively participate

in the learning process to construct knowledge themselves, rather than just receiving ready

made information from their instructor. Even teachers will be able to get enough opportunities

to play the role of facilitators and guides shifting themselves from the traditional methods

imposing instractions. The idea of the presentation of every mathematical item is directly or

indirectly re lected from the writer's long expenena, more than two decades, of teaching optional

mathematics.

Each unit of Vedanta Excel in Opt. Mathematics series is provided with many more worked out

examples, arranged in the hierarchy of the learning objectives and they are re lective to the

corresponding exercises.

Vedanta Excel in Opt. Mathematics class 10 covers the latest syllabus of CDC, the government of

Nepal, on the subject. My honest efforts have been to provide all the essential matter and practice

materials to the users. I believe that the book serves as a staircase for the students of class 10.

The book contains practice exercises in the form of simple to complex including the varieties of

problems. I have tried to establish relationship between the examples and the problems set for

practice to the maximum extent.

In the book, every chapter starts with review concepts of the same topic that the students have

studied in previous classes. Discussion questions in each topic are given to warm up the students

for the topic. Questions in each exercise are catagorized into three groups - Very Short Questions,

Short Questions and Long Questions.

The project works are also given at the end of exercise as required. In my exprerience, the students

of class 10 require more practices on Trigonometry and Vector Geometry, the examples and the

exercise questions are given to ful ill it in the corresponding topics. The latest syllabus of the

subject speci ication grid and a model question issued by CDC are given at the end of the book.

My hearty thanks goes to Mr. Hukum Pd. Dahal, Tara Bahadur Magar and P.L. Shah, the series

editors, for their invaluable efforts in giving proper shape to the series. I am also thankful to my

colleage Mr. Gyanendra Shrestha who helped me a lot during the preparation of the book.

I am also thankful to my respected parents and my family members for their valuable support to

bring the book out in this form. I would also like to express my hearty gratitude to all my friends,

colleagues and beloved students who always encouraged me to express my knowledge, skill and

experience in the form of books. I am highly obliged to all my known and unknown teachers who

have laid the foundation of knowledge upon me to be such a person.

Last but not the least, I am hearty theankful to Mr. Pradeep Kandel, the computer and designing

senior of icer of the publication for his skill in designing the series in such an attractive form.

Efforts have been made to clear the subject matter included in the book. I do hope that teachers

and students will best utilize the series.

Valuable suggestions and comments for its further improvement from the concerned will be

highly appreciated. Piyush Raj Gosain

CONTENT

Unit 1 Functions 5
25
Unit 2 Polynomials 44
84
Unit 3 Sequence and Series 97
113
Unit 4 Linear Programming 128
154
Unit 5 Quadratic Equations and Graphs 199
269
Unit 6 Continuity 299
347
Unit 7 Matrices

Unit 8 Co-ordinate Geometry

Unit 9 Trigonometry

Unit 10 Vectors

Unit 11 Transformations

Unit 12 Statistics

Syllabus
Specification Grid
Model Questions

Relations Relations

1

1.1 Order

The arrangement of numbers or elements in a specific manner is known as an order.
For example:
(i) 12, 22, 32, 42, 52

i.e. 1, 4, 9, 16, 25
This is an order of square of the first five natural numbers.
(ii) I, II, III, IV and V
The first five Roman numerals.
(iii) 2, 4, 6, 8, 10
The first five multiples of 2.

1.2 Pair

A set of two elements is known as pair. For example, a pair of shoes, a pair of
slippers, a pair of gloves, Ram and Sita, Shiva and Parvati, Nepal and Kathmandu,
Bangladesh and Dhaka etc. In pair of elements, the order of occurrence of elements
is not important.

1.3 Ordered Pair

Ordered pair is the combination of two numbers or elements in a fixed order in such
a way that the first element represents the x-component and the second element
represents the y-component. The elements of an ordered pair are always separated
by comma and closed between round or small brackets.

For example: (4, 8), (Shiva, Parvati), (sun, moon), etc.

(2, 5) and (5, 2) are different ordered pairs we can understant it plotting there points
on a graph paper.

vedanta Excel in Additional Mathematics - Book 7 7

Relations

For example :
The following are some ordered pairs which represent the nations and their capital
cities.

(Nepal, Kathmandu)
(India, Delhi)
(USA, Washington D.C.)
The x-component is also called antecedent and y-component is called the
consequence.
We can also use the diagram to represent the ordered pairs.

is a capital of

Kathmandu Nepal
New Delhi India
Washington DC
USA

? (Kathmandu, Nepal) and (New Delhi, India, USA, Washington DC) are ordered
pair.

Note :

(a, b) and (b,a) are not equal ordered pairs as in (a, b) the first component is a and
the second is b and whereas in (b, a), the first components is b and the second
component is a.

1.4 Equality of ordered pair

The ordered pairs (a, b) and (p, q) are said to be equal if their corresponding elements
are equal. That is, the first component a should be equal to the first component p
and the second component b should be equal to the second component q.
Mathematically,

If (a, b) = (p, q) then
Ÿ a = p and b = q
For example: Are the ordered pairs (4, 5) and (4, 5) equal?
Yes, the ordered pairs are equal as their corresponding components are equal.

8 vedanta Excel in Additional Mathematics - Book 7

Relations

Worked Out Examples

Example 1. Find out the x-component and y-component of the following ordered
Solution: pairs:

(a) (4, 7) (b) (–3, 4)

(a) Here, (4, 7)

x-component = 4 y-component = 7

(b) Here, (–3, 4)

x-component = –3 y-component = 4

Example 2. Check which of the following ordered pairs are equal. Give reason
Solution: for your answer.

(a) (2, 3) and (2, 3)

(b) (4, 5) and (7, 8)

(a) Here, in (2, 3) and (2, 3)

The x-components; 2 = 2 (which is true)

y-components; 3 = 3 (which is true)

So, (2, 3) and (2, 3) are equal ordered pairs.

(b) Here, in (4, 5) and (7, 8)

The x-components; 4 = 7 (which is false)

y-components; 5 = 8 (which is false)

So, (4, 5) and (7, 8) are not equal ordered pairs.

Example 3. Find the values of x and y from the following equal ordered pairs.
Solution:
(a) (x, y) = (4, 5)

(b) (2x, 4) = 6, y
2
x y
(c) 2 , 3 = (3, 2)

(d) (3x – 1, 2) = (4, y + 2)

(a) Here, (x, y) = (4, 5)

Since, the ordered pairs are equal, the corresponding elements
are equal.

So, x = 4, y = 5

vedanta Excel in Additional Mathematics - Book 7 9

Relations

(b) Here, (2x, 4) = 6, y
2

Since the ordered pairs are equal, the corresponding elements

are equal.

So, 2x = 6

or, x = 6 = 3
2
y
and 4 = 2

or, y = 4 × 2 = 8

Hence, x = 3, y = 8

(c) Here, x , y = (3, 2)
2 3

Since the ordered pairs are equal the corresponding elements

are equal.

So, x = 3
2

or, x = 2 × 3 = 6

and y = 2
3

or, y = 3 × 2 = 6

Hence, x = 6 and y = 6.

(d) Here, (3x – 1, 2) = (4, y + 2)

Since the ordered pairs are equal the corresponding elements
are equal.

So, 3x – 1 = 4

or, 3x = 4 + 1

or, 3x = 5

or, x = 5
3

and 2 = y + 2

or, 2 – 2 = y

or, 0 = y

or, y = 0

Hence, x = 5 , y = 0
3

10 vedanta Excel in Additional Mathematics - Book 7

Relations

1.5 Cartesian Product

Let A and B be two non empty sets. Then the product A × B read as A cross B is
defined as the set of all ordered pairs in such a way that the first element is taken
from set A and the second element is taken from set B.

Symbolically,

A × B = {(x, y) : x A and y B }
For example: Let A = {x, y}, B = {a, b}

Then, A × B can be calculated as in the following table:

Ax B b
y a (x, b)
(x, a) (y, b)
(y, a)

Now, A × B in the set of all above ordered pairs and we write it as,

A × B = {(x, a), (x, b), (y, a), (y, b)}

Similarly, B × A can be written as,

Ba A y
b x (a, y)
(a, x) (b, y)
(b, x)

So, B × A = {(a, x), (a, y), (b, x), (b, y)}

From the example, it is also observed that

Since (x, a) ≠ (a, x), (y,a ) ≠ (a, y) (x, b) ≠ (b, x), (y, b) ≠ (b, y)

A×B≠B×A

The cartesian product can be also obtained by using an arrow diagram:
AB

xa

yb

It is an arrow diagram representing A × B.

From this arrow diagram also, we can see that
A × B = {(x, a), (x, b), (y, a), (y, b)}

vedanta Excel in Additional Mathematics - Book 7 11

Relations

Similarly for B × A, we place B first and A second and draw arrows to represent it.

So, B × A = {(a, x), (a, y), (b, x), (b, y)} A
B

ax

by
An arrow diagram representing B × A

Worked Out Examples

Example 4. If A = {a, b} and B = {x}, find A × B using table.

Solution: Here, A = {a, b} and B = {x}

Now, A × B is given by

B

x

Aa (a, x)

b (b, x)

? A × B = {(a, x), (b, x)}

Example 5. If P = {6, 7} and Q = {8, 9}, find P × Q using an arrow diagram.

Solution: Here, P = {6, 7} and Q = {8, 9}

Now, P × Q is given by Q
P

68

79

So, P × Q = {6, 7} × {8, 9} = {(6, 8), (6, 9), (7, 8), (7, 9)}

Example 6. If A = { 1, 2, 3} and B = {4, 5}, find:

(a) A × B (b) B × A

(c) Show that A × B z B × A

Solution: Here, A = {1, 2, 3} and B = {4, 5}

12 vedanta Excel in Additional Mathematics - Book 7

Relations

(a) Now, A × B is given by

A 1 B 5
2 4 (1, 5)
(1, 4) (2, 5)
3 (2, 4) (3, 5)
(3, 4)

Here, A × B = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

Similarly, B × A is given by

B4 1 A 3
5 (4, 1) 2 (4, 3)
(5, 1) (5, 3)
(4, 2)
(5, 2)

So, B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}

Again,

Since, (1, 4) ≠ (4, 1), (2, 4) ≠ (4, 2), ..., (5, 3) ≠ (3, 5)

A×BzB×A

Example 7. If P = {2, 4} find P × P. P
Solution: Here, P = {2, 4} 2

Now, P × P is given by
P
2

44

? P × P = {(2, 2), (2, 4), (4, 2), (4, 4)}

Example 8. If P × Q = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}, find the sets P and Q.
Solution: Here, P × Q = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

By the definition of Cartesian product P × Q, in ordered pair, the first
elements are from set P and the second elements are from set Q.
? P = {1, 2, 3, 4, 5} and Q = {2, 3, 4, 5, 6}

vedanta Excel in Additional Mathematics - Book 7 13

Relations

Exercise 1.1

Short Questions :

1. (a) Define an ordered pair.

(b) Define equal ordered pairs.

(c) Define cartesian product of two sets.

2. Determine x - component (antecedent) and y - component (consequence) from
the following ordered pairs:

(a) (4, 5) (b) (6, 8)

(c) (–2, 4) (d) –2, – 1
(e) (Nepal, Mt. Everest) 2

(f) (Pokhara, Phewa Lake)

3. Check which of the following ordered pairs are equal. Give reason to support
your answer:

(a) (4, 5) and (4, 5) (b) (6, 5) and (6, 5)

(c) (4, 5) and (4, – 5) (d) (3 + 4, 2 – 1) and (7, 1)

(e) 25 , 3 and 5, 9 (f) (p + q, q – p) and (q + p, p – q)
5 3

4. Find the values of x and y from the following equal ordered pairs:

(a) (x, y) = (4, 5) (b) (x, – 5) = (– 6, – y)

(c) (5x, 2y) = (10, 10) (d) x , 7y = (2, 14)
(e) (10x + 1, 2y – 2) = (11, – 6) 7

(f) (x, 3) = (4, y – 1)

(g) (x – 4, y + 6) = (3, 6) (h) x – 7, y = (4, 5)
(i) (x – 5, 9) = (4x – 5, y + 3) 2

5. Using table find A × B if

(a) A = {2, 3} and B = {4, 5} (b) A = {a} and B = {p, q}

(c) A = {6, 7} and B = {p} (d) A = {4, 6, 8} and B = {1, 2, 3}

6. Using an arrow diagram calculate B × A from the questions of Q. No 5.

7. If A = {a, b, c} and B = {p, q}. Find

(a) A × B (b) B × A

(c) Show that A × B ≠ B × A

8. If A = {d, o, g} and B = {g, o, d}, Show that A × B = B × A.

14 vedanta Excel in Additional Mathematics - Book 7

Relations

9. If A = {p, q} find A × A.

10. If A = {p, q} and B = {r, s}, find the following :

(a) A × B (b) B × A

(c) A × A (d) B × B

11. (a) If A × B = {(p, q), (r, s), (t, u), (u, v)}, find set A and set B.

(b) If P × Q = {(2, 4), (4, 16), (5, 25)}

Find (i) set P and set Q (ii) P × P

Project Work

12. Give any four examples of pair of object with relation. Write they are ordered
pair or not. Also write their uses.

1. and 2. Show to your teacher

3. In (a), (b), (d), (e) are equal ordered pairs

4. (a) x = 4, y = 5 (b) x = –6, y = 5 (c) x = 2, y = 5 (d) x = 14, y = 2

(e) x = 1, y = – 2 (f) x = 4, y = 4 (g) x = 7, y = 0 (h) x = 11, y = 10

(i) x = 0, y = 6 5.(a) {(2, 4), (2, 5), (3, 4), (3, 5)}

(b) {(a, p), (a, q)} (c) {(6, p), (7, p)}

(d) {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2)}, (6, 3), (8, 1), (8, 2), (8, 3)}

6. (a) {(4, 2), (4, 3), (5, 2), (5, 3)} (b) {(p, a), (q, a)} (c) {(p, 6), (p, 7)}

(d) {(1, 4), (1, 6), (1, 8), (2, 4), (2, 6), (2, 8), (3, 4), (3, 6), (3, 8)}

7. (a) {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}

(b) {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)} 9. {(p, p), (p, q), (q, p), (q, q)}

10. (a) {(p, r), (p, s), (q, r), (q, s)} (b) {(r, p), (r, q), (s, p), (s, q)}

(c) {(p, p), (p, q), (q, p), (q, q)} (d) {(r, r), (r, s), (s, r), (s, s)}

11. (a) A = {p, r, t, u}, B = {q, s, u, v}

(b) (i) P = {2, 4, 5}, Q = {4, 16, 25}

(ii) {(2, 2), (2, 4), (2, 5), (4, 2), (4, 4) (4, 5), (5, 2), (5, 4), (5, 5)}

vedanta Excel in Additional Mathematics - Book 7 15

Relations

1.6 Relation - Introduction

There may be various types of relations in our daily life. Let us consider some of them,
(i) Kathmandu is the capital of Nepal.
(ii) Dasharath was father of Ram.
(iii) 20 is greater than 10.
In above relations, (i) shows the relation of "the capital and country," (ii) shows the
relation of "father and son" and (iii) shows "greater than" relation.
Hence, the word relation shows an association of two objects, or people, or numbers,
and so on. Here, we focus our attention to mathematical relation.
Let A and B be any two non-empty sets. Then a relation R from A to B is a non-
empty subset of the cartesian product A × B. It is defined by

R = {(x, y) : x  A and y  B} Ž A × B
For example : Let A = {1, 2} and B = {3, 4},

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Then, let us define the following relations from A × B.
(i) R1 = {(x, y) : x + y = 5}

= {(1, 4), (2, 3)}
(ii) R2 = {(x, y) : x < y}

= {1, 3), (1, 4), (2, 3), (2, 4)}
(iii) R3 = {(x, y) : x > y} = I
In all of above relations R1, R2 and R3, all of them are subsets of A × B.
Hence a relation is a subset of cartesian product A × B.

Worked Out Examples

Example 1. If A = {1, 2, 3} and B = {1, 3, 5}, find the set of ordered pairs in the
Solution: relation "is greater than" from A to B.
Here, A = {1, 2, 3} and B = {1, 3, 5}
Then,
A × B = {1, 2, 3} × {1, 3, 5}

= {(1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)}
Let R1 = the set of the ordered pair whose first component is greater

than the second components

16 vedanta Excel in Additional Mathematics - Book 7

Relations

i.e. R1 = {(x, y) : x > y}
= {(2, 1), (3, 1)}

1.7 Ways of Representation of a Relation

There are various ways in which a relation can be expressed. They are as follows:

(a) Roster form (b) Set builder form

(c) Arrow or mapping diagram (d) Tabulation method

(e) Graphical method

Example 2. If A = {1, 4, 9, 16}, B = {1, 2, 3, 4} and a relation R : A o B is defined
by "is square of", represent the relation by the following ways:

(a) Roster method (b) Arrow diagram method

(c) Tabulation (d) Graphical method

Solution: Here, A = {1, 4, 9, 16} and B = {1, 2, 3, 4},

A × B = {1, 4, 9, 16} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (4, 1), (4, 2), (4, 3), (4, 4),

(9, 1), (9,2), (9,3), (9,4), (16, 1), (16, 2), (16, 3), (16, 4)}

Now, R : A o B i.e. R is a relation from A to B.

Let us represent above relations in the following ways

(a) Roster method

R = {(x, y) : x is square of y}

= {(x, y) : x = y2}

= {(1, 1), (4, 2), (9, 3), (16, 4)}

(b) Arrow diagram method
R

11
42
93
16 4

R{(x, y) : x = y2} 17

vedanta Excel in Additional Mathematics - Book 7

Relations

(c) Tabulation method

x149 16
y123 4

(d) Graphical method

Here, R = {(1, 1), (4, 2), (9, 3), (16, 4)}

Let us show each point of R in a graph :
Y

(9, 3) (16, 4)

(4, 2)
(1, 1)

X' O X

1.8 Y'

Domain and Range of a Relation

Let R : A o B be a relation from A to B.

Suppose R = {(1, 1), (2, 4), (3, 9), (4, 16)}

In an arrow diagram, we can show R as in A R B
diagram given below :

Then the set all the first component of R is 1 1
called domain. 2 4

? Domain of R = {1, 2, 3, 4} 3 9

Its elements are called pre-images. 4 16

The set of all the second components of R is
called range.

? Range of R = {1, 4, 9, 16}

Its elements are also called images.

1.9 Type of Relations

Let A and B be two non-empty sets. Then, a relation R from A to B may be any one
of the following types.

(a) One to one relation
A relation R from set A to set B is called one to one if different element set A are
related to different elements of set B.

18 vedanta Excel in Additional Mathematics - Book 7

Relations

Example : Let R = A o B be defined by

R = {(1, 5), (2, 6), (3, 7) A R B
Here, R is a one to one relation because different 1 5
element of set A has different image in set B. 2 6
3 7
(b) Many to one relation :

A relation R from set A set B is called many to one

relation if two or more elements A are associated with unique element of set B.

R = {(a, x), (b, x), (c, x)}

R B
A x
a
b
c

Hence, R is called many to one relation as three elements of A are associated
with unique element x of A.

(c) One to many relation

A relation R : A o B is called one to many relation if at least one element of set
A is associated with two or more elements of set B.

Here, R = {(a, x), (a, y), (b, y), (b, z)} A R B
a x
"a of A is associated with two elements x and y of B y
and b is associated with y and z of B". Hence, R is b z
called one to many relation.

1.10 Inverse Relation

Let R = {(a, x), (b, y), (c, z)}, then interchange the element of each of the ordered
pair, we get a new relation R–1 = {(x, a), (y, b), (z, c)}.

Thus new relation R–1 is called inverse relation of R.

If R = {(x, y) : x  A and y  B}

then, R–1 = {(y, x) : y  B and x  A}

R B R–1 B
A x A a
a y x b
b z y c
c z

vedanta Excel in Additional Mathematics - Book 7 19

Relations

Example 3. If R = {(1, 2), (2, 3), (3, 4), (4, 5)}, find the inverse relation of R.

Solution: Here, R = {(1, 2), (2, 3), (3, 4), (4, 5)}
Inverse relation of R is given by

R–1 = {(2, 1), (3, 2), (4, 3), (5, 4)}

Exercise 1.2

Short Questions :
1. (a) Define a relation.

(b) State any five ways of representation of a relation.
(c) Define one to one relation with an example.
2. Write the domain and range of the following relations:
(a) R1 = {(1, 4), (2, 5), (3, 6), (4, 7)}
(b) R2 = {(1, 1), (2, 8), (3, 27), (4, 64)}
(c) R3 = {(m, a), (n, b), (p, c)}
(d) R4 = {(4, 8), (5, 10), (6, 12), (7, 14)}
3. Find the inverse relation of the following relations:
(a) R1 = {(a, p), (b, q), (c, r), (d, t)}
(b) R2 = {(1, 4), (2, 5), (3, 6), (5, 8)}
(c) R3 = {(10, 5), (8, 4), (6, 3), (2, 1)}
(d) g = {(1, 1), (4, 2), (9, 3), (16, 4)}
Long Questions :
4. Find the domain, range and inverse of relation the following relations.

(a) R (b) R

ap 1 2
bq 2 4
cr 3 6
4 8

5. Let A = {1, 2, 3} and B = {4, 5, 6}, find A × B, find the relation from A to B
determined by the conditions.

(a) x < y (b) x + y = 5

(c) x > y (d) y = 2x

20 vedanta Excel in Additional Mathematics - Book 7

Relations

6. If A = {1, 2, 3} and B = {1, 8, 27} and a relation R : A o B is defined by "cube
of". Represent R by the following wags:

(a) Set of ordered pairs

(b) Arrow diagram

(c) Tabulation method R

7. If R is a relation from P to Q as shown in the 1 1
adjoining diagram, then, 2 4
3 9
(a) Find the relation R in set of ordered pair 4 16

(b) Express R in tabular form

(c) Express R in the rule method

2. (a) D = {1, 2, 3, 4}, R = {4, 5, 6, 7} (b) D = {1, 2, 3, 4}, R = {1, 8, 27, 64}

(c) D = {m, n, p}, R = {a, b, c} (d) D = {4, 5, 6, 7}, R = {8, 10, 12, 14}

3. (a) {(p, a), (q, b), (r, c), (t, d)} (b) {(4, 1), (5, 2), (6, 3), (8, 5)}

(c) {(5, 10), (4, 8), (3, 6), (2, 1)} (d) {(1, 1), (2, 4), (3, 9), (4, 16)}

4. (a) D = {a, b, c}, R = {p, q, r}, R–1 = {(p, a), (q, b), (r, c)}

(b) D = {1, 2, 3, 4}, R = {2, 4, 6, 8}, R–1 = {(2, 1), (4, 2), (6, 3), (8, 4)}

5. A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(a) {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(b) {(1, 4)} (c) { } (d) {(1, 2), (2, 4), (3, 6)}

6. (a) {(1, 1), (2, 8), (3, 27)}

(b) x 1 2 3
y 1 8 27

7. (a) R = {(1, 1), (2, 4), (3, 9), (4, 16)}

(b) x 1 2 3 4 (c) R = {(x, y) : y = x2}
y 1 4 9 16

vedanta Excel in Additional Mathematics - Book 7 21

Number System and Surds 2

Number System and
Surds

2.0 Introduction

Mathematics is the science of numbers, quantity, and space. So, studying
mathematics should start from the number system. Here in this level, we study
about real numbers. Let’s observe the chart given below.

Real Numbers

Rational Numbers Irrational Numbers

Integers Fractions

Negative Integers Zero Positive integers (Natural numbers)

2.1 Natural Numbers

123456789

From the above number line, we have a set of numbers and all are positive. We move
towards the right side making such numbers, the process never ends if we continue
it. The numbers so marked are called natural numbers and denoted by ‘N’. Hence,
the natural numbers are 1, 2, 3, 4, 5, .............
(a) The first natural number is 1 which is also the least natural number.
(b) The next natural number can be obtained by adding 1 to any preceding number.
(c) Natural numbers are also called counting numbers.
(d) Natural numbers are also called positive integers.

22 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

2.2 Zero

The number 0 (zero) is the smallest non-negative integer. The natural numbers
following 0 is 1. The number 0 is neither positive nor negative. It is usually displayed
as the central number in a number line. Zero is a number which quantifies a
non-count or an amount of null size.

2.3 Whole Numbers (W)

The set of all the natural numbers with zero is known as whole numbers. It is denoted
by W. Hence whole numbers W = {0, 1, 2, 3, .......}
(a) The first and the least whole number is 0.
(b) The next whole number can be obtained by adding 1 to any whole number.

2.4 Integers (Z)

We know natural numbers are : 1, 2, 3, 4, 5, ............ and
negative of natural numbers are : –1, – 2, – 3, –4, –5, ................

Therefore, the set of all positive natural numbers and negative of natural numbers
including of zero is known as integers. It is denoted by Z.
So, Z = {......., – 3, – 2, – 1, 0 , 1, 2, 3, .......}

Integers

Negative Integers Zero Positive Integers

Let us define the different types of integers

Note:

1. Positive Integers : (Z+)

The set of all positive natural numbers are called positive integers.

It is denoted by Z+
? Z+ = {1, 2, 3, 4, ..........................}

Positive integers are natural numbers.

2. Negative Integers : (Z–)
The set of all negative of natural numbers are called negative integers.

It is denoted by Z–.

? Z– = {– 1, – 2, – 3, ..............}

vedanta Excel in Additional Mathematics - Book 7 23

Number System and Surds

3. Zero

Zero does not have a positive or negative value. However, zero is considered
as a whole number. Zero is commonly accepted as one of counting numbers.
Hence, it is also considered as natural number.

2.5 Rational Number (Q)

A rational number is a number that can be expressed as a fraction in the form of
rpqa,tiwonhaelrenubmotbhetrhceannnuomtebreatzoerroa.nd denominator are integers, the denominator in a

The decimal of rational numbers are terminating or recurring. That is the terminating
or recurring decimal numbers are rational numbers.

Example: 1 , 4 , 7 , 2, 7, etc.
2 5 4

Note:

The decimal numbers of rational members are either terminating or recurring.

Any rational number can be written in a ratio. When the real numbers are
divided, the decimals are either terminating or repeating.

For example : (a) 3 (b) 7 (c) 8 (d) 22
2 8 3 7
3
Solution: (a) Here, 2

2 3 1.5

–2

10

– 10

×

? 3 = 1.5 which is terminating decimal.
2
7
(b) Here, 8

8 70 0.875

– 64

60

– 56

40

– 40

×

? 7 = 0.875 which is also terminating decimal.
8

24 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

(c) Here, 8
3

3 8 2.666

–6

20

– 18

20

– 18

20

– 18

2

? 8 = 2.666 .............
3

which is repeated non-terminating decimal number

(d) Here, 22
7

7 22 3.142857...

– 21

10

–7

30

– 28

20

– 14

60

– 56

40

– 35

50

– 49

10

? 22 = 3.142857 .........
7

It is a terminating and repeated decimal number where 142857

repeats after the decimal. So, it is a rational number.

vedanta Excel in Additional Mathematics - Book 7 25

Number System and Surds

Irrational Numbers (Q)

The numbers which cannot be expressed in the form p where p and q both are
q
integers and q z o are called irrational numbers.

For example : 7, 5, 5 11, 7 + 2, 2 1 3 , S, ...........
+
The decimals of irrational numbers are neither terminating nor repeating.

In an Irrational Number na Radicand

Let n a be an irrational number. Then,
Order / Degree

Radical sign

In 5, the order of 5 is 2.
In 3 2, the order of 3 2 is 3.

2.6 Surds

When we cannot simplify a number to remove a square root (or cube root etc.), it is
a surd. Surds are used to write irrational numbers.

For example : 7

Here, 7 is the rational number. When square root of 7 is taken out it becomes
irrational. Hence, 7 is a surd.

2.7 Operation of Surds

Let us do the basic operations addition, subtraction, multiplication and division of
irrational numbers.

Addition or Sum

We can add irrational numbers if they have the same order, and the same radicand.

For example : (a) 5 + 2 5 = 3 5 [ a + 2a = 3a]

Here, 5 is added with 2 5 as the radicand is 5 as well as the order is 2.

Again, (b) 7 + 3 = 7 + 3 [ a + b = a + b]

Here, we cannot add 7 with 3 as the radicand are different even though the order
is same.

Similarly, (c) 3 7 + 3 = 3 7 + 3 7 are different although they

Here, addition is not possible as the order of 3 7 and
have same radicand.

26 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

Subtraction or Difference

We subtract irrational numbers if they have the same order and the same radicand
as in addition.

For example : (a) 3 7 – 7 = 2 7 ( 3a – a = 2a)

(b) 7 – 3, it cannot be subtracted as they have different

radicands.
(c) 3 5 – 5, it cannot be subtracted as they have different

orders although they have the same radicand.

Multiplication

Multiplication is possible with irrational numbers if the order of irrational number
is same product of a and b is a × b = ab
For example: (a) 5 × 7

= 5×7
= 35
Here, the order of the both surds is 2, multiplication is possible. As the order is
same, the radicands are multiplied together to give the result.
(b) 3 5 × 3 3
= 35×3
= 3 15
(c) 2 × 3 5

This cannot be multiplied as the orders are different.

(d) 3 × 3 = 3 × 3 = 9 = 3

Division

We can divide the irrational numbers if they have the same order.
If a and b are real numbers.

a = a
b b

For example:

(a) 27 ÷ 3= 27 = 32.3 = 3 3 = 3
3 3 3

vedanta Excel in Additional Mathematics - Book 7 27

Number System and Surds

(b) 5 32 ÷ 5 125 = 5 32
5 125

= 5 25 = 2
125 125
5

(c) 37 ÷ 2 = 3 7
2
Here, the operation is not possible as the orders are different.

2.8 Square Root and Cube Root of Rational and
Irrational Number

The square root and cube root of irrational numbers are not possible, but the square
numbers have square root and cube numbers have cube root in rational numbers.
If possible, we try to take out some factors of square root or cube root of irrational
numbers.

For example : (a) 25 (b) 125 (c) 3 125 (d) 3 343

(a) 25 = 5×5

= 52

= 5 which is a rational number.

Here, 25 can be written as, the product of 5 and 5. We usually reduce the
number in their primes. We have 5 twos. So, square root of it is 5.

(b) 125 = 5 × 5 × 5

= 5×5× 5

= 52 × 5

= 5 5 which is an irrational numbers.

Here, 125 can be written as the product of 5, 5 and 5. We then can split it as
shown in the example as 5 × 5 × 5. Hence, 5 × 5 = 5 and 5 remains as it
is.

(c) 3 125 = 35×5×5
= 3 53

= 5 (Rational)

Here, 125 is written as the product of 5, 5 and 5. As this problem is based on
cube root we require a triplet inside the root to get a number.

It 3 125 = 3 53 = 5.

? 3 125 = 5

28 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

(d) 3 343 = 37×7×7
= 3 73
= 7 which is a rational number.

Here, 343 can be written as 7 × 7 × 7.
? 3 343 = 7

Worked Out Examples

Example 1. Determine whether the following are terminating or repeating
Solution: decimals:

(a) 3 (b) 22
4 7
3
(a) Here, 4

4 30 0.75

– 28

20

– 20

×

Here, the quotient is 0.75. It is terminating after 2 decimals.

(b) Here, 22
7

7 22 3.142857...

– 21

10

–7

30

– 28

20

– 14

60

– 56

40

– 35

50

– 49

10

vedanta Excel in Additional Mathematics - Book 7 29

Number System and Surds

? 22 = 3.142857 .........
7

It is a terminating and repeated decimal number in which

142857 repeats after the decimal. So, it is a rational number.

Example 2. Simplify:
Solution:
(a) 5 3 + 3 3 (b) 7 7 + 4 7+ 12 7

(c) 8 5 – 3 5 + 4 5 (d) 15 2 – 7 2 + 2 2 – 6 2

(a) 5 3 + 3 3 = (5 + 3) 3

=8 3

(b) 7 7 + 4 7 + 12 7 = (7 + 4 + 12) 7

= 23 7

(c) 8 5 – 3 5 + 4 5 = (8 – 3 + 4) 5

= (12 – 3) 5

=9 5

(d) 15 2 – 7 2 + 2 2 – 6 2 = (15 – 7 + 2 – 6) 2

= (17 – 13) 2

=4 2

Example 3. Multiply : (b) 3 3 and 2 5
Solution: (a) 8 5 × 2 5 = 16 5 × 5
(a) Here, 8 5 × 2 5 = 16 × 52
= 16 × 5
(b) Here, 3 3 × 2 5 = 80
=3 3×2 5
=6 3×5
= 6 15

Example 4. Divide : (b) 3 15 ÷ 2 5
(a) 6 3 by 3
= 6 3
Solution: (a) 6 3 by 3 3

=6 × 3 = 6
3

30 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

(b) 3 15 ÷ 2 5 = 3 15
2 5

= 3 × 15
2 5

= 3 × 15
2 5

= 3 3
2

Example 5. Reduce the following surds into the simplest form:
Solution:
(a) 81 (b) 27 (c) 3 729 (d) 4 10000

(a) 81 = 3 × 3 × 3 × 3

= 32 × 32

=3×3

=9

(b) 27 = 3 × 3 × 3

= 32 × 3

=3 3

(c) 3 729 = 3 3 × 3 × 3 × 3 × 3 × 3

= 3 33 × 33

=3×3

=9

(d) 4 10000 = 4 10 × 10 × 10 × 10

= 4 104

= 10

Example 6. Simplify the following: (b) (3 + 7) (3 – 7)
Solution: (a) (3 + 5) (2 + 3) = (3 + 5) (2 + 3)
(a) (3 + 5) (2 + 3) = 3(2 + 3) + 5(2 + 3)
=6+3 3 +2 5 + 3×5
(b) (3 + 7) (3 – 7) = 6 + 3 3 + 2 5 + 15
= (3 + 7) (3 – 7)
= (3)2 – ( 7)2
=9–7
=2

vedanta Excel in Additional Mathematics - Book 7 31

Number System and Surds

Exercise 2.1

Short Questions :

1. (a) Define a rational number with two examples.

(b) Define an irrational number with two examples.

(c) Define integers.

(d) Define whole numbers with examples.

(e) Define a real number system with chart.

(f) Define natural numbers with examples.

2. Determine whether the following are terminating or repeating decimals:

(a) 1 (b) 16
4 45

(c) 7 (d) 12
2 7

3. Simplify :

(a) 2 5 + 4 5 (b) 4 3 + 10 3

(c) 5 7 – 2 7 (d) 15 3 – 8 3

(e) 8 3 + 5 3 + 4 3 – 5 5 (f) 13 5 – 6 3 + 12 3 – 8 5

(g) 11 + 4 7 – 8 3 + 3 3 – 2 11 + 3 7

4. Multiply :

(a) 2 5 × 3 3 (b) 6 5 × 3 125

(c) 5 7 × 2 3 (d) 2 7 × 2 5

5. Divide :

(a) 4 3 by 2 3 (b) 6 10 by 2 5

(c) 3 81 by 3 3 (d) 5 64 by 5 2

6. Reduce the following surds into the simplest surds:

(a) 27 (b) 128

(c) 32 (d) 3 125
(e) 3 2401 (f) 3 4096

7. Simplify :

(a) (5 + 3) (3 + 3) (b) ( 7 + 3) ( 7 – 3)

(c) ( 7 + 5)2 (d) ( 7 – 3)2

32 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

1. Show to your teacher 2.(a) terminating (b) repeating

(c) terminating (d) non-terminating repeated

3, (a) 6 5 (b) 14 3 (c) 3 7 (d) 7 3

(e) 17 3 – 5 5 (f) 5 5 + 6 3 (g) 7 7 – 11 – 5 3

4. (a) 6 15 (b) 450 (c) 10 21 (d) 4 35

5. (a) 2 (b) 3 2 (c) 3 (d) 2

6. (a) 3 3 (b) 8 2 (c) 4 2 (d) 5

(e) 737 (f) 16

7. (a) 18 +8 3 (b) 4 (c) 12 + 2 35 (d) 10 – 2 21

2.9 Rationalizing Factor

Two irrationals are said to be rationalizing factors of each other if the product of
these two irrationals is a rational number.

For example:

(a) 5 5 and 2 5 are rationalizing factors of each other as

5 5 × 2 5 = 10 5 × 5

= 10 × 52

= 10 × 5

= 50, which is a rational number.

Hence, 2 5 is a rationalizing factor of 5 5.

(b) 5 5 × 5 =5 5×5

= 5 52

=5×5

= 25 which is a rational number.

Hence, 5 is a simplest rationalizing factor.

Note :

There may be more than one rationalizing factor of an irrational factor. We try to
find the simplest rationalizing factor.

vedanta Excel in Additional Mathematics - Book 7 33

Number System and Surds

2.10 Rationalizing Factor for the Surd of the Form a – b

For the given surd of the form a – b, we multiply it with another surd a + b so
that it turns into rational number. Then a + b and a – b are called rationalizing
factors of each other. They are called conjugate of each other.

Here, ( a – b) ( a + b) = ( a)2 – ( b)2

= a – b (which is a rational number)

For example:

(a) ( x – y) × ( x + y) = ( x – y) ( x + y)

= ( x)2 – ( y)2

= x – y (which is rational)

(b) Find the rationalizing factor of 5 – 2.

5 + 2 is the rationalizing factor of 5 – 2 as,

( 5 – 2) × ( 5 + 2) = ( 5)2 – (2)2

=5–4

2.11 = 1 (which is rational)

Rationalization of Denominator

The process of multiplying a given surd by its rationalizing factor to get a rational
number as a product is called rationalization of a given surd.

In this case, we try to rationalize the denominator of given surds by multiplying the
rationalizing factor of the denominator.

For example : Rationalize the denominator (a) 1 (ii) 1
3 5–1
1
(a) 3

Here, the simplest rationalizing factor of 3 is 3.

So, multiplying the numerator and denominator by 3, we get

1 × 3 = 3
3 3 3×3

= 3
9

= 3
32

= 3
3

Hence, the denominator 3 is a rational number.

34 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

(b) 1
5–1
Here, the rationalizing factor for 5 – 1 is its conjugate is 5 + 1.

So, multiplying the numerator and denominator by 5 + 1, we get,

1 1 × 5+1 =( 5+1
5– 5+1 5)2 – 1

= 5+1
5–1

= 5+1
4

As 4 is a rational number, the denominator is rationalized.

Example 7. Write the simplest rationalizing factor of :
Solution:
(a) 12 (b) 3 125

(a) 12 = 2 × 2 × 3

= 22 × 3

=2 3

Now, if we multiply 2 3 by 3 we will get a rational number.

So, 3 is the simplest rationalizing factor of 12 as,

2 3× 3 =2 3× 3

= 2 32

=2×3

= 6 which is a rational number.

(b) 3 125 =3 5×5×5

= 3 52 × 5

=3×5 5

= 15 5

The simplest rationalizing factor of 3 125 is 5 as,

15 5 × 5 = 15 5 × 5

= 15 52

= 15 × 5

= 75 which is a rational number.

vedanta Excel in Additional Mathematics - Book 7 35

Number System and Surds

Example 8. Rationalize the denominator:
Solution:
(a) 2 (b) 1
5 7+1
2
(a) 5

Here, denominator = 5

rationalizing factor = 5

Multiplying the numerator and denominator by 5, we get,

2 × 5 = 25
5 5 ( 5)2

= 25
5

Hence, the denominator is rationalized.

(b) 1
7+1

Here, denominator = 7 + 1

rationalizing factor = 7 – 1

Multiplying the numerator and denominator by 7 – 1, we get

7 1 1 × 7–1 =( 7–1
+ 7–1 7)2 – 12

= 7–1
7–1

= 7–1
6

Hence, the denominator in rationalized.

Exercise 2.2

Short Questions :

1. Simplify :

(a) ( x – y) ( x + y) (b) ( m – n) ( m + n)

(c) ( 3 – 2) ( 3 + 2) (d) ( 5 – 3) ( 5 + 3)

2. Write the simplest rationalizing factor of-

(a) 28 (b) 75 (c) 10 2 (d) 3 98

3. Rationalise the denominators of the following irrational numbers.

(a) 3 (b) 4 (c) 3 (d) 35
3 5 23 27

36 vedanta Excel in Additional Mathematics - Book 7

Number System and Surds

(e) 7 (f) 1 (g) 2 (h) 7
55 3+1 7–1 5+ 3

(i) 53 (j) 22
2 2– 3 2 2+2 3

Long Questions :

4. Simplify :

(a) 6 2 + 2 2 – 3 2 + 32 (b) 4 12 + 5 3 – 7 3 + 75
(d) 9 32 – 6 50 + 2 2
(c) 4 20 + 3 45 – 5 5 (f) 6 35 ÷ 3 28

(e) 3 20 – 2 45 + 80 – 5

(g) 9 24 ÷ 18

5. Simplify :

(a) 1 2+ 1 2 (b) 1 3+ 1 3
3– 3+ 5– 5+

(c) 2 3+ 3 3 (d) 8 3+ 16 3
5– 6– 11 – 11 +

Project Work

6. Write a short note on 'Real Number' system. (Include chart, definitions,
applications)

1. (a) x – y (b) m – n (c) 1 (d) 2

2. (a) 7 (b) 5 (c) 2 (d) 2

3. (a) 3 (b) 45 (c) 3 (d) 3 35
5 2 14
35 3–1 ( 7 + 1) 7( 5– 3)
(e) 25 (f) 2 (g) 3 (h) 2

(i) 2 6 + 3 (j) 6 – 2

4. (a) 9 2 (b) 11 3 (c) 12 5 (d) 8 2

(e) 3 5 (f) 5 (g) 6 3

5. (a) 2 3 + 5 + 6 (b) 2 5 (c) 2 3 + 6 + 5 (d) 3 11 – 3

vedanta Excel in Additional Mathematics - Book 7 37

Trigonometry

3Trigonometry

3.0 Introduction

The word trigonometry is derived from Greek word 'trigonon' which means triangle
and 'metron' means measure. i.e. 'trigonometry = trigonon + metron'. Hence
trigonometry refers to the measurement of triangles. Nowadays, trigonometry is
used in Engineering, Astronomy, Geology, Survey, etc.

3.1 Triangles

A geometrical, closed figure bounded by three sides with three angles is known as
triangle.
Triangles are primarily divided into two types:

(a) On the basis of sides.
(b) On the basis of angles.

3.2 Types of Triangles on the Basis of Sides

By using a scale, measure the sides of 'PQR in different three figure (i), (ii) and (iii)
and tabulate in the given table.

P PP

QR Q R R
Fig (i) Fig (ii)
Side PR Q Fig (iii)
Fig. Side PQ Side QR
(i) Remarks
(ii)
(iii) vedanta Excel in Additional Mathematics - Book 7

38

Trigonometry

Triangles on the basis of sides

(i) Equilateral triangle : In figure (i), we have observed a triangle with all three
sides are equal. Such a triangle is known as an Equilateral triangle.

(ii) Isosceles triangle : Figure (ii) is observed as a triangle with two equal sides.
Such triangle is known as an isosceles triangle.

(iii) Scalene triangle : Figure (iii) shows a triangle with no sides equal to each
other. Such triangle is known as scalene triangle.

3.3 Triangles on the basis of angles

Let us measure the angles of triangles given below and fill the observation chart that
follow.

P

PP

Q R Q RQ R
Fig (i)
‘PQR Fig (ii) Fig (iii)
Fig.
‘QPR ‘PRQ Remarks
(i)

(ii)

(iii)

(i) Acute angled triangle

In figure (i), we have observed a triangle whose all angles are less than 90°
(Acute). Hence, a triangle whose all angles are acute angles is known as acute
angled triangle.

(ii) Obtuse angled triangle

In figure (ii), it is seen that one of the angle is more than 90° (obtuse). So, such
triangle having one angle obtuse is called obtuse angled triangle.

(iii)Right angled triangle

In figure (iii), it is observed that one angle of the triangle is 90°. So, the triangle
whose one angle is 90° is known as a right angled triangle.

vedanta Excel in Additional Mathematics - Book 7 39

Trigonometry

3.4 Different Components of a Right Angled Triangle

As we have seen that a triangle having one angle 90° is known as P
a right angled triangle at Q = 90°.

In the figure alongside, ‘PQR is a right angled triangle, right
angle at Q.

Reference Angle QR

An angle that is taken into consideration in a right angled triangle P
is known as reference angle. Let ‘PRQ be the reference angle.

Now, ‘P is known as remaining angle. h
bR
So, the side in front of right angle is known as hypotenuse. It is p

denoted by h. Q
? PR = hypotenuse (h). It is the longest side in the triangle.

The side in front of reference angle in known as perpendicular. It is
denoted by p. Therefore, PQ = perpendicular (p), the remaining side except
hypotenuse and perpendicular is known as base. It is denoted by b. In figure,
QR = base (b).

In trigonometry, we represent the reference angles by various Greek alphabets.
A list of most commonly used Greek alphabets with their pronunciation is given
below:

S.N. Pronunciation Symbol
(i) Alpha D
(ii) Beta E
(iii) Gamma J
(iv) Theta T
(v) Pi S
(vi) Phi I

P

For example: h p
Here, ‘PQR is a right angled triangle, right angled at R. R

Let, ‘PQR = T be the reference angle. T
Q
b
Then, side PQ = hypotenuse (h) [ It is opposite to right angle]

side QR = base (b) [ It is side opposite to remaining angle]

side PR = perpendicular (b) [ It is opposite to reference angle]

40 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

Worked Out Examples

Example 1. Find hypotenuse (h), perpendicular (p), and base (b) from the given
right angled triangles:

(a) A (b) P

D E

C B Q R
A
Solution: (a) Here, 'ABC is a right angled triangle where, B

‘ABC = 90° and ‘ACB = D (reference angle) R

? Side AB = perpendicular (p)

Side AC = Hypotenuse (h) D
Side CB = base (b)
C

(b) Here, 'PQR is a right angled triangle right angle at P.

i.e., ‘QPR = 90° P

‘PRQ = E (reference angle)

? Side PQ = perpendicular (p)

Side QR = Hypotenuse (h) E

Side PR = base (b) Q

3.5 Pythagoras Theorem

Pythagoras theorem is a relation among the sides of a right angled triangle. This
theorem is established by a famous Greek mathematician, Pythagoras.

Pythagoras theorem is a relation developed on the basis of the elements of right
angled triangle, i.e., perpendicular, base and hypotenuse. It is established by a
famous mathematician named Pythagoras. To honour him, this relation is known
as Pythagoras relation / theorem.

Let’s do the following activity to generalize what pythagoras theorem states:

vedanta Excel in Additional Mathematics - Book 7 41

Trigonometry

Graphical work for Pythagoras Theorem.

1. In the figure given below, squares are constructed on the sides of the triangles.
By counting the number of squares presents in the square paper note down in
the box given below and express your result.

PP P P
QR QR
Q RQ R
Fig (iii) Fig (iv)
Fig (i) Fig (ii)

Figure Square Square on Square on QR2 + PR2 PQ2 Remarks
on QR PR PQ

(i)

(ii)

(iii)

(iv)

Conclusion of the above activity:

Pythagoras theorem states that the sum of squares of the perpendicular and base of
a right angled triangle is equal to the square of the hypotenuse.

In simpler language, the square made on the hypotenuse of a right angled triangle
is always equal to the sum of squares on the remaining two sides.

Mathematically, h2 = p2 + b2

Corollary p2 = h2 – b2 Ÿ p = h2 – b2

b2 = h2 – p2 Ÿ b = h2 – p2

42 vedanta Excel in Additional Mathematics - Book 7

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Example 2. Find the unknown side from the given right angled triangles:

(a) P (b) A

3 cm 13cm 5cm
3 cm 5cm
Solution: Q 4 cm R
BC

(a) Here, PQR is a right angled triangle right angled at Q.

So, PQ = 3 cm P

QR = 4 cm

PR = ?

We know, in right angled PQR,

h2 = p2 + b2 Q 4 cm R
So, PR2 = PQ2 + QR2 A

or, PR2 = 32 + 42 C

or, PR2 = 9 + 16

or, PR2 = 25

or, PR2 = 52

? PR = 5 cm

Hence, the unknown side PR is 5 cm

(b) Here, 'ABC is a right angled triangle, 13cm
right angled at C.

So, AB = 13 cm

AC = 5 cm B
BC = ?

We know, in right angled 'ABC,

h2 = p2 + b2

or, AB2 = AC2 + BC2

or, 132 = 52 + BC2

or, 169 = 25 + BC2

or, 169 – 25 = BC2

or, 144 = BC2

or, BC2 = 122

? BC = 12 cm

Hence, the unknown side BC = 12 cm.

vedanta Excel in Additional Mathematics - Book 7 43

Trigonometry

3.6 Converse of Pythagoras Theorem

It states that in any triangle if h2 = p2 + b2. Then, the triangle is called right
angled, where h is the longest side of the triangle and p and b are the remaining
sides.

As we have observed that in a right angled triangle, h2 = p2 + b2 which is known
as Pythagoras theorem. On the contrary, the converse of Pythagoras relation
holds true i.e. if h2 = p2 + b2 holds true then the triangle is right angled. If the
relation is false, the triangle is not right angled triangle.

Example 3. Check whether the given triangles are right angled.

(a) C (b) P 4 cm Q

6 cm 10 cm 6 cm 4 cm

B 8 cm A R

Solution: (a) Here, AC = 10 cm

BC = 6 cm

AB = 8 cm
Let the longest side be hypotenuses,
i.e., AC = h and BC = p, AB = b

102 = 62 + 82
or, 100 = 36 + 64

or, 100 = 100, which is true.

Hence, 'ABC is a right angled triangle.

(b) Here, let the longest side PR be hypotenuse,

i.e. PR = h = 6 cm

Also, let PQ = p = 4 cm

and QR = b = 4 cm

Now, h2 = p2 + b2

or, 62 = 42 + 42

or, 36 = 16 + 16

or, 36 = 32 [which is false]

? 'PQR is not a right angled triangle.

44 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

Exercise 3.1

Short Questions :
1. Determine perpendicular, base, and hypotenuse in the triangles given below for

the given angle of reference:
(a) P (b) B C

R T Q D
P A
(c) Q
(d) A
D

E B C
R

2. Find the length of the unknown side in the following triangles.

(a) P (b) A

10 cm 13 cm

R 8 cm Q

B C

12 cm

(c) P (d) M

2 cm 10 cm

Q 2 cm R NP
2 2 cm

vedanta Excel in Additional Mathematics - Book 7 45

Trigonometry (f) X
Z
(e) P 4 cm TR 25 cm
Y 20 cm
3 cm
Q

2. (a) 6 cm (b) 5 cm (c) 2 cm (d) 2 cm
(e) 5 cm (f) 15 cm

3.7 Trigonometric Ratios

A right angled triangle consists of three sides, i.e. perpendicular, base, and the
hypotenuse. Six trigonometric ratios can be obtained from these three sides which
are as follows:

(a) p (b) b (c) p
h h b

(d) b (e) h (f) h
p b p

These six ratios are called basic or fundamental trigonometric ratios and they

are given certain names. Detailed information about trigonometric ratios are given

below.

Let 'ABC be a right angled triangle where ‘ABC = 90°, A

‘ACB = D be the reference angle.

Then, Side AB = perpendicular (p)

Side AC = hypotenuse (h)

Side BC = base (b) D B
C
Now, we can get six ratios

(a) p = AB .................... (i) (b) b = BC ...................... (ii)
h AC h AC

(c) p = AB ...................... (iii) (d) b = BC ...................... (iv)
b BC p AB

(e) h = AC ...................... (v) (f) h = AC ...................... (vi)
b BC p AB

46 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

Now, lets introduce the names for these ratios.

S.N. Ratio Name of ratios Abbreviation
1. sine D sin D
2. p cosine D cos D
3. h tangent D tan D
4. cotangent D cot D
5. b secant D sec D
6 h cosecant D cosec/csc D

p
b

b
p

h
b

h
p

Now, from the above ratios we get,

To remember trigonometric ratios of sin, cos, and tan, we can make a sentence
"Pandit Badri Prasad Hari Hari Bola".

Now, in the earlier statement if we try to link these ratios, we get.

sin D = p = AB cos D = b = BC
h AC h AC

tan D = p = AB cosec D = h = AC
b BC p AB

sec D = h = AC cot D = b = BC
b BC p AB

Worked Out Examples

Example 1. In 'DEF given alongside find, all trigonometric ratios for reference
Solution:
angle D. D

Here, in 'DEF,

‘DEF = 90° D

‘EDF = D be the reference angle.

Then, Side EF = perpendicular (p)

Side DF = hypotenuse (h)

Side DE = base (b) E F

Now, sin D = p = EF cos D = b = DE
h DF h DF
p EF h DF
tan D = b = DE cosec D = p = EF

vedanta Excel in Additional Mathematics - Book 7 47

Trigonometry

sec D = h = DF cot D = b = DE
b DE p EF

Example 2. Find all six trigonometric ratios for the adjoining figure with respect
Solution:
to D and T: A
Example 3.
Solution: Here, 'ABC is a right angled triangle, D
‘ABC = 90°

Taking ‘ACB = T as the reference angle,

Perpendicular (p) = AB T
Base (b) = BC
BC

Hypotenuse (h) = AC

So, sin T = p = AB cos T = b = BC
h AC h AC

tan T = p = AB cosec T = h = AB
b BC p AC

sec T = h = AC cot T = b = BC
b BC p AB

Now,

Taking ‘BAC= D as the reference angle,

Perpendicular (p) = BC

Base (b) = AB

Hypotenuse (h) = AC

So, sin D = p = BC cos D = b = AB
h AC h AC

tan D = p = BC cot D = b = AB
b AB p BC

sec D = h = AC cosec D = h = AC
b AB p BC

In the right angled PQR,

(i) Find PQ. R
Q
(ii) Find all trigonometric ratios for reference angle T.

Here, in right angled 'PQR,

‘PQR = 90° 7 cm

‘RPQ = T be the reference angle 25 cm

So, Perpendicular (p) = QR = 7 cm

Hypotenuse (h) = PR = 25 cm T
Base (b) = PQ = ? P

48 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

We know from Pythagoras theorem

h2 = p2 + b2

or, (PR)2 = (QR)2 + (PQ)2

or, 252 = 72 + (PQ)2

or, 625 = 49 + (PQ)2

or, 625 – 49 = (PQ)2

or, (PQ)2 = 576

or, (PQ)2 = 242

? PQ = 24

? p = QR = 7 cm, h = PR = 25 cm and b = PQ = 24 cm

So, sin T = p = 7 cos T = b = 24
h 25 h 25

tan T = p = 7 cosec T = h = 25
b 24 p 7

sec T = h = 25 cot T = b = 24
b 24 p 7

Exercise 3.2

Short Questions :
1. Find all six trigonometric ratios with respect to the reference angles in the given

figures:

(a) A (b) P M

C T B D

(c) M N

(d) X

N TR Y OZ

vedanta Excel in Additional Mathematics - Book 7 49

Trigonometry F (f) U

(e) D

E V DW

E

2. Find the six trigonometric ratios with respect the given reference angles in the
figures:

(a) P (b) M 12 N
D
3 5
Q 5
P
13

T R
4

(c) D 2 cm (d) X 12 cm
2 cm 4 5 cm
E T Y T Z
2 cm 8 cm
F

(e) U (f) P

2 cm 8 cm

3 cm T W 4 cm T R
V 1 cm Q 4 3 cm

50 vedanta Excel in Additional Mathematics - Book 7

Trigonometry

3. From the given triangles, find the length of the missing sides and then find the
six fundamental trigonometric ratios:

(a) P (b) M 12 N
D

5

5

Q T R
4
P

4. (a) Given 'PQR is a right angled triangle. Find the following ratios:

(i) sinT P 5 R
T
(ii) cosE E
(iii) cotT 3 4

(iv) tanE Q
(v) cosecT

(vi) cotE

(b) From the given right angled triangle ABC, find the following ratios:

(i) sinD A

(ii) tanD D 2 C
(iii) cosecD 1
B E
(iv) cosE 3
(v) cotE A
(vi) secE

5. (a) In right angled triangle ABC, 6 8
C
(i) Find BC

(ii) Find all trigonometric ratio for T B T

(b) In right angled triangle PQR, find all the trigonometric ratios for the
reference angle T.

P

13

R T Q

12

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Trigonometry

1. Show to your teacher.

2. (a) sinT = 3 , cosT = 4 , tanT = 3 , cosecT = 5 , secT = 5 , cotT = 4
5 5 4 3 4 3
5 12 5 12 13 12
(b) sinD = 12 , cosD = 13 , tanD = 12 , cosecD = 5 , secD = 12 , cotD = 5

(c) sinD = 1 , cosD = 1 , tanD = 1, cosecD = 2 , secD = 2 , cotD = 1
2 2

(d) sinT = 5 , cosT = 2 , tanT = 5 , cosecT = 3 , secT = 3 , cotT = 2
3 3 2 5 2 5

(e) sinT = 3 , cosT = 1 , tanT = 3, cosecT = 2 , secT = 2 , cotT = 1
2 2 3 3

(f) sinT = 1 , cosT = 3 , tanT = 1 , cosecT = 2 , secT = 2 , cotT = 3
2 2 3 3

3. (a) PQ = 3 cm, sinT = 53, cosT = 45, tanT = 43, cosecT = 35, secT = 5 , cotT = 4
4 3
153, 1123, 152, 153, 1132, 12
(b) MP = 13 cm, sinD = cosD = tanD = cosecD = secD = cotD = 5

4. (a) (i) 4 (ii) 4 (iii) 3 (iv) 3
5 5 4 4
5 4
(v) 4 (vi) 3

(b) (i) 3 (ii) 3 (iii) 3
2 2

(iv) 3 (v) 2
5. (a) (i) BC = 10 3
4 3 4 5
(b) (i) PR = 5 (ii) sinT = 5 , cosT = 5 , tanT = 3 , cosecT = 4 ,

secT = 5 , cotT = 3
3 4
5 12 5 13
(ii) sinT = 13 , cosT = 13 , tanT = 12 , cosecT = 5 ,

secT = 13 , cotT = 12
12 5

52 vedanta Excel in Additional Mathematics - Book 7