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Vedanta Excel in Opt. Maths Book 6 Final (2078)

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Published by diyomath2021, 2021-06-25 21:27:53

Vedanta Excel in Opt. Maths Book 6 Final (2078)

Vedanta Excel in Opt. Maths Book 6 Final (2078)

vEMexdcAaenlTtaiHn AEdMdAitTiIoCnSal

6Book

Authors

Hukum Pd. Dahal Piyush Raj Gosain

Editors

Tara Bdr. Magar P. L. Shah

vedanta

Vedanta Publication (P) Ltd.
jb] fGt klAns];g k|f= ln=

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
info@vedantapublication.com.np
www.vedantapublication.com.np

EvMexdcAaenlTtaiHn AEMddAiTtiIoCnSal

6Book

Authors
Hukum Pd. Dahal and 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
Price: Rs 95.00
Printed in Nepal

Published by:

Vedanta Publication (P) Ltd.
jb] fGt klAns;] g kf| = ln=

Vanasthali, Kathmandu, Nepal
+977-01-4982404, 01-4962082
info@vedantapublication.com.np
www.vedantapublication.com.np

Preface

Vedanta Excel in Additional Mathematics for class 6 is completely based on the
contemporary pedagogical teaching learning activities and methodologies. It is
an innovative and unique in the sense that the contents of the book are written
and designed to ful ill the need of integrated teaching learning approaches.

Vedanta Excel in Additional Mathematics has incorporated applied constructivism
the latest trend of learner centered teaching pedagogy. 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
experience, more than two decades, of teaching optional mathematics.

Each unit of Vedanta Excel in Additional 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 Additional Mathematics is an independent kind in its contents
as the Curriculum Development Centre (CDC) does not consider it under its
curriculum. It helps the students of class 6 to lay foundation for class 9 and 10
in compulsory mathematics and optional mathematics. My honest efforts have
been to provide all the essential matters and practice materials to the students.
It is believed that the book serves as a staircase for the students of class 6. 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.

The book aims to give an elementary knowledge of Measurement of Angles,
Trigonometry, Ordered Pair, Coordinate Geometry, Matrices. Special emphasis

has been given to all the chapters as all of them are entirely new to the students.
Questions in each exercise are catagorized into two groups - Short Questions and
Long Questions.

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 colleague 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 thankful 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 Measurement of Angles ..................................................................7-19

1.0 Review .......................................................................................................... 7
1.1 Different Systems of Measurement of Angles ............................................. 8
1.2 Conversion of degree measure into grade measure ................................. 10
1.3 Conversion of grade measure into degree measure ................................. 10
1.4 Conversion of degree measure into radian measure ................................ 11
1.5 Conversion of grade measure into radian measure .................................. 11

Unit 2 Trigonometry ................................................................................. 20-45

2.0 Introduction ............................................................................................... 20
2.1 Right angled triangle ................................................................................. 20
2.2 Perpendicular (p), base (b) on the basis of reference angle ..................... 21
2.3 Ratios of the sides of a right angled triangle ............................................ 23
2.4 Trigonometric Ratios ................................................................................. 25
2.5 Operations on Trigonometric Ratio .......................................................... 29
2.6 Relation between trigonometric ratios ..................................................... 34
2.7 Trigonometric Identity .............................................................................. 36
2.8 Pythagoras Theorem .................................................................................. 39

Unit 3 Ordered Pair ..................................................................................... 46-50

3.1 Pair ............................................................................................................. 46
3.2 Ordered Pair .............................................................................................. 46
3.3 Equal ordered pairs ................................................................................... 47

Unit 4 Coordinate Geometry ................................................................... 51-63

4.0 Introduction ............................................................................................... 51
4.1 Number lines and coordinate axes ........................................................... 51
4.2 Coordinates of a point in different quadrants .......................................... 53
4.3 Plotting points in different quadrants ...................................................... 54
4.4 Distance between two points .................................................................... 57
4.5 Distance formula ....................................................................................... 59

Unit 5 Matrices ............................................................................................. 64-74

5.1 Matrix - Introduction ................................................................................ 64
5.2 Rows and columns of a matrix ................................................................. 65
5.3 Order of a Matrix ....................................................................................... 66
5.4 Equal Matrices ........................................................................................... 67
5.5 Operations on Matrices.............................................................................. 67

Model Questions ................................................................................................... 75

Measurement of Angles

1Measurement of Angles

1.0 Review

Let us review some basic terms related to an angle.

(a) Ray A
In geometry, a ray can be defined as a part of a line
that has a fixed starting point but no end point. It O B
can extend infinitely in one direction. A

In the figure, OA is a ray.

On its way to infinity, a ray may pass through more than one point.

(b) Vertex ray 2
ray 1
The common end point of two or more rays or line
segments is called vertex.

In the figure, O is the vertex. OA and OB are two O
rays, that cut at a point O.

(c) Angle

An angle is the figure formed by two rays sharing a common end point. The
common end point is the vertex. In the above figure AOB is an angle.

(d) Arms of an angle Q
P
The two rays which form an angle are called arms of

an angle. In the figure OP and OQ are the arms of the

angle POQ. O

Note :

(i) The symbol '‘' is used to represent an angle. Thus, ‘POQ represents the
angle POQ.

(ii) An angle takes three capital letters in such a way that the letter at the middle
is always the vertex of the angle.

vedanta Excel in Additional Mathematics - Book 6 7

Measurement of Angles
For example:

Q

OP

‘POQ = ‘O

(e) Protractor
A protractor is an instrument for measuring angles. It is typically in the form of
a flat semi-circle marked with degrees along the curved edge.

Steps for measuring an angle

The following are the steps to measure an angle. Let the angle to be measured be
POQ
(i) Place the protractor in such a way that its centre point is exactly at O.
(ii) Adjust the protractor so that the 0°-180° line is along the same arm PS.

Q

SO P

(iii) Start counting from the zero which is above OP and read the number of degrees
at the point from where other arm OQ passes.

In the figure, ‘POQ = 50°

1.1 Different Systems of Measurement of Angles

There are different system of measurement of angles. P

(a) Degree Measure (Sexagesimal System)
In this system, 1 right angle is divided into 90 R
equal parts and each part is called 1 degree. The
'Degree' measurement is denoted by the symbol
(°) and written at the top and right side of the
given value of angle.

The figure alongside is a part of a protractor in
which is a right angle is divided into 90 equal

Q

8 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

parts. So, 1 right angle = 90° and 2 right angles = 180°. R

In this system of measurement of angles, the
sum of angles of a triangle is 180°.

In the figure, PQR is a triangle. Q P
‘P + ‘Q + ‘R = 180°

i.e., ‘QPR + ‘PQR + ‘PRQ = 180°

(b) Grade Measure (Centesimal System)
In this system, 1 right angle is divided into 100 R
equal parts and each part is called 1 grade. The
'Grade' measurement is denoted by (g) and written
at the top and right side of the given value of
angle.

The adjoining figure is a part of a protractor in

which a right angle is divided into 100 equal

parts. So, 1 right angle = 100g and 2 right angles Q P
= 200g .

In this system of measurement, the sum of angles of a triangle is 200g.

In the figure, PQR is a triangle. R

Sum of angle of triangle PQR = 200g

i.e., ‘P + ‘Q + ‘R = 200g

or, ‘QPR + ‘PQR + ‘PRQ = 200g

(c) Radian System (Circular Measure) Q P

An angle subtended at the centre of any circle by an arc whose length is equal

to the radius of the circle is called a radian. B

Let OA = r be the radius of the circle with centre at O. O A
Take an arc AB equal in length to the radius.

i.e. AB = OA = r

Join OB. Then, by definition of radian,

‘AOB = 1c (read as 1 radian)

In circular measure, radian is taken as the standard unit of measurement. It is

symbolized by 1c. In this system of measurement, R

the sum of angles of a triangle is Sc.

In 'PQR,

‘P + ‘Q + ‘R = Sc Q P
or, ‘QPR + ‘PQR + ‘PRQ = Sc

vedanta Excel in Additional Mathematics - Book 6 9

Measurement of Angles

1.2 Conversion of degree measure into grade measure

In degree measurement, 1 right angle = 90°

In grade measurement, 1 right angle = 100g

Thus, 90° and 100g both of them represent a right angle.

90° = 100g

1° = 100 g
90

= 10 g
9

x° = 10 × x g
9

where x is the given angle in degree and to be converted into grade.

For example: Convert 45° into grade.

we have,

1° = 10 g
9

45° = 10 × 45 g
9

? 45° = 50g

1.3 Conversion of grade measure into degree measure

In this case, 100g = 90°

1g = 90 °
100

= 9°
10

xg = 9 × x °
10

where x is the given angle in grade and asked to convert into degree.

For example: Convert 60g into degree.

we have,

1g = 9°
10
9 °
60g = 10 × 60

? 60g = 54°

10 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

1.4 Conversion of degree measure into radian measure

Since, sum of angles of a triangle = 180°

Also, sum of angles of a triangle = Sc

? 180° = Sc

i.e. 1° = Sc
180

For example : Convert 60° into radian.

We have 1° = Sc
180
Sc
Now, 60° = 180 × 60

= Sc
3

1.5 Conversion of grade measure into radian measure

Since, sum of angles of a triangle = 200g

Also, sum of angles of a triangle = Sc

? 200g = Sc

i.e. 1g = Sc
200

For example: Convert 150g into radian.

We have, 1g = Sc
200
Sc
Now, 150g = 200 × 150g

= 3Sc
4

Worked out Examples

Example 1. Convert 72° into grade measurement.
Solution:
We know that

90° = 100g

1° = 100 g= 10 g
90 9

72° = 10 × 72 g
9

= 80g

vedanta Excel in Additional Mathematics - Book 6 11

Measurement of Angles

Example 2. Convert 80g into degree measurement .
Solution:
We know that,

100g = 90°

1g = 90 °= 9°
100 10

80g = 9 × 80 °
10

= 72°

Example 3. Convert 45° into radian.
Solution:
We have, 1° = Sc
Example 4. 180
Solution: Sc
Now, 45° = 180 × 45
Example 5.
Solution: = Sc
4

Convert 120g into radian.

We have, 1g = Sc
200
Sc
Now, 120g = 200 × 120

= 3Sc
5
Sc
Convert 4 into

(a) degree measure (b) grade measure

(a) Since 1c = 180°
S
Sc 180° S
Now, 4 = S × 4

= 45°

(b) Since 1c = 200g
S
Sc 200g S
Now, 4 = S × 4

= 50g

Example 6. Find the unknown angles in the following figures.

(a) in degrees (b) in grades
R
P

x° Q

20° xg 80g M
S O P QN

12 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

Solution: (a) Here, x° + 20° = 90° (complementary angle)
or, x° = 90° – 20° (straight angle = 200g)
? x° = 70°

(b) Here, xg + 80g = 200g
or, xg = 200g – 80g
? xg = 120g

Example 7. If the degree measurement of one of the complementary angles is
Solution: 63°, find the other angle in grade measurement.

We know that,

90° = 100g

1° = 100 g= 10 g
90 9
10
63° = 9 × 63 g

= 70g

Let the other complementary angle be xg.

Now, x + 70g = 100g

or, x = 100g – 70g

? x = 30g

The required angle is 30g.

Example 8. If the grade measurement of one of the supplementary angles is 75g,
Solution: find the other angle in degree measurement.

We know that

100g = 90°

1g = 90 ° = 9°
100 10
9 °
75g = 10 × 75 = 67.5°

Let the other supplementary angle be x.

Now, x + 67.5° = 180°

or, x = 180° – 67.5°

? x = 112.5°

So, the required angle is 112.5°.

vedanta Excel in Additional Mathematics - Book 6 13

Measurement of Angles

Example 9. Find the unknown angles in the following triangle in degree:

(a) P (b) A

x
30°

Q 30° 70° R

B xC

Solution: (a) We have,
the sum of angles of a triangle = 180°
i.e. x + 30° + 70° = 180°
or, x + 100° = 180°
or, x = 180° – 100°
? x = 80°

(b) Here 'ABC is a right angled triangle.
So, ‘B = 90°
The sum of angles of a triangle = 180°
or, 90° + x + 30° = 180°
or, 120° + x = 180°
or, x = 180° – 120°
? x = 60°

Example 10. Find the unknown angles in grade in the following triangles.

(a) P (b) C
90g 70g

60g x A x
Q R B

Solution: (a) We have the sum of angles of a triangle PQR is 200g

‘Q + ‘P + ‘R = 200g

i.e. 60g + 90g + x = 200g

or, 150g + x = 200g

14 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

or, x = 200g – 150g
? x = 50g
(b) Here, 'ABC is a right angled triangle
i.e., ‘A = 100g
The sum of angles of triangle is 200g
i.e. ‘A + ‘B + ‘C = 200g
or, 100g + x + 70g = 200g
or, 170g + x = 200g
or, x = 200g – 170g
? x = 30g

Example 11. Find the each of angles of given triangles in radian:

(a) P (b) A

2x 3x

3x xR x x
Q B C

Solution: (a) Here, the sum of angles of triangle in radian is Sc

In triangle PQR,

‘P + ‘Q + ‘R = Sc

or, 2x + 3x + x = Sc

or, 6x = Sc

? x = Sc
6
Sc Sc
Now, ‘P = 2x = 2 × 6 = 3

‘Q = 3x = 3 × Sc = Sc
6 2
Sc
‘R = x = 6

(b) Here, the sum of angles of a triangle in radian is Sc.

In 'ABC,

‘A + ‘B + ‘C = Sc

or, 3x + x + x = Sc

or, 5x = Sc

vedanta Excel in Additional Mathematics - Book 6 15

Measurement of Angles

? x = Sc
5
Sc 3Sc
Now, A = 3x = 3 × 5 = 5

and, ‘B = ‘C = x = Sc
5

Exercise 1

Short Questions : In degree In grade In radian
1. Complete the table.
Sc
SN Given angle 2
(a) Right angle

(b) Two right angle 200g

or

Sum of angles of a triangle

(c) Half of right angle 1 × 90° = 45°
2

(d) 2 × right angle 2 × 90° = 36°
5 5

2. Convert the following degree into grade:

(a) 45° (b) 36° (c) 54°

(d) 9° (e) 72° (f) 81°

3. Convert the following grade measure into degree:

(a) 70g (b) 20g (c) 50g

(d) 40g (e) 60g (f) 30g

4. Convert the following degree measure into radian:

(a) 10° (b) 20° (c) 30°

(d) 45° (e) 90° (f) 36°

5. Convert the following grade measure into radian:

(a) 20g (b) 40g (c) 50g

(d) 100g (e) 75g (f) 150g

16 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

6. Convert the following radian measure into degree:

(a) Sc (b) Sc (c) Sc
2 3 5
Sc Sc 4Sc
(d) 6 (e) 8 (f) 5

7. Convert the following radian measure into grade:

(a) Sc (b) 2Sc (c) Sc
5 5 4
3Sc Sc 4Sc
(d) 5 (e) 2 (f) 5

8. Find the unknown angles in degree from the following figures:

(a) (b)
x° 45°

40° x°

(c) (d)

x° 70° x°
60°

9. Find the unknown angles in grade from the following grade:

(a) (b)
xg
60g 72g
xg

(c) (d)
150g xg xg 108g

10. (a) If the degree measurement of one of the complementary angles is 27°, find
the other angle in grade measurement.

(b) If the grade measurement of one of the complementary angles is 50g, find
the other angle in degree measurement.

(c) If the degree measurement of one of the supplementary angles is 81°, find
the other angle in grade measurement.

(d) If the grade measurement of one of the supplementary angles is 32g, find
the other angle in degree measurement.

vedanta Excel in Additional Mathematics - Book 6 17

Measurement of Angles

Long Questions :
11. Find the unknown angles in degree from the following triangle:

(a) (b)
x° 30°

55° 70° 110° x°

(c) 70° 35° (d)
50°



12. Find the unknown angles in grade from the following figure:

(a) 70g (b)
x° 80g x°

120g

20g

(c) 40g (d)
x° 70g



13. Find the measure of each angle in radian from the following figure:

(a) A (b) E
5x 2x

B 4x 3x

C F xG

(c) X (d) P
x x

Y xZ QR
vedanta Excel in Additional Mathematics - Book 6
18

Measurement of Angles

14. (a) If 3x°, 5x°, and 2x° are angles of a triangle, find the each angle in degree.
(b) If 4x°, 6x°, and 5x° are the angles of a triangle, find the each angle is degree.
(c) If 5xg, 2xg, and 3xg are the angles of a triangle, find the each angle in grade.
(d) If 3xg, 4xg, and 3xg are the angles of a triangle, find the each angles in grade.
(e) If 2xc, 3xc, and 4xc are the angles of a triangle, find each angle in radian.
(f) If 4xc, 6xc, and 10xc are the angles of a triangle, find each angle in radian.

2. (a) 50g (b) 40g (c) 60g (d) 10g

(e) 80g (f) 90g

3. (a) 63° (b) 18° (c) 45° (d) 36°

(e) 54° (f) 27°

4. (a) Sc (b) Sc (c) Sc (d) Sc
18 9 6 4
Sc Sc
(e) 2 (f) 5

5. (a) Sc (b) Sc (c) Sc (d) Sc
10 5 4 2
3Sc 3Sc
(e) 8 (f) 4

6. (a) 90° (b) 60° (c) 36° (d) 30°

(e) 22.5° (f) 144°

7. (a) 40g (b) 80g (c) 50g (d) 120g

(e) 100g (f) 160g

8. (a) 50° (b) 45° (c) 120° (d) 110°

9. (a) 40g (b) 28g (c) 50g (d) 92g

10. (a) 70g (b) 45g (c) 110g (d) 151.2°

11. (a) 55° (b) 40° (c) 75° (d) 40°

12. (a) 50g (b) 60g (c) 60g (d) 30g

13. (a) 5Sc , Sc , Sc (b) Sc , Sc , Sc (c) S4c, Sc , Sc (d) Sc , Sc , Sc
12 3 4 3 2 6 2 4 3 3 3
14. (a) 54°, 90°, 36° (b) 48°, 72°, 60° (c) 100g, 40g, 60g (d) 60g, 80g, 60g

(e) 29Sc, Sc , 4Sc (f) Sc , 3Sc , Sc
3 9 5 10 2

vedanta Excel in Additional Mathematics - Book 6 19

Measurement of Angles 2

Trigonometry

2.0 Introduction

Trigonometry is a branch of mathematics that studies relationships between side
lengths and angles of triangles.

Trigonometry is used in real life directly or indirectly. Trigonometry spreads its
applications into various fields such as architecture, survey, astronomy, physics,
engineering, etc.

(a) Trigonometry can be used to measure the height of buildings or mountains.

(b) It is used in navigation to find the distance of the shore from a point in the sea.

(c) It is used in oceanography in calculating the height of tides in oceans.

(d) It is used in finding the distance between celestial bodies.

(e) Architects use trigonometry to calculate structural load, roof slopes, ground
surface and many other aspects, including sun shading and light angles.

2.1 Right angled triangle

A triangle with one angle a right angle is the right-angled triangle. C
B
In the figure, 'ABC is a right-angled triangle and ‘B is a right angle.

C
A

Side AC is the opposite side of the right angle (‘B).

It is called the hypotenuse (h) of the right-angled triangle. AC
B

Side BC is the opposite side of ‘A.
It is called the perpendicular (p) of the right-angled triangle.

AB

20 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

C

The side BA is the opposite side of ‘C.

It is called the base (b) of the right angled triangle.

B A

2.2 Perpendicular (p), and base (b) on the basis of
reference angle

In a right angled triangle, there are two acute angles and one right angle.

Hypotenuse (h) of a right-angled triangle is always the opposite side of the right
angle. But the naming of perpendicular and base depends on the reference of two
acute angles. In this case, any one acute angle can be the reference angle.

Let us consider a right angled triangle with ‘B = 90°.

From two acute angles we can consider one angle at a time. C
This considered angle is called reference angle.
h

Let's take ‘A as the reference angle. The side opposite to ‘A is BC. p

So, BC is the perpendicular. The remaining side BA is the base or

adjacent side. Bb A
C

Let's take ‘C as the reference angle. The side opposite to ‘C is BA. h b
So, BA is the perpendicular. The remaining side BC is the base.

A pB

Exercise 2.1

Short Questions :

1. From the given right angled triangles, taking reference angle T as indicated in
the figures, identify the perpendicular (p), base (b), and hypotenuse (h). Fill in
the table.

(a) A (b) C

T T A
BC B

vedanta Excel in Additional Mathematics - Book 6 21

Measurement of Angles (d) P
(c) P

TR T R
Q
R
Q T (f) Y
(e) Q P
X T
(g) M Z
(h) X
T

N T Y Z
(i) O (j) T
T
P

QT R U V
(k) Z S
(l)

T Y T T
X R
Perpendicular (p) Base (b)
Fig. AB BC Hypotenuse (h)
(a) AC
(b)
(c)
(d)
(e)

22 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

(f)
(g)
(h)
(i)
(j)
(k)
(l)

2. From the given right angled triangle PQR, taking reference angle T and D

complete the table (‘QPR = T, ‘PRQ = D). R

Reference Perpendicular Base Hypotenuse D

angle (p) (b) (h)

T
DT

PQ

Show your teacher.

2.3 Ratios of the sides of a right angled triangle

In the figure, ABC is a right angled triangle with, ‘B = 90°. Take reference angle A.
C
Then, AC is the hypotenuse (h).

BC is the perpendicular (p). hp
AB is the base (b).

There are six ratios formed from p, b, and h.

Now, the ratio of perpendicular to hypotenuse is BC = p Ab B
AC h
AB b
The ratio of base to hypotenuse is AC = h

The ratio of perpendicular to base is BC = p
AB b
AC h
The ratio of hypotenuse to perpendicular is BC = p

The ratio of hypotenuse to base is AC = h
AB b
AB b
The ratio of base to perpendicular is BC = p

Thus, p , b , p , h , h , and b are the six ratios of the sides of the right-angled triangle
ABC. h h b p b p

They are also called six fundamental trigonometric ratios.

vedanta Excel in Additional Mathematics - Book 6 23

Measurement of Angles

Exercise 2.2

1. The reference angle marked as T. Fill the table given below with ratios :

(a) C (b) C

T

A T A B
(c) B (d)
R
R

T Q T
P PQ

Fig. p b p h h b
h h b p b p

(a) AC AB AC BC BC AB
BC BC AB AC AB AC

(b)
(c)
(d)

2. In the given right angled triangle ABC, ‘B = 90°, ‘A = T and ‘C = D. Taking
the reference angle T and D successively, complete the table: C

Reference p b p h h D
h h b p b
Angle

TT
D BA

Show your teacher.

24 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

2.4 Trigonometric Ratios

There are six ratios of the three sides of a right-angled triangle. C
tTrhigeosneormateiotrsicarreathpio,shb. p h hb, b
, b , p , and p . These six ratios are called p
B
In a right angled triangle, hypotenuse is the opposite side of the A h
right angle. Let ‘A = T. Perpendicular is the opposite side of the
reference angle and base is the adjacent side of the reference angle. T
b
The name of the six ratios are given as follows.

p = perpendicular = opposite side is called sine of reference angle T.
h hypotenuse hypotenuse

b = base = adjacent side is called cosine of reference angle T.
h hypotenuse hypotenuse

p = perpendicular = opposite side is called tangent of reference angle T.
b base adjacent side

b = base = adjacent side is called cotangent of reference angle T.
p perpendicular opposite side

h = hypotenuse = hypotenuse is called secant of reference angle T.
b base adjacent side

h = hypotenuse = hypotenuse is called cosecant of reference angle T.
p perpendicular opposite side

The above ratios can also be written in the following ways :

(i) Sine of reference angle T, written as sinT = perpendicular = p
hypotenuse h

(ii) Cosine of reference angle T, written as cosT = base = b
hypotenuse h

(iii) Tangent of reference angle T, written as tanT = perpendicular = p
base b

(iv) Cotangent of reference angle T, written as cotT = base = b
perpendicular p

(v) Secant of reference angle T, written as secT = hypotenuse = h
base b

(vi) Cosecant of reference angle T, written as cosecT = hypotenuse = h
perpendicular p

Thus,

sinT = p cosT = b tanT = p
h h b
h h b
cosecT = p secT = b cotT = p

vedanta Excel in Additional Mathematics - Book 6 25

Measurement of Angles

Here, you see that cosecT is the opposite ratio of sinT, secT is the opposite
ratio of cosT and cotT is the opposite ratio of tanT.

Worked Out Examples

Example 1. In the adjoining right-angled triangle ABC, right angled at B, find six
Solution: C
trigonometric ratios in terms of its sides.

In the given figure, ‘B is the right angle.

When ‘A is the reference angle,

AC = h, BC = p and AB = b.

Now, sinA = p = BC cosA = b = AB A B
h AC h AC

tanA = p = BC cosecA = h = AC
b AB p BC

secA = h = AC cotA = b = AB
b AB p BC

When ‘C is the reference angle, AC = h, AB = p and BC = b.

Now, sinC = p = AB cosC = b = BC
h AC h AC

tanC = p = AB cosecC = h = AC
b BC p AB

secC = h = AC cotC = b = BC
b BC p AB

Example 2. In the right-angled triangle PQR right-angled at Q, the marked acute

angle is the reference angle. Find the values of six R

trigonometric ratios.

Solution: In the given right-angled triangle PQR. 5 cm 4 cm

‘Q is the right angle and ‘P is the

reference angle.

So, PR = h = 5 cm, QR = p = 4 cm and P 3 cm Q

PQ = b = 3 cm

Now, sinP = p = QR = 4 cosP = b = PQ = 3
h PR 5 h PR 5

tanP = p = QR = 4 cosecP = h = PR = 5
b PQ 3 p QR 4

secP = h = PR = 5 cotP = b = PQ = 3
b PQ 3 p QR 4

26 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

Example 3. In a right-angled triangle XYZ right-angled at X, the refernce angle
is Y. Find the trigonometric ratios sinY, cosY and Z
tanY in terms of its sides.

Solution: In the right-angled triangle XYZ, ‘X is the right angle

and ‘Y is the reference angle.

? YZ = h, XZ = p and XY = b XY

Now, sinY = p = XZ
h YZ

cosY = b = XY
h YZ

and tanY = p = XZ
b XY

Example 4. In the figure, PQR is a right angled triangle, ‘P = 60°. Find six
Solution:
fundamental trigonometric ratios of 60°. R

In the given right angled triangle PQR,

‘P = 60°, PQ = base (b) = 1 cm 2 cm 3 cm

QR = perpendicular (p) = 3 cm 60°
and PR = hypotenuse (h) = 2 cm P 1 cm Q

Now, taking 60° as the reference angle, we get the following
trigonometric ratios

sin60° = p = 3 cm = 3
h 2 cm 2

cos60° = b = 1 cm = 1
h 2 cm 2

tan60° = p = 3 cm = 3
b 1 cm

cosec60° = h = 2 cm = 2
p 3 cm 3

sec60° = h = 2 cm = 2
b 1 cm

cot60° = b = 1 cm = 1
p 3 cm 3

vedanta Excel in Additional Mathematics - Book 6 27

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Exercise 2.3

Short Questions : h R
T p
1. In a right angled triangle PQR if hypotenuse = h, Q
b C
perpendicular = p, and base = b. Then complete the table
filling ratios of sides taking reference angle T: T B
4 cm P
Trigonometric sinT cosT tanT cosecT secT tanT P
ratios

Ratio of sides

2. In a right angle triangle ABC, let ‘A = T be the reference angle,

p = 6 cm, b = 4 cm, h = 10 cm. Then, complete the table.

sinT p = 6 = 3
h 10 5
cosT
tanT 10 cm 6 cm
cosecT
secT A
cotT

3. (a) In the right-angled triangle PQR given alongside, find all

six trigonometric ratios in terms of its sides taking ‘P as

the reference angle. Q R
E C
B
(b) In the adjoining right-angled triangle CDE, right angled at

D, find all six trigonometric ratios in terms of its sides by

taking each acute angle as reference angle separately. D 4
C
(c) In the given right-angled triangle ABC right-angled
at C, the reference angle is marked. Find the values 3

of all six trigonometric ratios. A5

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4. (a) In the figure PQR is a right angled triangle, PQ = 1 cm, P1 cm
QR = 1 cm, PR = 2 cm. Reference angle ‘R = 45°. Find 2 cm

six fundamental trigonometric ratios of 45°. 45° 1 cm
R 1 cm Q
(b) In the figure ABC is a right angled triangle,
AB = 3 cm, BC = 1 cm, AC = 2 cm. Reference C
2 cm

angle ‘A = 30°. Find six fundamental trigonometric 30°
A 3 cm B

ratios of 30°.

2. sinT = 3 , cosT = 2 , tanT = 3 , cosecT = 5 , secT = 5 , cotT = 2
5 5 2 3 2 3
QPQR, PR QR PQ PPQR , PR
3. (a) sinP = cosP = PQ , tanP = PR , cosecP = QR , secP = cotP = QR

(b) sinC = ED , cosC = DECC, tanC = ED , cosecC = EC , secC = EC , cotC = DC
EC DC ED DC ED
DC EEDC, DC EC EC ED
sinE = EC , cosE = tanE = ED , cosecE = DC , secE = ED , cotE = DC

(c) sinB = 3 , cosB = 4 , tanB = 3 , cosecB = 5 , secB = 5 , cotB = 4
5 5 4 3 4 3
1 1
4. (a) sin45° = 2 , cos45° = 2 , tan45° = 1,

cosec45° = 2, sec45° = 2, cot45° = 1

(b) sin30° = 1 , cos30° = 3 , tan30° = 1 ,
2 2 3

cosec30° = 2, sec30° = 2 , cot30° = 3
3

2.5 Operations on Trigonometric Ratio

(a) Addition and subtraction of trigonometric ratios
We can add or subtract the same trigonometric ratios as the addition and
subtraction of like terms in algebra.

vedanta Excel in Additional Mathematics - Book 6 29

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For examples :

1. 2sinT + sinT = 3sinT [Same as 2x + x = 3x]
2. 4sinT + 3sinT = 7sinT [Same as 4x + 3x = 7x]
3. 4cosT – cosT = 3cosT [Same as 4x – x = 3x]
4. 2tanT – tanT = tanT [Same as 2x – x = x]

(b) Multiplication and division of trigonometric ratios

We can multiply and divide the same trigonometric ratios as the algebraic
operations.

1. sinT × sinT = sin2T (same as x . x = x2)

2. 4sinT × 2sinT = 8sin2T (same as 4x . 2x = 8x2)

3. 4cosT × 3cosT = 12cos2T (same as 4x . 3x = 12x2)

4. 16sin2T = 2sinT same as 16x2 = 2x
8sinT 8x

5. 28tan3T = 4tan2T same as 28x3 = 4x2
7tanT 7x
Note :

1. sinT . sinT = sin2T 2. sin2T = (sinT)2

Worked Out Examples

Example 1. Add : (b) 8tanT + 2tanT
Solution: (a) sinT + 6sinT [same as x + 6x = 7x]
(a) Here, sinT + 6sinT [same as 8x + 2x = 10x]

= 7sinT [same as 10x – 7x = 3x]
(b) Here, 8tanT + 2tanT [same as 12x – 10x = 2x]

= 10tanT

Example 2. Subtract :
Solution: (a) 10sinT – 7sinT
(b) 12tanT – 10tanT
(a) Here, 10sinT – 7sinT

= 3sinT
(b) Here, 12tanT – 10tanT

= 2tanT

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Example 3. Multiply :
Solution:
(a) sinT . sin2T (b) sinT . cosT . cos2T . sin3T

(c) 4sin2T . 3sinT

(a) Here, sinT . sin2T

= sin3T [same as x . x2 = x3]

(b) Here, sinT . cosT . cos2T . sin3T

= (sinT . sin3T) (cosT . cos2T)

= sin4T . cos3T

(c) Here, 4sin2T . 3sinT

= 4 . 3 . sin2T . sinT

= 12sin3T

Example 4. Divide :
Solution:
(a) tan3T ÷ tanT (b) 27sin2T ÷ 9sinT

(a) Here, tan3T ÷ tanT

= tan3T same as x3 = x2
tanT x

= tan2T

(b) Here, 27sin2T ÷ 9sinT

= 27sin2T same as 27x2 = 3x
9sinT 9x

= 3sinT

Example 5. Simplify :

Solution: (a) 4sinT – 3sinT + 2sinT In algebra
(b) 5sinT – cosT – 4sinT + 3cosT 4x – 3x + 2x
(a) Here, 4sinT – 3sinT + 2sinT = 4x + 2x – 3x
= 6x – 3x
= 4sinT + 2sinT – 3sinT = 3x
= 6sinT – 3sinT
= 3sinT In algebra
(b) Here, 5sinT – cosT – 4sinT + 3cosT 5x – y – 4x + 3y
= 5sinT – 4sinT – cosT + 3cosT = 5x – 4x – y + 3y

= sinT + 2cosT = x + 2y

vedanta Excel in Additional Mathematics - Book 6 31

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Exercise 2.4

Short Questions :
1. Add :

S.N. In Trigonometry In Algebra
(a) 2sinA + sinA = 3sinA 2x + x = 3x
(b) 2sinB + 3sinB = 2y + 3y =
(c) 4cos2A + 5cos2A = 4y2 + 5y2 =
(d) 2tanA + 3tanA = 2z + 3z =
(e) 3sec2B + 7sec2B = 3x2 + 7x2 =
(f) 2cos3A + 4cos3A = 2x3 + 5x3 =

2. Subtract :

S.N. In Trigonometry In Algebra
(a) 3sinA – sinA 3x – x = 2x
(b) 4cosA – 2cosA 4y – 2y =
(c) 8tan2A – 3tan2A 8x2 – 3x2 =
(d) 7secA – 3secA 7x – 3x =
(e) 8cot2B – 4cot2B 8y2 – 4y2 =
(f) 9cosecA – 4cosecA 9x – 4x =

3. Multiply :

S.N. In Trigonometry In Algebra
(a) sinA × sinA = x × x = x2
(b) 2cosA × 3cosA = 2x × 3x =
(c) 3tanA × 6tanA = 3x × 6x =
(d) 3sin2A × 2sinA = 3x2 × 2x =
(e) 4secA × sec2A = 4x × x2 =
(f) 6cot2A × 2cot2A = 6x2 × 2x2 =

4. Divide :

S.N. In Trigonometry In Algebra

(a) sin3A = x3 = x2
sinA x

(b) 4cos2A = 4x2 =
cosA x

32 vedanta Excel in Additional Mathematics - Book 6

(c) 8tan3A = 8x3 = Measurement of Angles
2tanA 2x 33

(d) 16sec2A = 16x2 =
2sec2A 2x2

(e) 8cotA = 8x =
4cot2A 4x2

(f) 24cosec2A = 24x2 =
6cosec3A 6x3

5. Simplify.

(a) 6sinA – 2sinA + 3sinA

(b) 8sinB – 2sinB – 2sinB

(c) 8cosC + 3cosC – 4cosC

(d) 2tanA – 5tanA + 4tanA

(e) 2sin2B + 4sin2B – 3sin2B

(f) 6cos2A – 4cos2A – 2cos2A

6. Simplify.

(a) 3sinA + 4cosA – 2cosA + 2sinA

(b) 8sinA – 3cosA + 6cosA – 2sinA

(c) 8tanA – 2cotA – 2tanA + 3cotA

(d) 7secA + 8cosecA – 5secA – 5cosecA

(e) 8sin2A + 2cos2A – 3sin2A + 4cos2A

(f) 8tan2A – 4sec2A – 4sec2A – 6tan2 A

7. Find the products.

(a) 2sinA × 3sinA

(b) 4cosB × 3cosB

(c) 2sinA × 4sinA × sinA

(d) 2cosA × 2cosA × 3cosA

(e) 4tanA × 2tan2A

(f) 3sec2A × 4secA

8. Simplify.

(a) 8sin2A (b) 3sin3A
sinA sin2A

vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

(c) 6cosA (d) 8cos2A
2cos2A 2cos3A

(e) 8tan4B (f) 16sec4B
4tan2B 8sec3B

1. (b) 4sinB (c) 9cos2A (d) 5tanA

(e) 10sec2B (f) 6cos3A

2. (a) 2sinA (b) 2cosA (c) 5tan2A (d) 4secA

(e) 4cot2B (f) 5cosecA

3. (a) sin2A (b) 6cos2A (c) 18tan2A (d) 6sin3A

(e) 4sec3A (f) 12cot4A

4. (a) sin2A (b) 4cosA (c) 4tan2A (d) 8secA
(c) 7cosC (d) tanA
(e) 2 (f) 4
cotA cosecA
5. (a) 7sinA (b) 4sinB

(e) 3sin2B (f) 0

6. (a) 5sinA + 2cosA (b) 6sinA + 3cosA (c) 6tanA + cotA

(d) 2secA + 3cosecA (e) 5sin2A + 6cos2A (f) 2tan2A – 8sec2A

7. (a) 6sin2A (b) 12cos2B (c) 8sin3A (d) 12cos3A

(e) 8tan3A (f) 12sec3A (c) 3 (d) 4
cosA cosA
8. (a) 8sinA (b) 3sinA
(e) 2tan2B (f) 2secB

2.6 Relation between trigonometric ratios

In the given right-angled triangle ABC, right angle at A. Let's take ‘B = T as the

reference angle.

Then, AC = p, BC = h and AB = b C

Also, sinT = p cosT = b
h h
p h
tanT = b cosecT = p

secT = h cotT = b T A
b p B

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Now, sinT = p = 1 ÷ h = 1 = 1
h p h cosecT
p

cosT = b = 1 ÷ h = 1 = 1
h b h secT
b

tanT = p = 1 ÷ b = 1 = 1
b p b cotT
p

Similarly,

cosecT = h = 1 ÷ p = 1 = 1
p h p sinT
h

secT = h = 1 ÷ b = 1 = 1
b h b cosT
h

cotT = b = 1 ÷ p = 1 = 1
p b p tanT
b

Thus,

1. sinT . cosecT = 1 sinT = 1 cosecT = 1
cosecT sinT

2. cosT . secT = 1 cosT = 1 secT = 1
secT cosT

3. tanT . cotT = 1 tanT = 1 cotT = 1
cotT tanT

Furthermore, we know that sinT = p and cosT = b
h h
p
csoinsTT = h p h p
Now, b = h × b = b = tanT

h

scionsTT = b b h b
h h p p
Also, p = × = = cotT

h

Thus,

tanT = sinT and cotT = cosT
cosT sinT

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2.7 Trigonometric Identity

Let's consider, x + 3 = 5.

Now, let's replace x by putting different values.

When x = 0, 0 + 3 = 5 i.e. 3 = 5 which is false.

When x = 1, 1 + 3 = 5 i.e. 4 = 5 which is false.

When x = 2, 2 + 3 = 5 i.e. 5 = 5 which is true.

When x = 3, 3 + 3 = 5 i.e. 6 = 5 which is false.

When x = 4, 4 + 3 = 5 i.e. 7 = 5 which is false.

Thus, x + 3 = 5 is true only for the fixed value of x, i.e. x = 2.

So, x + 3 = 5 is an equation.

On the other hand, let (x + 1)2 = x2 + 2x + 1.

When x = 0, (0 + 1)2 = 02 + 2 . 0 + 1 = 1 i.e. 1 = 1 (true)

When x = 1, (1 + 1)2 = 12 + 2 . 1 + 1 = 4 i.e. 4 = 4 (true)

When x = 2, (2 + 1)2 = 22 + 2 . 2 + 1 = 9 i.e. 9 = 9 (true)

When x = 3, (3 + 1)2 = 32 + 2 . 3 + 1 = 16 i.e. 16 = 16 (true)

Thus (x + 1)2 = x2 + 2x + 1 is true for every value of x.

So, (x + 1)2 = x2 + 2x + 1 is an identify.

When trigonometric ratios are involved in an identity, it is called the trigonometric identity.

For example: tanT × cotT = 1 is an trigonometric identity.

From an equation, we find the value of the variable. But, in the case of identity, we
prove that the Left Hand Side (LHS) is equal to Right Hand Side (RHS) or RHS is
equal to LHS.

Worked out Examples

Example 1. Prove the following :
Solution:
(a) tanT . cotT = 1 (b) cosecT . sinT = 1

(a) Here, tanT . cotT = 1 Alternate Method

LHS = tanT . cotT LHS = tanT . cotT

= p . b = sinT . cosT
b p cosT sinT

=1 =1

= RHS Proved. = RHS Proved.

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(b) Here, cosecT . sinT = 1 Alternate Method

LHS = cosecT . sinT LHS = cosecT . sinT

= h . p = 1 . sinT
p h sinT

=1 =1

= RHS Proved. = RHS Proved.

Example 2. Prove that : cosT + sinT . cotT = 2cosT
Solution:
Here, LHS = cosT + sinT . cotT

= b + p . b
h h p
b b
= h + h

= 2 b
h

= 2cosT

= RHS Proved.

Alternate Method

Here, LHS = cosT + sinT . cotT

= cosT + sinT . cosT
sinT
= cosT + cosT

= 2cosT

= RHS Proved.

Exercise 2.5

1. Prove the following using trigonometric ratios of p, b and h.

(a) sinT . cosecT = 1

(b) cosT . secT = 1

(c) tanT . cotT = 1

(d) sinT = tanT
cosT

(e) scionsTT = cotT

(f) 1 = cotT
tanT

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2. Prove the following :
(a) sinT × coscT = 1
(b) cosT × secT = 1
(c) tanT × cotT = 1
(d) sinT × cotT = cosT
(e) cosT × tanT = sinT
(f) cosecT × tanT = secT
(g) secT × cotT = cosecT
(h) cosT × cosecT × tanT = 1
(i) sinT × cotT × secT = 1
(j) cosecT × cosT × tanT = 1
(k) secT × cotT × cosT × sinT = cosT

3. Prove that

(a) sinT = cosT
tanT

(b) ccoostTT = sinT

(c) cocsoetTcT = secT

(d) tsaencTT = cosecT

(e) cosesceTcT = cotT

(f) ccoostTT = cosecT

4. Prove that

(a) 2sinT + cosT . tanT = 3sinT

(b) 3cosT + sinT . cotT = 4cosT

(c) 2cosecT + secT . cotT = 3cosecT

(d) 2secT + cosecT . tanT = 3secT

(e) sinT . cotT + cosT . tanT = cosT + sinT

(f) tanT . cosecT + tanT . cotT = secT + 1

5. (a) If sinT = 4 and cosT = 53, find the value of tanT and cotT.
5

38 vedanta Excel in Additional Mathematics - Book 6

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(b) If sinT = 5 and cosT = 12 , find the value of tanT + cotT.
13 13

(c) If sinT = 3 and cosT = 4 , show that sin2T + cos2T = 1.
5 5

(d) If sinT = 4 and cosT = 3 , show that sin2T + cos2T = 1.
5 5

(e) If sinT = 12 and cosT = 5 , find the value of sec2T – tan2T.
13 13

(f) If tanT = 3 and cotT = 1 , find the value of tan2T – cot2T.
3

5. (a) 4 , 3 (b) 169 (c) 1 (e) 1 (f) 8
3 4 60 3

2.8 Pythagoras Theorem

Pythagoras theorem is a relation between hypotenuse, perpendicular and base of a
right angled triangle. Before discussing about Pythagoras theorem, let us review to
find area of squares.

XY
MN
PQ

CA CB BA
(i) (ii) (iii)

Find the areas of square ACXY, BCMN, and ABPQ, drawn the graph paper as shown
in the figure.

Obviously, area of square ACXY = 52 = 25 sq. units.

area of square BCMN = 42 = 16 sq. units.

area of square ABPQ = 32 = 9 sq. units.

Now, let us arrange these squares on right angled triangle ABC whose perpendicular
is 4 units, base 3 units, and hypotenuse 5 units as shown in the graph given below :

vedanta Excel in Additional Mathematics - Book 6 39

Measurement of Angles

X

CM

h2
Y

5 4 p2

A3 B N
b2

QP
(iv)

In right angled triangle ABC,
square on hypotenuse = h2 = 52 = 25 sq. units.
square on perpendicular = p2 = 42 = 16 sq. units.
square on base = b2 = 32 = 9 sq. units.

Now, let us check,
h2 = p2 + b2

i.e., 25 = 16 + 9
? 25 = 25 (Ture)
? h2 = p2 + b2
This is a relation between the perpendicular (p), base (b) and hypotenuse (h) of a
right angled triangle.

X

X h2 C M
Y
MC h2 Y
p2 p h h p p2

N Bb A Ab B N
b2 b2

PQ QP
(v) (vi)

Again, let us draw two right angled triangle ABC with ‘B = 90° as in the graph
given above.

h2 is the area of square ACXY along AC (h).

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p2 is the area of square BCMN along BC (p).
b2 is the area of square ABPQ along AB (b).
Now, let's count the number of smallest square units inside ACXY, ABPQ and BCMN
and write the result in the following table.

Remember that in the above graph, some of the square units are bisected by
their diagonals. So, two bisected squares make a whole square.

Fig ACXY = h2 Area of ABPQ = b2 p2 + b2 Result
(iv) 25 sq. units BCMN = p2 9 sq. units
(v) 8 sq. units 16 sq. units 4 sq. units 16 + 9 = 25 sq. units h2 = p2 + b2
(vi) 18 sq. units 4 sq. units 9 sq. units 4 + 4 = 8 sq. units h2 = p2 + b2
9 sq. units 9 + 9 = 18 sq. units h2 = p2 + b2

Thus, h2 = p2 + b2 is the relation between hypotenuse (h), perpendicular (p) and
base (b) of a right-angled triangle. This relation was discovered by Pythagoras for the
first time. So, it is universally known as Pythagoras Theorem.

Note :

(i) h2 is the area of a square made along the length of hypotenuse
(ii) p2 is the area of a square made along the length of perpendicular
(iii) b2 is the area of a square made along the length of base of a right-angled

triangle.

As we know that,
h2 = p2 + b2

Then, p2 = h2 – b2
Also, b2 = h2 – p2
Thus, if we know p and b, h2 is found by using h2 = p2 + b2.

if we know h and p, b2 is found by using b2 = h2 – p2.
if we know h and b, p2 is found by using p2 = h2 – b2.

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Worked out Examples

Example 1. Show that the triangle ABC given alongside is a right-angled triangle.
Solution:
Here, AC = 5 cm, AB = 4 cm and BC = 3 cm A

Now, AC2 = (5 cm)2 = 25 cm2
4 cm
5 cmAB2 = (4 cm)2 = 16 cm2

4 cmBC2 = (3 cm)2 = 9 cm2

Again, AB2 + BC2 = 16 cm2 + 9 cm2 C 3 cm B

= 25 cm2 whcih is AC2

? p2 + b2 = h2

So, 'ABC is a right-angled triangle right-angled at B.

Example 2. Find the unknown sides of the following right-angled triangles:

(a) C (b) R

6 cm
5 cm

AB P 8 cm Q
(a) Here, BC = h = 5 cm
Solution:
AC = p = 4 cm
AB = b = ?
We know that,
h2 = p2 + b2
or, 52 = 42 + b2
or, 25 = 16 + b2
or, b2 = 25 – 16
or, b2 = 9
or, b2 = 32
or, b = 3 cm
? b = AC = 3 cm

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(b) Here, QR = p = 6 cm

PQ = b = 8 cm

PR = h = ?
We know that

h2 = p2 + b2
or, h2 = 62 + 82
or, h2 = 36 + 64
or, h2 = 100

or, h2 = 102

or, h = 10
? h = PR = 10 cm

Example 3. Show that

(a) sin2T + cos2T = 1

(b) cosec2T = 1 + cot2T

Solution: (a) Here, sin2T + cos2T = 1

LHS = sin2T + cos2T

= p2 + b2
h2 h2

= p2 + b2 [ p2 + b2 = h2]
h2

= h2
h2

= 1 = RHS

? LHS = RHS proved.

(b) Here, cosec2T = 1 + cot2T

RHS = 1 + cot2T

=1+ b2
p

= 1 + b2
p2

= p2 + b2 [ p2 + b2 = h2]
p2

= h2
p2

vedanta Excel in Additional Mathematics - Book 6 43

Measurement of Angles

= h2
p

= cosec2T = LHS

? LHS = RHS proved.

Exercise 2.6

1. In a right angled triangle, if hypotenuse = h, perpendicular = p, and base = b.
Then, fill in the blanks by using Pythagoras theorem.

(a) h2 = p2 + .......... (b) p2 = h2 – .......... (c) b2 = ......... – p2

2. From the given table, find unknown length of sides by using Pythagoras
Theorem.

S.N. p (cm) b (cm) h (cm)
(a) 4 ........ 5
(b) 8
(c) 12 6 ........
(d) 9 ........ 13
(e) 8
12 ........
(f) 20 ........ 10

15 ........

3. Show that the following triangles are right-angled triangles.

(a) P (b) M (c) U

4 cm 10 cm 12 cm
6 cm

9 cm
6 cm
Q 3 cm R5 cm 5 cm N 8 cm P V 15 cm W
Q
4. Find the unknown sides of the following right-angled triangles.

(a) C (b) Z (c) P 15 cm

A 4 cm B X 8 cm Y 12 cm
R

44 vedanta Excel in Additional Mathematics - Book 6

Measurement of Angles

5. Prove the following :
(a) sin2T + cos2T = 1
(b) sin2T = 1 – cos2T
(c) cos2T = 1 – sin2T
(d) sec2T – tan2T = 1
(e) sec2T = 1 + tan2T
(f) tan2T = sec2T – 1
(g) cosec2T – cot2T = 1
(h) cosec2T = 1 + cot2T
(i) cot2T = cosec2T – 1
(j) (1 – cos2T)sec2T = tan2T
(k) cos2T . cot2T = sin2T

2. (a) 3 cm (b) 10 cm (c) 5 cm (d) 15 cm
(e) 6 cm (f) 25 cm (c) 9 cm
(b) 10 cm
4. (a) 3 cm

vedanta Excel in Additional Mathematics - Book 6 45

Ordered Pair

3Ordered Pair

3.1 Pair

In daily life, we use word a pair. A set of two things used together is known as a pair.
Examples: (i) a pair of gloves

(ii) a pair of sneakers
(iii) a pair of shoes
(iv) a pair of earrings
(v) a pair of scissors
(vi) a pair of beautiful blue eyes, etc.
In mathematics a pair is a set of two object or numbers written in the form of {a, b}.
i.e. a set of two objects a and b with no particular relation between them.
Example: Pair of natural numbers whose sum is 5 are {1, 4} and {2, 3}.
We can write {1, 4} = {4, 1} and {2, 3} = {3, 2}, in the sense of a pair of objects.

3.2 Ordered Pair

Let us take two pair of numbers written in the form of (1, 5) and (5, 1).

What is difference between them? Y X

Obviously, they are different pair of numbers. For 6 (1, 5)
clarity, we can plot them on a graph paper as shown 5
in the figure. 4
3
In ordered pair, the order of writing two elements 2 (5, 1)
or numbers is important. If the components of the 1
ordered pairs are interchanged, it changes the sense X'
of pair also. So, (1, 5) and (5, 1) are quite different O 123456
ordered pairs.
Y'

46 vedanta Excel in Additional Mathematics - Book 6

Ordered Pair

Definition: A pair of two objects having one object as the first and the second object
as the second element is called an ordered pair. An ordered pair having 'a' as the first
element and 'b' as the second element is denoted by (a, b). The ordered pairs (a, b)
and (b, a) are different.

3.3 Equal ordered pairs

Let us consider two ordered pairs (2, 5) and 4 , 10 . Are they equal or different
ordered pairs? 2 2

Here, we have, (2, 5) and 4 , 10 = (2, 5).
2 2

Thus above two ordered pairs are same. Such type of ordered pairs are called equal

ordered pair.

We can write, (2, 5) = 4 , 10
2 2

Again, let us consider two ordered pairs (1, 4) and (4, 1). Are they equal or different?

Here, the ordered pairs have different meanings.

Thus (1, 4) and (4, 1) are not equal ordered pairs we write (1, 4) z (4, 1) as they are
different ordered pairs.

Definition: Two ordered pairs are said to be equal of their corresponding components
are equal.

If (a, b) = (x, y), we write

a = x and b = y

Worked out Examples

Example 1. Which of the following ordered pairs are equal? Write them.

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

(b) (x, y) and (y, x)

(c) (2, 4) and (2, 4)

(d) (6, 3) and (6, 3)

(e) (5, 4) and 10 , 8
2 2

Solution: (a) Here, (2, 3) and (3, 2) are different ordered pairs.

(b) Here, (x, y) and (y, x) are different ordered pairs.

vedanta Excel in Additional Mathematics - Book 6 47

Ordered Pair

(c) Here, (2, 4) and (2, 4) are equal ordered pairs.

(d) Here, (6, 3) and (6, 3) are equal ordered pairs.

(e) Here, (5, 4) and 10 , 8 = (5, 4) are equal ordered pairs.
2 2

Example 2. Find the values of x and y from the following equal ordered pairs.

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

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

(c) x , y = (1, 2)
2 3

(d) x , y = 1 , 1
3 2 3 2

Solution: Equating the corresponding components of equal ordered pairs, we
get the values of x and y.

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

? x=4 and y = 5
(b) Here, (2x, y) = (2, 4)

? 2x = 2 and y = 4

or, x = 2
2

? x=1

(c) Here, x , y = (1, 2)
2 3

? x = 1 and y = 2
2 3

or, x = 2 or, y = 6

? x=2 and y = 6

(d) Here, x , y = 1 , 1
3 2 3 2

x = 1 and y = 1
3 3 2 2

or, x = 1 × 3 or, y = 1 × 2
3 2

? x=1 ? y=1

48 vedanta Excel in Additional Mathematics - Book 6

Ordered Pair

Exercise 3

1. Which of the following ordered pairs are equal? Write them.

(a) (6, 7) and (7, 6)

(b) (8, 9) and (8, 9)

(c) (4, 5) = 16 , 20
4 4

(d) 1 , 4 = 2, 5
2 5 4

(e) 3 , 4 = 1.5, 12
2 8

(f) (a, b) and (b, a)

2. From the given ordered pairs, write their first and second elements components
in the table.

S.N. Ordered pair First component Second component
5 10
(a) (5, 10)

(b) (2, 4)

(c) (x, y)

(d) (m, n)

(e) x , y
2 2

3. From the given equal ordered pairs, write the values of x and y in the table
below:

S.N. Equal ordered pairs Value of x Value of y

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

(b) (x, y) = (7, 8)

(c) x , y = (1, 2)
2 2

(d) x , y = (2, 3) x = 2 or, x = 8 y = 3 or, y = 9
4 3 4 3

(e) (x, y) = 4 , 6
2 3

(f) (x, 2) = (4, y)
(g) (7, y) = (x, 2)

vedanta Excel in Additional Mathematics - Book 6 49

Ordered Pair

4. Complete the given table.

SN First component Second component Ordered pair
(p, q)
(a) p q

(b) 4 5

(c) 6 8

(d) 10 4

(e) Kathmandu Nepal

(f) Sita Ram

(g) Krishna Kiran

(h) 4 64

5. Find the values of letters given in the following equal ordered pairs.

(a) (c, 10) = (5, d)

(b) (p, 10) = (4, q)

(c) (x + 1, 9) = (4, 9 + y)

(d) (4x, 1) = (8, y)

(e) (4x – 1, 2) = (7, y)

(f) (4x + 5, 7y + 8) = (2x, 15)

(g) (7x + 8, 9y + 6) = (6x, 6)

1. (a) Not equal (b) Equal (c) Equal (d) Not equal

(e) Not equal (f) Not equal

3. (a) x = 4, y = 5 (b) x = 7, y = 8 (c) x = 2, y = 4 (d) x = 8, y = 9

(e) x = 2, y = 2 (f) x = 4, y = 2 (g) x = 7, y = 2

4. (a) (p, q) (b) (4, 5) (c) (6, 8) (d) (10, 4)

(e) (Kathmandu, Nepal) (f) (Sita, Ram)

(g) (Krishna, Kiran) (h) (4, 64)

5. (a) c = 5, d = 10 (b) p = 4, q = 10 (c) x = 3, y = 0 (d) x = 2, y = 1

(e) x = 2, y = 2 (f) x = 5/2 (g) x = – 8, y = 0

50 vedanta Excel in Additional Mathematics - Book 6


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