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Introduction to Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 8 Notes - Download PDF

Introduction to Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 8 Notes - Download PDF

Updated on Apr 19, 2025 12:18 PM IST

The study of trigonometry investigates how angles together with triangle side lengths relate within right triangles. Multiple practical applications rely heavily on trigonometry for their operation, including physics together with engineering and navigation. The chapter focuses on explaining trigonometric ratios of sine, cosine, tangent and their mathematical characteristics and demonstrates how these functions solve problems related to heights and distances.

This Story also Contains
  1. NCERT Notes Class 10 Maths Chapter 8 Introduction to Trigonometry
  2. Opposite and Adjacent sides in a right-angled triangle:
  3. Trigonometry Ratios:
  4. Trigonometric Ratios of Some Specific Angles:
  5. Trigonometric Identities:
  6. Class 10 Chapter Wise Notes
  7. NCERT Exemplar Solutions for Class 10
  8. NCERT Solutions for Class 10
  9. NCERT Books and Syllabus
Introduction to Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 8 Notes - Download PDF
Introduction to Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 8 Notes - Download PDF

Learning trigonometry leads to significant progress in advanced mathematics as well as to its useful applications throughout astronomy and architecture, and the technology field. Master Maths with a clear understanding of each chapter provided through the NCERT class 10th maths notes. And, expanding your comprehension through NCERT Notes, you should study related concepts from classes 9 to 12.

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NCERT Notes Class 10 Maths Chapter 8 Introduction to Trigonometry

Trigonometry:
The word trigonometry is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the

sides and angles of a triangle. Mainly, it deals with right-angled triangles, where one angle is always 90°.

Components of Right-Angled Triangle:

In a right-angled triangle, there are three sides, which are known as follows:

  • Hypotenuse: The longest side opposite the right angle.
  • Base: The horizontal side.
  • Perpendicular: The vertical side.

These sides help define trigonometric ratios, which relate angles with sides.

Opposite and Adjacent sides in a right-angled triangle:

Here, in the diagram below, ∠CAB (or ∠A) is an acute angle. Note the position of the side BC to ∠A. It faces ∠A. We call it the side opposite to angle A. Side AC is the hypotenuse of the right triangle shown below, and the side AB is a part of ∠A. So, call it the side adjacent to angle A.

1743555027475

Trigonometry Ratios:

The ratios of right-angled triangle sides relative to its acute angles form the basis of Trigonometric Ratios. The ratios serve to express how angle measurements relate to the lengths of triangle segments.

Six Trigonometric Ratios:

The six fundamental trigonometric ratios are:

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  • Sine (sin θ) = Perpendicular / Hypotenuse
    Sin⁡θ = ph

  • Cosine (cos θ) = Base / Hypotenuse
    Cosθ = bh

  • Tangent (tan θ) = Perpendicular / Base
    Tanθ = pb

  • Cosecant (cosec θ) = Hypotenuse / Perpendicular (Reciprocal of sine)
    Cosecθ = hp

  • Secant (sec θ) = Hypotenuse / Base (Reciprocal of cosine)
    Secθ = hb

  • Cotangent (cot θ) = Base / Perpendicular (Reciprocal of tangent)
    Cotθ = bp

  • Many practical problems in distances, heights and navigation, along with physics, require solutions by using ratios. The ratios serve as principles to understand both trigonometric identities and functions.

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Many practical problems in distances, heights and navigation, along with physics, require solutions by using ratios. The ratios serve as principles to understand both trigonometric identities and functions.

Trigonometric Ratios of Some Specific Angles:

In trigonometry, certain angles commonly appear in problems, such as 0°, 30°, 45°, 60°, and 90°. Trigonometric ratios of standard angles maintain constant values, which enable smooth solutions to mathematical and practical applications.

  1. Trigonometric Ratios of 45°

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In triangle ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠A = ∠C = 45°.

So, BC = AB (sides opposite to equal angles are equal)

Now, Suppose BC = AB = a.

Then by Pythagoras' Theorem, AC2 = AB2 + BC2
a2+a2=2a2

Hence, AC = √2 a

Using the definitions of the trigonometric ratios:

Sin 45° = PerpendicularHypotenuse=ABAC=a2a=12

Cos 45° = BaseHypotenuse=BCAC=a2a=12

Tan 45° = PerpendicularBase=ABBC=aa = 1

Cosec 45° = 1Sin45°=2

Sec 45° = 1Cos45°=2

Cot 45° = 1Tan45°= 1

  1. Trigonometric Ratios of 30° and 60°

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Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, ∠A = ∠B = ∠C = 60°.

Draw the perpendicular AD from A to the side BC.

From the Triangles Theorem, ABD ≌ ACD

And, BD = DC, ∠BAD = ∠CAD = 30°

Now, let's suppose that AB = 2a.

BD = 12BC=a

And, AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2

So, AD = √3 a

Using the definitions of the trigonometric ratios:

Sin 30° = PerpendicularHypotenuse=BDAB=a2a=12

Cos 30° = BaseHypotenuse=ADAB=3a2a=32

Tan 30° = PerpendicularBase=BDAD=a3a=13

Cosec 30° = 1Sin30° = 2

Sec 30° = 1Cos30°=23

Cot 30° = 1Tan30° = √3

Same for 60°

Sin 60° = 32

Cos 60° = 12

Tan 60° = √3

Cosec 60° = 1Sin60° = 23

Sec 60° = 1Cos60° = 2

Cot 60° =1Tan60° =13

  1. Trigonometric Ratios of 0° and 90°

When the angle is 0°, the perpendicular side becomes 0, and the hypotenuse is equal to the base.

So, in this case,

Sin 0° = PerpendicularHypotenuse = 0

Cos 0° = BaseHypotenuse = 1

Tan 0° = PerpendicularBase = 0

Cosec 0° = 1Sin0° = Not defined

Sec 0° =1Cos0° = 1

Cot 0° =1Tan0° = Not defined

When the angle is 90°, the base becomes 0, and the hypotenuse is equal to the perpendicular.

So, in this case,

Sin 90° = PerpendicularHypotenuse = 1

Cos 90° = BaseHypotenuse = 0

Tan 90° = PerpendicularBase = Not defined

Cosec 90° = 1Sin90° = 1

Sec 90° =1Cos90° = Not defined

Cot 90° =1Tan90° = 0

1

Trigonometric Identities:

Trigonometric identities are fundamental equations that hold for all values of an angle. The three most important identities are:

  1. cos2 A + sin2 A = 1
  2. 1 + tan2 A = sec2 A
  3. cot2 A + 1 = cosec2 A

Proof:

  1. cos2 A + sin2 A = 1

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In ABC, right-angled at B,

AB2 + BC2 = AC2

Dividing each term by AC2,

AB2AC2+BC2AC2=AC2AC2

I.e. (ABAC)2+(BCAC)2=(ACAC)2

And, from trigonometry ratios, this becomes:

(cos A)2 + (sin A)2 =1

cos2 A + sin2 A = 1

  1. 1 + tan2 A = sec2 A

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In ABC, right-angled at B,

AB2 + BC2 = AC2

Dividing each term by AB2,

AB2AB2+BC2AB2=AC2AB2

I.e. (ABAB)2+(BCAB)2=(ACAB)2

And, from trigonometry ratios, this becomes:

1 + (tan A)2 = (sec A)2

1 + tan2 A = sec2 A

  1. cot2 A + 1 = cosec2 A

1743555400812

In ABC, right-angled at B,

AB2 + BC2 = AC2

Dividing each term by BC2,

AB2BC2+BC2BC2=AC2BC2

I.e. (ABBC)2+(BCBC)2=(ACBC)2

And, from trigonometry ratios, this becomes:

(cot A)2 + 1 = (cosec A)2

cot2 A + 1 = cosec2 A

Class 10 Chapter Wise Notes

Students must download the notes below for each chapter to ace the topics.


NCERT Exemplar Solutions for Class 10

Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.

NCERT Solutions for Class 10

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

NCERT Books and Syllabus

To learn about the NCERT books and syllabus, read the following articles and get a direct link to download them.



Frequently Asked Questions (FAQs)

1. What is trigonometry in Class 10 Maths?

The word trigonometry is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle. Mainly, it deals with right-angled triangles, where one angle is always 90°.

2. What are the basic trigonometric ratios?

The six fundamental trigonometric ratios are:

Sine (sin θ) = Perpendicular / Hypotenuse
Sinθ = ph
Cosine (cos θ) = Base / Hypotenuse
Cosθ = bh
Tangent (tan θ) = Perpendicular / Base
Tanθ = pb
Cosecant (cosec θ) = Hypotenuse / Perpendicular (Reciprocal of sine)
Cosecθ = hp
Secant (sec θ) = Hypotenuse / Base (Reciprocal of cosine)
Secθ = hb
Cotangent (cot θ) = Base / Perpendicular (Reciprocal of tangent)
Cotθ = bp

3. What are the applications of trigonometry in real life?
  • Architecture & Engineering: Individuals from the professions of Architecture and Engineering create plans for constructing buildings as well as bridges alongside roofing solutions.
  • Navigation: Navigation provides a method to determine both ship-to-ship distance and physical locations.
  • Astronomy: Calculating distances between celestial bodies.
  • Physics & Mechanics: The analysis of wave motion, together with oscillations along with forces, forms part of Physics & Mechanics.
  • Geography & Mapping: Geography & Mapping: Measuring heights of mountains and distances.
4. How to memorize trigonometric ratios easily?

Use the mnemonic:
"Some People Have Curly Brown Hair, Through Proper Brushing."

  • Sinθ =  ph
  • Cosθ = bh
  • Tanθ = pb

For reciprocals:

  • Cosecθ = 1Sinθ
  • Secθ = 1Cosθ
  • Cotθ = 1Tanθ
5. What are the important formulas in Class 10 Trigonometry?

Trigonometric Ratios:

Sine (sin θ) = Perpendicular / Hypotenuse
Sinθ = ph
Cosine (cos θ) = Base / Hypotenuse
Cosθ = bh
Tangent (tan θ) = Perpendicular / Base
Tanθ = pb
Cosecant (cosec θ) = Hypotenuse / Perpendicular (Reciprocal of sine)
Cosecθ = hp
Secant (sec θ) = Hypotenuse / Base (Reciprocal of cosine)
Secθ = hb
Cotangent (cot θ) = Base / Perpendicular (Reciprocal of tangent)
Cotθ = bp

Trigonometric Identities:

cos2 A + sin2 A = 1
1 + tan2 A = sec2 A
cot2 A + 1 = cosec2

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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