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# Introduction To Trigonometry Class 10 Notes

## Class 10 Maths Chapter 8 Introduction to Trigonometry Notes

The notes for trigonometry class 10 Maths is provided here. In Maths, trigonometry is one of the branches, where we learn the relationships between angles and sides of a triangle. Trigonometry is derived from Greek wordsΒ  βtri’ (means three), βgonβ (means sides) and βmetronβ (means measure). In this chapter, we will learn the basics of trigonometry. Get the complete concept of trigonometry which is covered in Class 10 Maths. Also, get the various trigonometric ratios for specific angles, the relationship between trigonometric functions, trigonometry tables, various identities given here.

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

## Trigonometric Ratios

### Opposite & Adjacent Sides in a Right Angled Triangle

In the ΞABC right-angled at B, BC is the side opposite to β A, AC is the hypotenuse and AB is the side adjacent to β A.

### Trigonometric Ratios

For the right ΞABC, right-angled at β B, the trigonometric ratios of the β A are as follows:

• sin A=oppositeΒ side/hypotenuse=BC/AC
• cosec A=hypotenuse/oppositeΒ side=AC/BC

### Relation between Trigonometric Ratios

• cosec ΞΈ =1/sin ΞΈ
• sec ΞΈ = 1/cos ΞΈ
• tan ΞΈ = sin ΞΈ/cos ΞΈ
• cot ΞΈ = cos ΞΈ/sin ΞΈ=1/tan ΞΈ

Example: Suppose a right-angled triangle ABC, right-angled at B such that hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm. Also, β ACB = ΞΈ. Find the trigonometric ratios tan ΞΈ, sin ΞΈ and cos ΞΈ.

Solution: Given, in βABC,

Hypotenuse, AC = 5cm

Base, BC = 3cm

Perpendicular, AB = 4cm

Then, by the trigonometric ratios, we have;

tan ΞΈ = Perpendicular/Base = 4/3

Sin ΞΈ = Perpendicular/Hypotenuse = AB/AC = β

Cos ΞΈ = Base/Hypotenuse = BC/AC = β

To know more about Trigonometric Ratios, visit here.

### Visualization of Trigonometric Ratios Using a Unit Circle

Draw a circle of the unit radius with the origin as the centre. Consider a line segment OP joining a point P on the circle to the centre which makes an angle ΞΈ with the x-axis. Draw a perpendicular from P to the x-axis to cut it at Q.

• sin ΞΈ=PQ/OP=PQ/1=PQ
• cos ΞΈ=OQ/OP=OQ/1=OQ
• tan ΞΈ=PQ/OQ=sin ΞΈ/cos ΞΈ
• cosec ΞΈ=OP/PQ=1/PQ
• sec ΞΈ=OP/OQ=1/OQ
• cot ΞΈ=OQ/PQ=cos ΞΈ/sin ΞΈ

## Trigonometric Ratios of Specific Angles

The specific angles that are defined for trigonometric ratios are 0Β°, 30Β°, 45Β°, 60Β° and 90Β°.

### Trigonometric Ratios of 45Β°

If one of the angles of a right-angled triangle is 45Β°, then another angle will also be equal to 45Β°.

Let us say, ABC is a right-angled triangle at B, such that;

β  A = β  C = 45Β°

Thus, BC = AB = a (say)

Using Pythagoras theorem, we have;

AC2 = AB2 + BC2

= a2 + a2

= 2a2

AC = aβ2

Now, from the trigonometric ratios, we have;

• sin 45Β° = (Opp. side to angle 45Β°)/Hypotenuse = BC/AC = a/aβ2 = 1/β2
• cos 45Β° = (Adj. side to angle 45Β°)/Hypotenuse = AB/AC = a/aβ2 = 1/β2
• tan 45Β° = BC/AB = a/a = 1

Similarly,

• cosec 45Β° = 1/sin 45Β° = β2
• sec 45Β° = 1/cos 45Β° = β2
• cot 45Β° = 1/tan 45Β° = 1

### Trigonometric Ratios of 30Β° and 60Β°

Here, we will consider an equilateral triangle ABC, such that;

AB = BC = AC = 2a

β A =Β β B =Β β C = 60Β°

Now, draw a perpendicular AD from vertex A that meets BC at D

According to the congruency of the triangle, we can say;

Ξ ABD β Ξ ACD

Hence,

BD = DC

Now, in triangle ABD,Β β  BAD = 30Β° and β  ABD = 60Β°

Using Pythagoras theorem,

= (2a)2 β (a)2

= 3a2

So, the trigonometric ratios for 30-degree angle will be;

sin 30Β° = BD/AB = a/2a = 1/2

cos 30Β° = AD/AB = aβ3/2a = β3/2

tan 30Β° = BD/AD = a/aβ3 = 1/β3

Also,

cosec 30Β° = 1/sin 30 = 2

sec 30Β° = 1/cos 30 = 2/β3

cot 30Β° = 1/tan 30 = β3

Similarly, we can derive the values of trigonometric ratios for 60Β°.

• sin 60Β° =Β β3/2
• cos 60Β° = 1/2
• tan 60Β° =Β β3
• cosec 60Β° = 2/β3
• sec 60Β° = 2
• cot 60Β° = 1/β3

### Trigonometric Ratios of 0Β° and 90Β°

If ABC is a right-angled triangle at B, ifΒ β A is reduced then side AC will come near to side AB. So, ifΒ β  A is nearing to 0 degree, then AC becomes almost equal to AB and BC get almost equal to 0.

Hence, Sin A = BC /AC = 0

and cos A = AB/AC = 1

tan A = sin A/cos A = 0/1 = 0

Also,

cosec A = 1/sin A = 1/0 = not defined

sec A = 1/cos A = 1/1 = 1

cot A = 1/tan A = 1/0 = not defined

In the same way, we can find the values of trigonometric ratios for a 90-degree angle. Here, angle C is reduced to 0, and the side AB will be nearing side BC such that angle A is almost 90 degrees and AB is almost 0.

### Range of Trigonometric Ratios from 0 to 90 degrees

For 0ββ€ΞΈβ€90β,

• 0 β€ sin ΞΈ β€ 1
• 0 β€ cos ΞΈ β€ 1
• 0 β€ tan ΞΈ < β
1 β€ sec ΞΈ < β
• 0 β€ cot ΞΈ < β
• 1 β€ cosec ΞΈ < β

tan ΞΈ and sec ΞΈ are not defined atΒ  90β.

cot ΞΈ and cosec ΞΈ are not defined at 0β.

### Variation of trigonometric ratios from 0 to 90 degrees

As ΞΈ increases from 0β to 90β

• sinΒ ΞΈ increases from 0 to 1
• cosΒ ΞΈ decreases from 1 to 0
• tanΒ ΞΈ increases from 0 to β
• cosecΒ ΞΈ decreases from β to 1
• secΒ ΞΈ increases from 1 to β
• cotΒ ΞΈ decreases from β to 0

### Standard values of Trigonometric ratios

 β A 0o 30o 45o 60o 90o sin A 0 Β 1/2 Β 1/β2 Β β3/2 Β 1 cos A 1 Β β3/2 1/β2 Β 1/2 0 tan A 0 Β 1/β3 Β 1 β3 Β not defined cosec A not defined Β 2 Β β2 Β 2/β3 Β 1 sec A 1 2/β3 β2 Β 2 Β not defined cot A not defined Β β3 Β 1 1/β3 0

To know more about Trigonometric Ratios of Standard Angles, visit here.

## Trigonometric Ratios of Complementary Angles

### Complementary Trigonometric ratios

If ΞΈ is an acute angle, its complementary angle is 90ββΞΈ. The following relations hold true for trigonometric ratios of complementary angles.

• sinΒ (90Β°βΒ ΞΈ)Β =Β cosΒ ΞΈ
• cosΒ (90Β°βΒ ΞΈ)Β =Β sinΒ ΞΈ
• tanΒ (90Β°βΒ ΞΈ)Β =Β cotΒ ΞΈ
• cotΒ (90Β°βΒ ΞΈ)Β =Β tanΒ ΞΈ
• cosecΒ (90Β°βΒ ΞΈ)Β =Β secΒ ΞΈ
• secΒ (90Β°βΒ ΞΈ)Β =Β cosecΒ ΞΈ

Example: Find the value of sin65Β°/cos25Β°.

Solution: Since,

cos A = sin (90Β° – A)

cos 25Β° = sin (90Β° – 25Β°)

= sin 65Β°

Hence, sin65Β°/sin65Β° = 1

To know more about Trigonometric Ratios of Complementary Angles, visit here.

## Trigonometric Identities

The three most important trigonometric identities are:

• sin2ΞΈ+cos2ΞΈ=1
• 1+cot2ΞΈ=coesc2ΞΈ
• 1+tan2ΞΈ=sec2ΞΈ

Example:Β Prove that sec A (1 β sin A)(sec A + tan A) = 1.

Solution: We will start solving for LHS, to get RHS.

sec A (1 β sin A)(sec A + tan A) = (1/cos A)(1 – sin A)(1/cos A + sin A/cos A)

= [(1 – sin A)(1 + sin A)]/cos2 A

= [1 – sin2A]/cos2A

= (cos2A)/(cos2A)

= 1

Hence proved.

To know more about Trigonometric Identities, visit here.

## Trigonometry for Class 10 Solved Problems

Example 1:

Find Sin A and Sec A, if 15 cot A = 8.

Solution:

Given that 15 cot A = 8

Therefore, cot A = 8/15.

We know that tan A = 1/ cot A

Hence, tan A = 1/(8/15) = 15/8.

Thus, Side opposite to β A/Side Adjacent to β A = 15/8

Let BC be the side opposite to β A and AB be the side adjacent to β A and AC be the hypotenuse of the right triangle ABC respectively.

Hence, BC = 15x and AB = 8x.

Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.

(i.e) AC2 = AB2 + BC2

AC2 = (8x)2+(15x)2

AC2 = 64x2+225x2

AC2 = 289x2

AC = 17x.

Therefore, the hypotenuse AC = 17x.

Finding Sin A:

We know Sin A = Side Opposite to β A / Hypotenuse

Sin A = 15x/17x

Sin A = 15/17.

Finding Sec A:

To find Sec A, find cos A first.

Thus, cos A = Side adjacent to β A / Hypotenuse

Cos A = 8x/17x

We know that sec A = 1/cos A.

So, Sec A = 1/(8x/17x)

Sec A = 17x/8x

Sec A = 17/8.

Therefore, Sin A = 15/17 and sec A = 17/8.

Example 2:

If tan (A+ B) =β3, tan (A-B) = 1/β3, then find A and B. [Given thatΒ  0Β° <A+B β€ 90Β°; A>B ]

Solution:

Given thatΒ

Tan (A+B) = β3.

We know that tan 60 = β3.

Thus, tan (A+B) = tan 60Β° = β3.

Hence A+B= 60Β° β¦(1)

Similarly, given that,

Tan (A-B) = 1/β3.

We know that tan 30Β° = 1/β3.

Thus, tan (A-B) = tan 30Β° = 1/β3.

Hence, A-B = 30Β° β¦(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60Β° + 30Β°

2A = 90Β°

A = 45Β°.

Now, substitute A = 45Β° in equation (1), we get

45Β° +B = 60Β°

B = 60Β°- 45Β°

B = 15Β°

Hence, A = 45 and B = 15Β°.

## Video Lesson on Trigonometry

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Test your Knowledge on Trigonometry For Class 10

1. Shashank V M

NCERT

2. Shalini kumari

Very nice notes .It’s really help me

3. Yugansh

Can u answer my questions of this chapter!?

• Pramit Ranjan

What’s the question

4. found these points becoming most helpful to solve my confusion
thank you byju’s

5. SolangeloIsGOALSFIGHTME

Thank you!! This helped!

6. Piyush Singh

SinA=45Β°
CosA=35Β°
TanA=69Β°