NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

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NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are given in the article. Trigonometry studies relationships between side lengths and angles of triangles. Chapter 8 Introduction to Trigonometry solutions cover key concepts such including Trigonometric ratios of an acute angle of a right-angled triangle & their proofs. Values of the trigonometric ratios of 30⁰, 40⁰, and 60⁰ and relationships between Ratios.

Class 10 Maths Chapter 8 Introduction to Trigonometry is a part of Unit 5 Trigonometry. This unit holds a weightage of 12 Marks in Class 10 Maths Examination 2022-23. NCERT Solutions of Class 10 Maths Chapter 8 are based on the following topics:

Download: NCERT Solutions for Class 12 Mathametics Chapter 8 pdf

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are provided below:

Also check: Introduction to Trigonometry

Important Topics: NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry

Important Topics of NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry are elaborated below:

Trigonometric Identities

There are 6 kinds of Trigonometric Idenities:

Sine
Cosine
Tangent
Secant
Cosecant
Cotangent

Example Prove the following identity using the trigonometric identities:

[(sin 3θ + cos 3θ)/(sin θ + cos θ)] + sin θ cos θ = 1

Solution: Let’s the following identity:

a³+b³ = (a+b)(a²-ab+b²)

Using Pythagoras Theorem,

L.H.S. = [(sin 3θ + cos 3θ)(sin θ + cos θ)] + sin θ cos θ

= [(sin θ + cos θ)(sin2θ - sin θ cos θ + cos2θ)(sin θ + cos θ) + sin θ cos θ

= (sin2θ - sin θ cos θ + cos2θ) + sin θ cos θ

= sin2θ + cos2θ = 1 = R.H.S.

Trigonometry Table

Trigonometry Table is a collection of values of trigonometric ratios for various standard angles including 0°, 30°, 45°, 60°, 90°, sometimes with other angles like 180°, 270°, and 360° included, in a tabular format.

Example: Calculate the exact value of sin15º using the trigonometric value for standard angles from a trigonometry table.

Solution: Using the trigonometric table, we know that sin45º = 1/√2, cos30º = (√3/2), cos45º = 1/√2, and sin30º = 1/2

sin15º = sin(45º - 30º) = sin45ºcos30º - cos45ºsin30º = (√2/2) • (√3/2) - (√2/2) • (1/2) = (√6 - √2)/4 = (√3 - 1)/2√2

Therefore, tThe value of sin15º = (√3 - 1)/2√2

Trigonometry Ratios

Trigonometric ratios are the “ratios of length of sides of a triangle”.

Trigonometric ratios relate the ratio of sides of a right triangle to the respective angle. Basic trigonometric ratios are sin, cos, and tan. Other important trigonometric ratios, cosec, sec, and cot, can be derived using the sin, cos, and tan respectively.

Sum, Difference, Product Trigonometric Ratios Identities

Sum, Difference, Product Trigonometric Ratios include the following formulas:

sin (A + B) = sin A cos B + cos A sin B
sin (A - B) = sin A cos B - cos A sin B
cos (A + B) = cos A cos B - sin A sin B
cos (A - B) = cos A cos B + sin A sin B
tan (A + B) = (tan A + tan B)/ (1 - tan A tan B)
tan (A - B) = (tan A - tan B)/ (1 + tan A tan B)
cot (A + B) = (cot A cot B - 1)/(cot B - cot A)
cot (A - B) = (cot A cot B + 1)/(cot B - cot A)
2 sin A⋅cos B = sin(A + B) + sin(A - B)
2 cos A⋅cos B = cos(A + B) + cos(A - B)
2 sin A⋅sin B = cos(A - B) - cos(A + B)

NCERT Solutions For Class 10 Maths Chapter 8 Exercises

The detailed solutions for all the NCERT Solutions for Introduction to Trigonometry under different exercises are as follows:

Also check:

Important Chapter Related Topics
Introduction to Trigonometry Important Questions	Introduction to Trigonometry Formula	Introduction to trigonometry Revision Notes
Trigonometry Table	Cos 0	Value of cos 60
Sin 30 Degrees	sec 90	30 60 90 formula
cosec cot formula	sec 60	Difference between Trigonometry and Geometry

Also check:

Maths Study Guides
Math MCQs	Trigonometry	Mensuration
Geometry	Algebra	Difference Between in Maths
Maths Study Material	Math Formula	Class 10 Maths Study Notes

CBSE X Related Questions

1.
Using prime factorisation, find the HCF of 144, 180 and 192.
View Solution
2.
Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).
View Solution
3.
In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]
View Solution
4.
There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.
View Solution
5.
Find the sum of first 20 terms of an A.P. whose n\(^{th}\) term is given by \(a_n = 5 + 2n\). Can 52 be a term of this A.P. ?
View Solution
6.
The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, then find the length of the corresponding side of the second triangle.
View Solution

View All

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NCERT Solutions for Class 10 Maths: Chapter-wise Solutions and Important Topics

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You can also check

NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry

Important Topics: NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry

Trigonometric Identities

Trigonometry Table

Trigonometry Ratios

NCERT Solutions For Class 10 Maths Chapter 8 Exercises

CBSE X Related Questions

1.
Using prime factorisation, find the HCF of 144, 180 and 192.

2.
Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

3.
In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

5.
Find the sum of first 20 terms of an A.P. whose n\(^{th}\) term is given by \(a_n = 5 + 2n\). Can 52 be a term of this A.P. ?

6.
The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, then find the length of the corresponding side of the second triangle.

Similar Mathematics Concepts

Comments

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NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry

Important Topics: NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry

Trigonometric Identities

Trigonometry Table

Trigonometry Ratios

NCERT Solutions For Class 10 Maths Chapter 8 Exercises

CBSE X Related Questions

1.Using prime factorisation, find the HCF of 144, 180 and 192.

2.Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

3.In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

5.Find the sum of first 20 terms of an A.P. whose n\(^{th}\) term is given by \(a_n = 5 + 2n\). Can 52 be a term of this A.P. ?

6.The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, then find the length of the corresponding side of the second triangle.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Using prime factorisation, find the HCF of 144, 180 and 192.

2.
Find length and breadth of a rectangular park whose perimeter is \(100 \, \text{m}\) and area is \(600 \, \text{m}^2\).

3.
In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

5.
Find the sum of first 20 terms of an A.P. whose n\(^{th}\) term is given by \(a_n = 5 + 2n\). Can 52 be a term of this A.P. ?

6.
The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, then find the length of the corresponding side of the second triangle.