NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry

Relearn the concepts in the Maths syllabus with our NCERT Solutions for CBSE Class 10 Mathematics Chapter 8 Introduction to Trigonometry. Your trigonometry skills will be extremely useful if you aspire to have a rewarding career in the engineering stream. Practise the solutions written by our Maths experts to understand the application of the Pythagoras’ theorem to solve trigonometry problems. 

To strengthen your trigonometry basics, explore TopperLearning’s CBSE Class 10 Maths concept videos and chapter notes. Our sample papers and Maths past years’ question papers are also useful for exam preparation.

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Chapter 8 - Introduction to Trigonometry Exercise Ex. 8.1

Solution 1

In Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryABC by applying Pythagoras theorem
AC2 = AB2 + BC2
      = (24)2 + (7)2
      = 576 + 49
      = 625
AC = Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry = 25 cm

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Solution 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 5
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 6

 

Since Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryA and Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryB are acute angles, then Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryC is a right angle.

cos A = cos B .... given

AC/AB = BC/AB

AC = BC

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryB =Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To TrigonometryA           .... angles opposite to equal sides are equal in length.





Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 8
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 10
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 11

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Chapter 8 - Introduction to Trigonometry Exercise Ex. 8.2

Solution 1
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 2

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 4

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Chapter 8 - Introduction to Trigonometry Exercise Ex. 8.3

Solution 1
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 3

Given that

tan 2A = cot (A - 18)

cot (90 - 2A) = cot (A -18)
90 - 2A = A - 18
108 = 3A
A = 36

Solution 4

Given that
tan A = cot B
tan A = tan (90 - B)
A = 90 - B
A + B = 90

Solution 5

Given that
Sec 4A = cosec (A - 20)
Cosec (90 - 4A) = cosec (A - 20)
90 - 4A = A - 20
110 = 5A
A = 22

Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 7

sin 67 + cos 75
= sin (90 - 23) + cos (90 - 15)
= cos 23 + sin 15

Chapter 8 - Introduction to Trigonometry Exercise Ex. 8.4

Solution 1
We know that
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 2
We know that

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Solution 4

(i) 9sec2A - 9tan2A
= 9(sec2A - tan2A)
= 9 (1)            (as sec2A - tan2A = 1)
= 9
Hence alternative (B) is correct.

(ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Hence alternative (C) is correct.


(iii) (secA + tanA) (1 - sinA)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

(iv)

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Solution 5

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry


Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry 

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

 
Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry

 

Ncert Solutions Cbse Class 10 Mathematics Chapter - Introduction To Trigonometry