NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Exercise 8.1 of Chapter 8 Introduction to Trigonometry are created here for your help in boosting your preparation of the class 10 maths CBSE board exam. All questions of class 10 NCERT maths exercise 8.1 Introduction to Trigonometry are solved by a step-by-step method so every student of class 10 can understand the solutions with the clearance of doubts. Here you can study NCERT solutions of other chapters also. These NCERT Solutions class 10 maths exercise 8.1 will clear all your doubts about basic Trigonometry which is required to solve the questions of higher class maths.

NCERT Solutions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Exercise 8.1- Introduction to Trigonometry

Exercise 8.2 -Introduction to Trigonometry

Exercise 8.3-Introduction to Trigonometry

Exercise 8.4-Introduction to Trigonometry

NCERT solutions of Important Questions-Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Exercise 8.1 of Chapter 8 Introduction to Trigonometry

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Q1. In ΔABC right angled at B,AB = 24 cm,BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

Ans. The given triangle is ΔABC in which AB = 24 cm, BC = 7 cm and ∠B =90°

Applying Pythagoras theorem

AC = √(AB² + BC²) =√(24² + 7²)=√(576 +49) =√(625) = 25 cm

(i)

(ii)

Q2. In the given figure, find tan P – cot R

 

 

 

 

 

 

Ans.In the given ΔPQR, we have PR = 13 cm, PQ =12 cm

Applying Pythagoras theorem

QR = √(PR² – PQ²) = √(13² – 12²) = √(169 -144) = √25 = 5

The value of tan P – cot R

tan P – cot R= 0

Q3.If sin A = 3/4, calculate cos A and tan A

Ans. We are given sin A = 3/4

Since sin A = Perpendicular/Hypotenuse

Let perpendicular = 3x and hypotenuse = 4x

Base = √[(4x)²- (3x)²] = √(16x²-9x²) =x√7

Q4.Given 15 cot A = 8,find sin A and sec A

Ans. We are given that

15 cot A = 8

Let Base is 8x and perpendicular is 15x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(8x)² + (15x)²]= √(64x² + 225x²)= √(289x²) =17x

Q5. Given sec θ = 13/12, calculate all other trigonometric ratios

Ans. We are given that

Let hypotenuse is 13x and base is 12x

Applying Pythagoras theorem

Perpendicular = √(hypotenuse² – base²) =√(13² – 12²)=√(169-144)=√25 = 5

Q6. If ∠ A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Ans. We are given that

cos A = cos B where ∠A and ∠B are acute angles, the ΔABC must be as follows

 

 

 

 

 

 

Since

cos A = cos B

AC = BC

∠A = ∠B (angles opposite to equal sides in a triangle)

You can see the video for Q1 to Q6 of exercise 8.1

Q7. If cot θ = 7/8,evaluate

(ii) cot²θ

Ans. (i) We are given cot θ = 7/8

Let the base = 7x and perpendicular = 8x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(7x)² + (8x)²]= √(49x² + 64x²)= √(113x²) =x√113

The given expression

Putting the value of sin θ and cos θ in the given expression

Applying the identity ( x + y)(x -y) = x ² – y²

(ii) The given expression is

cot²θ

Putting the value of cos θ and sin θ

Q8.If 3 cot A = 4, check whether ( 1-tan²A)/(1 +tan²A) = cos²A – sin²A or not

Ans. We are given

3 cot A = 4

Let base = 4x and perpendicular =3x

Applying pythogorus theorem

Hypotenuse = √(Base² + Perpendicular²) = √[(4x)² + (3x)²]= √(16x² + 9x²)= √(25x²) =5x

The value of tan A = p/b = 3x/4x = 3/4, cos A =b/h =4x/5x = 4/5,sin A =p/h=3x/5x = 3/5

Putting the value of tan x, cosx and of sinx in the following equation

LHS = RHS, The equation shown is OK

See the video for Q7 and Q8

Q9. In ΔABC ,right-angled at B ,if tan A = 1/√3 ,find the value of :

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Ans. We are given that tan A = 1/√3

 

Let BC =x and AB =√3 x

Applying Pythagoras theorem

AC = √(AB² + BC²) = √[(√3x)² + (x)²]= √(3x² + x²)= √(4x²) =2x

 

 

 

 

 

 

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Q10.In ΔPQR ,right angled at Q,PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P,cos P and tan P.

Ans. The given ΔPQR,in which ∠Q=90°, PR + QR = 25 cm and PQ = 5 cm

PR + QR = 25 cm

QR = 25 – PR

Applying Pythogorus theorem

PR² = QR² + PQ²

PR² = (25 – PR)² + PQ²

PR² = 625 + PR²- 50PR + 5²

50PR =650

PR = 13 cm, QR = 25 -13 = 12 cm

sin P = p/h = QR/PR = 12/13

cos P = b/h = PQ/PR = 5/13

tan P = p/h = QR/PQ = 12/5

See the video for Q9,Q10 and Q11

NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

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Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
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Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
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Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
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Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

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Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

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NCERT solutions of class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
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Chapter 8-Application of Integrals