**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**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

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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**NCERT solutions of class 11 maths**

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 |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**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 |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |