NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 8 **Introduction to Trigonometry Class 10 NCERT Solutions Ex 8.1.**

Here you get the CBSE Class 10 Mathematics chapter 8, Introduction to Trigonometry and its Applications: We are provides free comprehensive chapter wise class 10 Mathematics notes with proper images & diagram. Here you will find all the answers to the NCERT textbook questions of Chapter 8 Introduction to Trigonometry.

- Introduction to Trigonometry Class 10 Ex 8.2
- Introduction to Trigonometry Class 10 Ex 8.3
- Introduction to Trigonometry Class 10 Ex 8.4

Board | CBSE |

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Introduction to Trigonometry |

Exercise | Ex 8.1 |

Number of Questions Solved | 11 |

Category | NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

NCERT Solutions for Class 10 Maths

Page No: 181

**Question 1.**

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A

(ii) sin C, cos C

**Solution: **In ∆ABC by applying Pythagoras theorem

AC

^{2}= AB

^{2}+ BC

^{2}

= (24)

^{2}+ (7)

^{2}

= 576 + 49

= 625

AC = √625 = 25 cm

**Question 2.** In Figure, find tan P – cot R.

**Solution: **In ∆PQR by applying Pythagoras theorem

PR

^{2}= PQ

^{2}+ QR

^{2}

(13)

^{2}= (12)

^{2}+ QR

^{2}

169 = 144 + QR

^{2}

25 = QR

^{2}

QR = 5

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

**Solution: **Let ∆ABC be a right angled triangle, right angled at point B.

Given that

sin A = 3/4

BC/AC = 3/4

Let BC be 3 K so AC will be 4 K where K ¡s a positive integer.

Now applying Pythagoras theorem in ∆ABC

AC

^{2}= AB

^{2}+ BC

^{2}

(4 K)

^{2}= AB

^{2}+ (3 K)

^{2}

16 K

^{2}– g K

^{2}= AB

^{2}

7 K

^{2}= AB

^{2}

AB = √7k

**Question 4.**

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

**Solution: **Consider a right triangle, right angled at B

Let AB be 8 K so BC will be 15 K where K is a positive integer.

Now applying Pythagoras theorem in ∆ABC

AC

^{2}= AB

^{2}+ BC

^{2}

= (8K)

^{2}+ (15K)

^{2}

= 64 K

^{2}+ 225 K

^{2}

= 289 K

^{2}

AC = 17 K

**Question 5.** Given sec θ = 13/12, calculate all other trigonometric ratios.

**Solution: **Consider a right angle triangle ∆ABC right angled at point B.

**Question 6.**

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

**Solution: **Since cos A = cos B

So AB/AC = BC/AC

AB = BC

So ∠A = ∠B (Angles opposite to equal sides are equal in length)

**Question 7.** If cot θ =7/8, evaluate :

(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right) } \)

(ii) Cot^{2}θ

**Solution: **Consider a right angle triangle ∆ABC right angled at point B.

**Question 8.** If 3cot A = 4/3 , check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A } ={ cos }^{ 2 }A-{ sin }^{ 2 }A\) or not.

**Solution: **

**Question 9.** In triangle ABC, 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

**Solution: **

**Question 10.** In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

**Solution: **

**Question 11.** State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

**Solution: **

Let AC be 12 K, AB will be 5 K, where K is a positive integer

Now applying Pythagoras theorem in ∆ABC

AC

^{2}= AB

^{2}+ BC

^{2}

(12 K)

^{2}= (5 K)

^{2}+ BC

^{2}

144 K

^{2}= 25 K

^{2}+ BC

^{2}

BC

^{2}= 119 K

^{2}

BC = 109 K

We may observe that for given two sides AC = 12 K and AB = S K

BC should be such that –

AC – AB < BC < AC + AB

12 K – 5 K < BC <12 K + 5 K

7 K < BC < 17 K

But BC = 10.9 K.

Clearly such a triangle is possible and hence such value of sec A is possible. Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.

(iv) cot A is not the product of cot and A but it is cotangent of ∠A. Hence, the given statement is false.

(v) sine θ = 4/3

We know that in a right angle triangle

In a right angle triangle hypotenuse is always greater then the remaining two sides.

Hence such value of sin θ is not possible. Hence, the given statement is false.

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