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# NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.1

Last updated date: 12th Sep 2024
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## NCERT Solutions for Maths Class 11 Chapter 3 Exercise 3.1 - FREE PDF Download

Chapter 3 Trigonometric Functions, is a crucial part of the Class 11 Maths NCERT syllabus for 2024-25. This chapter explores various aspects of trigonometric functions, including angles, degree and radian measures, and their relationships. This exercise focuses on these foundational topics, providing a solid base for understanding trigonometry.

Table of Content
1. NCERT Solutions for Maths Class 11 Chapter 3 Exercise 3.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 3 Exercise 3.1 Class 11 | Vedantu
3. Formulas Used in Class 11 Chapter 3 Exercise 3.1
4. Access NCERT Solutions for Maths Class 11 Chapter 3 - Trigonometric Functions
5. Class 11 Maths Chapter 3: Exercises Breakdown
6. CBSE Class 11 Maths Chapter 3 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

NCERT Solutions for ex 3.1 class 11  are designed to help students excel in board exams by offering clear explanations and step-by-step solutions. Practising these exercises will enhance students' understanding and problem-solving skills. Use Class 11 Maths NCERT Solutions to master the concepts and perform well in your exams.

## Glance on NCERT Solutions Maths Chapter 3 Exercise 3.1 Class 11 | Vedantu

• Exercise 3.1 Class 11 explains the basics of trigonometric functions, laying the groundwork for further exploration of angles and their measurements.

• It covers angles, which are measured in degrees or radians and are fundamental in trigonometry.

• Angles are measured in degrees or radians and are a fundamental concept in trigonometry.

• Degree measure quantifies angles by dividing a full circle into 360 equal parts, with each part being one degree.

• Radian measure defines angles by the length of the arc divided by the radius of the circle, with one full circle being 2π radians.

• Radians are related to real numbers where the circumference of a unit circle (2π) corresponds to 2π radians, linking angles to real numbers.

• Degrees and radians are related through the formula where 180 degrees equals π radians, allowing conversion between these units.

• In Trigonometric Functions there are 7 fully solved questions of Maths, Chapter 3 ex 3.1 class 11.

## Formulas Used in Class 11 Chapter 3 Exercise 3.1

• Degree to Radian Conversion: $Radians=Degrees\times \frac{\pi }{180}$

• Radian to Degree Conversion: $Degrees=Radians\times \frac{180}{\pi}$

• Length of an Arc (in radians): $Length of arc=r\theta$

• Area of a Sector (in radians): $Area of sector=\frac{1}{2}r^{2}\theta$

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 11 Chapter 3 - Trigonometric Functions

### Exercise 3.1

1: Find the radian measures corresponding to the following degree measures:

(i) 25o

Ans: As we all know that ${180^\circ } = \pi$ radian. So, ${25^\circ } = \frac{\pi }{{180}} \times 25$ radian $= \frac{{5\pi }}{{36}}$ radian.

(ii) -47o30’

Ans: As we all know that, $- {47^\circ }{30^\prime } = \,\, - 47\frac{1}{2}$

$= \frac{{ - 95}}{2}$ degree

Because of the ${180^\circ } = \pi$ radian $\frac{{ - 95}}{2}$ degree $= \frac{\pi }{{180}} \times \left( {\frac{{ - 95}}{2}} \right)\operatorname{radian} = \left( {\frac{{ - 19}}{{36 \times 2}}} \right)\pi$ radian $= \frac{{ - 19}}{{72}}\pi$ radian.

So, $- {47^\circ }{30^\prime } = \frac{{ - 19}}{{72}}\pi$ radian.

(iii) 240o

Ans: As we all know that ${180^\circ } = \pi$ radian

So, ${240^\circ } = \frac{\pi }{{180}} \times 240$ radian $= \frac{4}{3}\pi$ radian

(iv) 520o

Ans: As we all know that ${180^\circ } = \pi$ radian. So, ${520^\circ } = \frac{\pi }{{180}} \times 520$ radian $= \,\frac{{26\pi }}{9}$radian.

2. Find the degree measures corresponding to the following radian measures. (Use $\pi = \frac{{22}}{7}$)

(i) $\frac{{11}}{{16}}$

Ans: As we all know that $\pi$ radian $= {180^\circ }$. So, $\frac{{11}}{{16}}$ radian $= \frac{{180}}{\pi } \times \frac{{11}}{{16}}$ degree $= \frac{{45 \times 11}}{{\pi \times 4}}$ degree

$= \frac{{45 \times 11 \times 7}}{{22 \times 4}}$ degree $= \frac{{315}}{8}$ degree

$= 36\frac{3}{8}$ degree

$= {39^\circ } + \frac{{3 \times 60}}{8}$ minutes $\quad \left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\circ } = {{60}^\prime }} \right]$

$= {39^\circ } + {22^\prime } + \frac{1}{2}$ minutes

$= {39^\circ }{22^\prime }{30^{\prime \prime }}\quad \left[ {{1^\prime } = {{60}^{\prime \prime }}} \right]$

(ii) -4

Ans: As we all know that $\pi$ radian $= {180^\circ }$. So, $- 4$ radian $= \frac{{180}}{\pi } \times ( - 4)$ degree $= \frac{{180 \times 7( - 4)}}{{22}}$ degree

$= \frac{{ - 2520}}{{11}}$ degree $= - 229\frac{1}{{11}}$ degree

$= - {229^\circ } + \frac{{1 \times 60}}{{11}}$ minutes

$\left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\circ } = {{60}^\prime }} \right]$

$= - {229^\circ } + {5^\prime } + \frac{5}{{11}}$ minutes

$= - {229^\circ }{5^\prime }{27^{\prime \prime }}\quad \left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\prime } = {{60}^{\prime \prime }}} \right]$

(iii) $\frac{{5\pi }}{3}$

Ans: As we all know that $\pi$ radian$= {180^\circ }$. So,$\frac{{5\pi }}{3}$ radian $= \frac{{180}}{\pi } \times \frac{{5\pi }}{3}$ degree $= {300^\circ }$

(iv) $\frac{{7\pi }}{6}$

Ans: As we all know that $\pi$ radian$= {180^\circ }$. So, $\frac{{7\pi }}{6}$ radian $= \frac{{180}}{\pi } \times \frac{{7\pi }}{6} = {210^\circ }$.

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Ans: The number of revolutions mades by the wheel in one minute$= 360$

As a result, the number of revolutions made by the wheel in one second$= \frac{{360}}{{60}} = 6$

The wheel rotates at a $2\pi$radian angle in one complete revolution. As a result, it will turn an angle of $6 \times 2\pi$radian in 6 complete revolutions, i.e., $12\pi$ radian

As a result, the wheel rotates at a $12\pi$radian angle in one second.

4: Find the degree measure of the angle subtended at the centre of a circle of radius $100\;cm$by an arc of length $22\;cm$. $\left( {} \right.$ Use $\left. {\pi = \frac{{22}}{7}} \right)$

Ans: As we all know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta$ radian at the centre, then $\theta = \frac{1}{r}$

Hence, for $r = 100\;{\text{cm}},l = 22\;{\text{cm}}$, we have $\theta = \frac{{22}}{{100}}$ radian $= \frac{{180}}{\pi } \times \frac{{22}}{{100}}$ degree $= \frac{{180 \times 7 \times 22}}{{22 \times 100}}$ degree

$= \frac{{126}}{{10}}$ degree $= 12\frac{3}{5}$ degree $= {12^\circ }{36^\prime }\quad \left[ {{1^\circ } = {{60}^\prime }} \right]$

As a result, the needed angle is ${12^\circ }{36^\prime }$.

5: In a circle of diameter$40\;cm$, the length of a chord is $20\;cm$. Find the length of minor arc of the chord.

Ans: The diameter of the circle $= 40\;{\text{cm}}$

So, Radius $(r)$ of the circle $= \frac{{40}}{2}\;{\text{cm}} = 20\;{\text{cm}}$

Let ${\text{AB}}$ be a chord ( length $= 20\;{\text{cm}})$ of the circle.

In $\Delta OAB,OA = OB =$ Radius of circle $= 20\;{\text{cm}}$

Also, ${\text{AB}} = 20\;{\text{cm}}$

As a result, $\Delta OAB$ is an equilateral triangle.

So, $\theta = {60^\circ } = \frac{\pi }{3}$ radian

We also know that in a circle of radius ${\text{r}}$ unit, if an arc of length $l$ unit subtends an angle $\theta$ radian at the centre then $\theta = \frac{l}{r}$

$\frac{\pi }{3} = \frac{{\widehat {AB}}}{{20}} \Rightarrow \widehat {AB} = \frac{{20\pi }}{3}\;{\text{cm}}$

As a result, the length of the minor arc of the chord is $\frac{{20\pi }}{3}\,{\text{cm}}.$

6. If in two circles, arcs of the same length subtend angles ${60^\circ }$ and ${75^\circ }$ at the centre, find the ratio of their radii.

Ans: The radii of the two circles should be ${r_1}$ and ${r_2}$. Whereas an arc of length $l$ subtend an angle of ${60^\circ }$ at the centre of the circle of radius ${r_1}$, while let an arc of length/subtend an angle of ${75^\circ }$ at the centre of the circle of radius ${r_2}$.

Now, ${60^\circ } = \frac{\pi }{3}$ radian and ${75^\circ } = \frac{{5\pi }}{{12}}$ radian. We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta$. radian at the centre then $\theta = \frac{l}{r}$ or $l = r\theta$

So, $l = \frac{{{r_1}\pi }}{3}$ and $l = \frac{{{r_2}5\pi }}{{12}}$

$\Rightarrow \frac{{{r_1}\pi }}{3} = \frac{{{r_2}5\pi }}{{12}}$

$\Rightarrow {r_1} = \frac{{{r_2}5}}{4}$

$\Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{5}{4}$

As a result, the ratio of the radii is 5: 4.

7. Find the angle in radian through which a pendulum swings if its length is $75\;cm$ and the tip describes an arc of length.

(i) 10cm

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta$ radian at the centre, then $\theta = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

Now, $l = 10\;{\text{cm}}$

$\theta = \frac{{10}}{{75}}$ radian $= \frac{2}{{15}}$ radian

(ii) 15cm

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta$ radian at the centre, then $\theta = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

Now, $l = 15\;{\text{cm}}$

$\theta = \frac{{15}}{{75}}$ radian $= \frac{1}{5}\operatorname{radian}$

(iii) 21cm

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta$ radian at the centre, then $\theta = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

Now, $l = 21\;{\text{cm}}$

$\theta = \frac{{21}}{{75}}$ radian $= \frac{7}{{25}}$ radian

## Conclusion

Class 11 Maths Chapter 3 Exercise 3.1 Maths is essential for understanding the basics of trigonometric functions. It covers key concepts like angles, degree and radian measures, and their conversions. Focus on mastering these conversions and the application of radian measures in calculating arc lengths and sector areas. Vedantu provides the best solutions, offering clear explanations and step-by-step guidance. By regularly practising with these expert solutions of ex 3.1 class 11, students can strengthen their problem-solving skills and prepare well for their exams.

## Class 11 Maths Chapter 3: Exercises Breakdown

 Exercise Number of Questions Exercise 3.2 10 Questions & Solutions Exercise 3.3 25 Questions & Solutions Miscellaneous Exercise 10 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.1

1. What will I learn in this chapter of class 11 maths chapter 3?

Here, you will learn about Degree of measure, Angles, Radian measure, Relations between Radian and Real numbers, Relation between Degree and Radian, Trigonometric Functions, Sign of Trigonometric Functions, Domain and Range of Trigonometric Functions, Trigonometric Functions of Sum and Difference of Two Angles and Trigonometric Equations. All the topics will help to learn how to find the radian measure of degree measures, how to find the degree measures of a radian measure, how to find the ratio of the radii of a circle, how to find trigonometric functions in quadrants and many more calculations.

2. Give anyone an illustrative example for trigonometry?

Illustrations:

1. Tan-1(−½) + Tan-1 (−⅓) = Tan-1 [(−½ − ⅓)/ (1− ⅙)]

= Tan-1(−1)

= −π/4

2. Tan-1(−2) + Tan-1(−3) = Tan-1[(−2+−3)/ (1−6)]

= Tan-1(−5/ −5) = Tan-1 1

= π/4

3. Tan-1 (−3) + Tan-1 (−⅓) = − ( Tan-1B) + Tan-1(⅓)

= −π/2

4. Tan-1(5/3) − Tan-1(¼) = Tan-1[(5/3−¼)/ (1+5/12)]

= Tan-1(17/17)

= Tan-11 = π/4

5. Tan-12x + Tan-13x = π/4

= Tan-1[(5x)/ (1−6x2)] = π/4

= 5x/ (1−6x2) = 1

⇒ 6x2 − 5x + 1 = 0

⇒ x = 1/6 or −1

∴ x = 16 as, x = −1

6. If Tan-1(4) + Tan-1(5) = Cot-1(λ). Find λ

Here,

Tan-1[9/ (1−20)] = Cot-1  λ

= Tan-1(-9/19) = Cot-1(λ)

= − Tan-1(9/19) = Cot-1(λ)

= − Cot-1 (19/9) = Cot-1(λ)

Or, λ = −19/9

3. What are trigonometric identities?

In earlier classes, we have studied the concept of ratio. We now define a particular ratio which involves the sides of a right angled triangle, and then call them as trigonometric ratios. This chapter also will introduce you to a few advanced concepts which are related to trigonometric identities, which are commonly the square terms of the functions. They are:

• Sin2 A + cos2 A = 1

• 1 + tan2 A = sec2 A

• 1 + cot2 A = cosec2 A

Three proofs are presented concerning these expressions. These three expressions have a vast list of applications in different forms.

4. What does property set 1 and property set 2 of trigonometry consists?

Below are the sets of property 1 and 2:

Property Set 1:

Sin-1(x) = cosec-1(1/x), x∈ [−1,1]−{0}

Cos-1(x) = sec-1(1/x), x ∈ [−1,1]−{0}

Tan-1(x) = cot-1(1/x), if x > 0 Or,

= cot-1(1/x) −π, if x < 0

Cot-1(x) = tan-1(1/x), if x > 0 Or,

= tan-1(1/x) + π, if x < 0

Property Set 2:

Sin-1(−x) = −Sin-1(x)

Tan-1(−x) = −Tan-1(x)

Cos-1(−x) = π − Cos-1(x)

Cosec-1(−x) = − Cosec-1(x)

Sec-1(−x) = π − Sec-1(x)

Cot-1(−x) = π − Cot-1(x)

• Sin-1(−x) = −Sin-1(x)

• Cos-1(−x) = π − Cos-1(x)

5. What are the important concepts required to solve Class 11 Maths Exercise 3.1?

The important concepts that you will require to learn in Class 11 Maths Chapter 3 Exercise 3.1 are the initial side and terminal side of an angle, different measures to calculate angle (degree measure and radian measure). If you have understood these concepts very clearly, you will be able to easily solve all the questions in Exercise 3.1. Use Vedantu’s official website or the Vedantu app for NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.1 to get comprehensive answers to the exercise at free of cost. These solutions have been written by experts in an easy to understand language to help you score good marks in exams.

6. What study plan to follow for Class 11 Maths Exercise 3.1?

Maths can seem like a tough subject, but if you have a study plan to strategize and organize your syllabus, you can overcome the biggest hurdles of time management and the vast syllabus of Class 11 Maths Exercise 3.1. The best study plan is to practise the sums every day and take the help of Vedantu for a detailed explanation. You can get access to Vedantu’s Study Plan for Class 11 Maths NCERT Solutions for Class 11 Maths Chapter 3 at free of  cost. For Exercise 3.1, all you need to do is practise all the questions once and keep revising them to ensure you don’t forget the concepts.

7. What is the initial side and terminal side?

The initial side is the original ray from which the angle originates and the terminal side is the ray, on which the angle after rotating is finally positioned. For more simple explanations, you should check out the NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.1 on Vedantu to get the answers to the full 3.1 Exercise. Comprehensive answers have been provided by experts with miscellaneous questions and answers as well.

8. Is class 11 Maths chapter 3 easy?

Class 11 Chapter 3 can become easy for you if you have the correct mindset. You need to have the correct guidance to show you the best path. The best guide that you will find today is Vedantu. You can improve your score with Vedantu’s NCERT Solutions. You should use NCERT Solutions Class 11 Maths Chapter 3 to be able to solve this chapter. The simplistic nature of the solution will help you practice the exercises smoothly.

9. How to score full marks in questions from Chapter 3 Class 11 Maths?

Getting top scores in Class 11 Chapter 3 Maths may look like a tough job, but with Vedantu, the journey will be a smooth ride. With Vedantu, you will get the best study plan to organize your syllabus around a routine and the best NCERT Solutions Class 11 Maths to give you a clearer idea of all the concepts and exercises, including Chapter 3. All the exercises have been solved step wise so that everything is graspable.

10. What are the key formulas to remember for class 11 maths 3.1?

Key formulas include degree to radian conversion $Radians=Degrees\times \frac{\pi }{180}$ and radian to degree conversion $Degrees=Radians\times \frac{180}{\pi}$.

11. How can NCERT Solutions help with class 11 maths ch 3 ex 3.1?

NCERT Solutions offer detailed explanations and step-by-step solutions for each question, simplifying the understanding and solving of problems. These solutions enhance learning and exam preparation.