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

Last updated date: 22nd Mar 2024
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## Class 11 Maths NCERT Solutions Chapter 3 – Trigonometric Functions

Class 11 Maths NCERT Solutions Chapter 3 is devised accurately as per the latest CBSE syllabus introduced. The Trigonometry Class 11 in Chapter 3 of the syllabus comprises of step by step shortcut techniques. Our study material is available for free in the online PDF free downloads for all subjects. Vedantu is the most preferred option for exam preparation as we offer the download option for all CBSE class 11 Maths NCERT Solutions without any cost. They can download NCERT Solution PDF to score high marks in their examination.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 3 - Trigonometric Functions Exercise: Exercise - 3.1 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

Competitive Exams after 12th Science

## NCERT Solutions for Class 11 Maths 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

### NCERT Solution Class 11 Maths of Chapter 3 All Exercises

 Chapter 3 - Trigonometric Functions Exercises in PDF Format Exercise 3.1 7 Questions & Solutions Exercise 3.2 10 Questions & Solutions Exercise 3.3 25 Questions & Solutions Exercise 3.4 9 Questions & Solutions Miscellaneous Exercise 10 Questions & Solutions

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

Exercise 3.1 Class 11 Maths NCERT solution provides an introductory approach to trigonometric functions. The two units of angular values – degree and radian, are explained in detail in the solution of Ex 3.1 Class 11. The process of conversion between the two units is also demonstrated in the latest Class 11 Maths NCERT Solutions.

Calculation of radius is also demonstrated under this exercise of the NCERT books PDF. Relations between various elements have to be kept in mind to solve these sums accurately. Several short cut techniques can be applied for ease of calculation through a formula cheat sheet available on Trigonometry Class 11 NCERT Solutions PDF.

Proper understanding of the various trigonometric functions such as sine, cosine, and tangent can also be grasped through Trigonometric Functions Class 11 PDF. Three fundamental equations demonstrating the relationship between these values are explained in detail, along with its proof.

Conversion of a trigonometric value to another form is also an integral component of this exercise. Based on these, several questions are framed in Exercise 3.1 Class 11 that ensure students understand the underlying concepts carefully.

### Important Concepts Covered in Exercise 3.1 of Class 11 Maths NCERT Solutions

Exercise 3.1 of Class 11 Maths NCERT Solutions is mainly based on the following concepts:

1. Measuring degrees of angles

3. Relation between radian and real numbers

4. Relation between degree and radian

Radians and degrees are the most common units for measuring an angle. So, learning the basic fundamentals of measuring angles in the study of trigonometry is a must for Mathematics and other subjects as well. The questions provided in this exercise are based on finding the degrees and radians of angles.

The solutions provided in this chapter are very helpful for the students to build a deep understanding of these concepts and to form a strong base for learning advanced topics in Mathematics. By practising all the questions present in this exercise, students can score well in the exam.

### Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions – Other Exercises

• Class 11 Maths Chapter 3 – Exercise 2

The second section of Chapter 3 deals with the angular and functional relationship of these trigonometric values. Procedures in which angles can be calculated given a function are specified and are clearly mentioned with several practice examples.

Similarly, the calculation of trigonometric functions with given angles is also specified in this chapter. Combined, these form important questions for exams in all major schools as well as CBSE board exams.

• Class 11 Maths NCERT Solutions Chapter 3 – Exercises 3 and 4

Finding the principal and general solutions of trigonometric functions are fall under the latest syllabus of Trigonometry Class 11. Usage of standard identities is taught online for or through PDF free downloads for this purpose. Domain and range values of all trigonometric functions can also be used to calculate and solve complex equations.

A formula chart is present in this exercise, which contains details on how to perform basic mathematic operations on trigonometric values. Separate formulas for each trigonometric function of sin, cos, and tan is present for addition, subtraction, multiplication, and division respectively.

Trigonometric Functions Class 11 also contains more detailed and complex problems based on the concepts learned in this exercise. Consequently, it is important to go through the various pointers learned in this exercise to complete the sums mentioned in Exercise 3.2 and 3.3.

Trigonometric functions have massive applications in calculus, one of the most complicated portions of Mathematics. Accordingly, it is essential to grasp the concepts of trigonometric functions to excel in this subject in the future. This is where our Class 11 Maths NCERT Solutions Trigonometry comes to play with their expertly strategized solutions that offer a deeper insight into the solutions.

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## 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.