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

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 Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 3 All Exercises

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 

2. Measuring radians 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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