**NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions** – Here are all the NCERT solutions for Class 11 Maths Chapter 3. This solution contains questions, answers, images, explanations of the complete chapter 3 titled Of Trigonometric Functions taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 3 Trigonometric Functions After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions in one place.

## NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 3 Trigonometric Functions , Maths, Class 11.

Class | 11 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 3 |

Chapter Name |
Trigonometric Functions |

### NCERT Solutions Class 11 Maths chapter 3 Trigonometric Functions

Class 11, Maths chapter 3, Trigonometric Functions solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Trigonometric Functions

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### Question & Answer

Q.1:Find the radian measures corresponding to the following degree measures: \(\begin{array}{lll}{\text { (i) } 25^{\circ}} & {\text { (ii) }-47^{\circ} 30^{\prime}} & {\text { (iii) } 240^{\circ}} & {\text { (iv) } 520^{\circ}}\end{array}\)

Ans :\(\begin{array}{l}{\text { (i) } 25^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 25^{\circ}=\frac{\pi}{180} \times 25 \text { radian }=\frac{5 \pi}{36} \text { radian }}\end{array}\) \(\begin{array}{l}{\text { (ii) }-47^{\circ} 30^{\prime}} \\ {-47^{\circ} 30^{\prime}=-47 \frac{1}{2}} \\ {=\frac{-95}{2}}\end{array}\) Since \(180^{\circ}=\pi\) radian \(\begin{array}{l}{\text { (iii) } 240^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 240^{\circ}=\frac{\pi}{180} \times 240 \text { radian }=\frac{4}{3} \pi \text { radian }}\end{array}\) \(\begin{array}{l}{\text { (iv) } 520^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 520^{\circ}=\frac{\pi}{180} \times 520 \text { radian }=\frac{26 \pi}{9} \text { radian }}\end{array}\)

Q.2:Find the degree measures corresponding to the following radian measures (Use \(\pi=\frac{22}{7} )\). \(\begin{array}{llll}{\text { (i) } \frac{11}{16}} & {\text { (ii) }-4} & {\text { (iii) } \frac{5 \pi}{3}} & {\text { (iv) } \frac{7 \pi}{6}}\end{array}\)

Ans :\(\begin{array}{l}{\text { (i) } \frac{11}{16}} \\ {\text { We know that } \pi \text { radian }=180^{\circ}}\end{array}\) \(\begin{aligned} \therefore \frac{11}{16} \operatorname{radain} &=\frac{180}{\pi} \times \frac{11}{16} \text { degree }=\frac{45 \times 11}{\pi \times 4} \text { deg ree } \\ &=\frac{45 \times 11 \times 7}{22 \times 4} \text { deg ree }=\frac{315}{8} \text { degree } \end{aligned}\) \(\begin{array}{l}{=39 \frac{3}{8} \text { deg ree }} \\ {=39^{\circ}+\frac{3 \times 60}{8} \text { min utes } \quad\left[1^{\circ}=60^{\circ}\right]} \\ {=39^{\circ}+22^{\prime}+\frac{1}{2} \text { min utes }} \\ {=39^{\circ} 22^{\prime} 30^{\prime \prime}} & {\left[1^{\prime}=60^{\prime \prime}\right]}\end{array}\) \(\begin{array}{l}{(ii)-4} \\ {\text { We know that } \pi \text { radian }=180^{\circ}} \\ {-4 \text { radian }=\frac{180}{\pi} \times(-4) \text { deg ree }=\frac{180 \times 7(-4)}{22} \text { degree }} \\ {=\frac{-2520}{11} \text { deg ree }=-229 \frac{1}{11} \text { degree }}\end{array}\) \(=-229^{\circ}+\frac{1 \times 60}{11} \text { min utes } \quad\left[1^{\circ}=60^{\circ}\right]\) \(=-229^{\circ} 5^{12} 27^{n} \quad \quad\left[1^{\prime}=60^{\prime \prime}\right]\) \(\begin{array}{l}{\text { (iii) } \frac{5 \pi}{3}} \\ {\text { We know that } \pi \text { radlan }=180^{\circ}} \\ {\therefore \frac{5 \pi}{3} \text { radian }=\frac{180}{\pi} \times \frac{5 \pi}{3} \text { degree }=300^{\circ}}\end{array}\) \(\begin{array}{l}{\text { (iv) } \frac{7 \pi}{6}} \\ {\text { We know that } \pi \text { radian }=180^{\circ}} \\ {\therefore \frac{7 \pi}{6} \text { radian }=\frac{180}{\pi} \times \frac{7 \pi}{6}=210^{\circ}}\end{array}\)

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

Ans :Number of revolutions made by the wheel in 1 minute = 360 \(\therefore \) Number of revolutions made by the wheel in 1 second = \(\frac{360}{60}=6\) Hence, In 6 complete revolutions, it will turn an angle of \(6 \times 2 \pi\) radian, i.e., 12 \(\pi \) radian Thus, In one second, the wheel turns and angle of 12 \(\pi \) radian.

Q.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 (Use \(\pi=\frac{22}{7} )\)

Ans :We know that in a circle of radius r unit, if an arc of length l unit subtends and angle \(\theta \) radian at the centre, then \(\theta=\frac{1}{r}\) Therefore, for r = 100 cm, l = 22 cm, we have \(\begin{aligned} \theta &=\frac{22}{100} \text { radian }=\frac{180}{\pi} \times \frac{22}{100} \text { deg ree }=\frac{180 \times 7 \times 22}{22 \times 100} \text { deg ree } \\ &=\frac{126}{10} \text { deg ree }=12 \frac{3}{5} \text { deg ree }=12^{\circ} 36^{\circ} \quad\left[1^{\circ}=60^{\prime}\right] \end{aligned}\) Thus, the required angle is \(12^{\circ} 36^{\prime}\).

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

Ans :Diameter of the circle = 40 cm Radius of the circle = \(\frac{40}{2} \mathrm{cm}=20 \mathrm{cm}\) Let AB be a chord (length = 20 cm) of the circle. \(\begin{array}{l}{\text { In } \triangle O A B, O A=O B=\text { Radius of circle }=20 \mathrm{cm}} \\ {\text { Also, } A B=20 \mathrm{cm}} \\ {\text { Thus, } \triangle O A B \text { is an equilateral triangle. }}\end{array}\) \(\therefore \theta=60^{\circ}=\frac{\pi}{3}\) We 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 \(\theta=\frac{l}{r}\) \(\frac{\pi}{3}=\frac{\widetilde{A B}}{20} \Rightarrow \widetilde{A B}=\frac{20 \pi}{3} \mathrm{cm}\) Thus, the length of the minor arc of the chord is \(\frac{20 \pi}{3} \mathrm{cm}\).

## NCERT / CBSE Book for Class 11 Maths

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### All NCERT Solutions Class 11

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