NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1:
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1 provide a detailed guide to understanding the foundational concepts of Trigonometric Functions.
This exercise focuses on topics such as angles, degree and radian measures, and their interrelation, along with the conversion between degrees and radians. These solutions are created to simplify complex concepts and provide step-by-step explanations, helping students strengthen their problem-solving skills.
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1 Overview
Exercise 3.1 in Chapter 3 of Class 11 Maths focuses on the fundamental concepts of angles and their measurements in different units, such as degrees and radians. It introduces the relationship between degrees and radians and how to convert between these units. Students also learn to express angles as multiples of π in radians and perform basic calculations involving angular measurements.
This exercise is foundational for understanding Trigonometric Functions, as it establishes the basis for working with angles in real-world applications and advanced mathematical problems. The solutions for this exercise provide clear step-by-step explanations, helping students grasp these essential concepts with ease.
Class 11 Maths Chapter 3 Exercise 3.1 Questions and Answers PDF
The PDF for NCERT Solutions to Class 11 Maths Chapter 3 Exercise 3.1 contains detailed answers to all the questions, focusing on key topics like angles, degree and radian measures, and their conversions. These step-by-step solutions simplify complex problems, making it easier for students to understand and practice effectively. The PDF can be accessed through the link provided below for convenient reference and offline study.
Class 11 Maths Chapter 3 Exercise 3.1 Questions and Answers PDF
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1
Below is the 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) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°
Solution:
(iv) 520°
2. Find the degree measures corresponding to the following radian measures (Use π = 22/7).
(i) 11/16
(ii) -4
(iii) 5π/3
(iv) 7π/6
Solution:
(i) 11/16
Here, π radian = 180°
(ii) -4
Here, π radian = 180°
(iii) 5π/3
Here, π radian = 180°
We get
= 300
o
(iv) 7π/6
Here, π radian = 180°
We get
= 210
o
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution:
It is given that
No. of revolutions made by the wheel in
1 minute = 360
1 second = 360/60 = 6
We know that
The wheel turns an angle of 2π radian in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian
Therefore, in one second, the wheel turns at an angle of 12π radian.
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 π = 22/7).
Solution:
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Solution:
The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB as the chord of the circle, i.e., length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = π/3 radian
In a circle of radius
r
unit, if an arc of length
l
unit subtends an angle
θ
radian at the centre,
We get θ = 1/r
Therefore, the length of the minor arc of the chord is 20π/3 cm.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Solution:
7. Find the angle in the radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
Solution:
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r
We know that r = 75 cm
(i) l = 10 cm
So we get
θ = 10/75 radian
By further simplification,
θ = 2/15 radian
(ii) l = 15 cm
So, we get
θ = 15/75 radian
By further simplification,
θ = 1/5 radian
(iii) l = 21 cm
So, we get
θ = 21/75 radian
By further simplification,
θ = 7/25 radian
Benefits of Solving NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.1
-
Concept Clarity
: This exercise helps students understand the basic concepts of angles, degree and radian measures, and their interrelations, laying a strong foundation for advanced trigonometry.
-
Improves Problem-Solving Skills
: By practicing step-by-step solutions, students develop the ability to solve problems systematically and accurately.
-
Exam Preparation
: The solutions align with the CBSE syllabus, making it easier for students to prepare for exams by focusing on relevant questions and concepts.
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Better Exam Performance
: Regular practice of these exercises helps students become familiar with different question types and improves their speed and accuracy in exams. This contributes to better performance in board exams and competitive exams.
