NCERT Solutions Class 11th Maths Chapter – 3 Trigonometric Functions Exercise 3.1

NCERT Solutions Class 11th Maths Chapter – 3 Trigonometric Functions Exercise 3.1

NCERT Solutions Class 11th Maths Chapter – 3 Trigonometric Functions Exercise 3.1

?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
= 300o

(iv) 7π/6
Here π radian = 180°

We get
= 210o

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 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 minor arc of the chord.

Solution: The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB be 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 radian though 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