Class 11 Maths Ncert Solutions Ex 3.1

Class 11 Maths Ncert Solutions Chapter 3 Ex 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(i).  25

(ii).  240

(iii).  4730'

(iv).  520

 

Sol:

(i).  25

As we know that, 180 = π radian

Therefore, 25 = π180×25 radian = 5π36 radian

Hence, 25 = 5π36 radian

 

(ii).  240

As we know that 180 = π radian

Therefore, 240 = π180×240 radian = 4π3 radian

Hence, 240 = 4π3 radian

 

(iii).  4730'

= 4712

= 952 degree

As we know that, 180 = π radian

Therefore, 952 degree = π180×952 radian = 1936×2π radian = 1972π

Hence, 4730' = 1972π

 

(iv).  520

As we know that, 180 = n radian

Therefore, 520 = π180×520 radian = 26π9 radian

Hence, 520 = 26π9 radian

 

 

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = 227]

(i) 1116

 

(ii) -4

 

(iii) 5π3

 

(iv) 7π6

 

Sol:

(i).  1116:

As we know, that Π Radian = 180°

Therefore, 1116 radian = 180π×1116 degree

= 45×11π×4 degree

= 45×11×722×4 degree

= 3158 degree

= 3938 degree

= 39+3×608 minute  [1 = 60’]

= 39+22'+12 minute

= 39+22'+30  [1’ = 60’’]

 

(ii).  -4:

As we know, that Π Radian = 180°

Therefore, -4 radian = 180π×(4) degree

= 180×7(4)22 degree

= 252011 degree

= 229111 degree

= 229+1×6011 minutes     [1 = 60’]

= 229+5'+511

= 229+5'+27 [1’ = 60’’]

 

(iii).  5π3

As we know, that Π Radian = 180°

Therefore, 5π3 radian = 180π×5π3 degree

= 300

 

(iv).  7π6

As we know, that Π Radian = 180°

Therefore, 7π6 radian = 180π×7π6

= 210

 

 

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Answer:

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = 36060 = 6

In one revolution, the wheel rotates an angle of 2π radian.

Therefore, in 6 revolutions, it will turn an angle of 12π radian.

 

 

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

[Use π = 227]

Answer:

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ=lr

Given:

r = 100 m and L = 22 m

We have,

θ=22100 radian

= 180π×22100 degree

= 180×7×2222×100 degree

= 12610 degree

= 1235 degree

= 1236' [1 = 60’]

Therefore, the req angle is 1236'

 

 

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

Answer:

13

Diameter of circle = 40 m

Radius of circle = 4020 m = 20 m

Let XY be the chord (length = 20 m) of the circle.

In ΔOXY, OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

ΔOXY is an equilateral triangle.

θ=60 = π3 radian

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = lr

π3 = arc(AB)20

arc(AB)20 = 20π3 m

The length of the minor arc of the chord is 20π3 m.

 

 

Q.6: In two circles, arcs which has same length subtended at an angle of 60 and 75 at the center. Calculate the ratio of their radii.

 Sol:

Let, the radii of two circles be r1 and r2.

Let, an arc of length l subtend an angle of 60 at the center of the circle of radius r1, while let an arc of length l subtend an angle of 75 at the center of the circle of radius r2.

60 = π3 radian

75 = 5π12 radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:

θ = lr

l = r θ

l = r1π3

l = r25π12

r1π3 = r25π12

r1 = r254

r1r2 = 54

Therefore, the ratio of radii is 5: 4

 

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(i) 10 cm

 

(ii) 15 cm

 

(iii) 21 cm

 

Sol:

As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle θ radian at the center, then:

θ=lr

Here, r = 75 cm

(i).  10 cm:

θ = 1075 radian = 215 radian

 

(ii).  15 cm:

θ = 1575 radian = 15 radian

 

(iii).  21 cm:

θ = 2175 radian = 725 radian