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Title: Mathematics class 11th R.S Aggarwal solutions of chapter 14.
Description: Mathematics class 11th R.s Aggarwal solutions of chapter 14.
Description: Mathematics class 11th R.s Aggarwal solutions of chapter 14.
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Measurement Of Angles
Exercise 14
Q
...
A
...
60°
Answer :
• Draw a straight line AB
...
This dot represents the vertex of the angle
...
• Find 60° on the scale and mark a small dot at the edge of the protractor
...
• Mark the angle with a small arc as shown below
...
1
...
Using a protractor, draw each of the following angles
...
• Place a dot at B
...
• Place the centre of the protractor at B and the baseline of the protractor along the arm
BA
...
• Join the vertex B to the small dot with a ruler to form the second arm, BC, of the angle
...
Q
...
C
...
300°
Answer :
• Draw a straight line AB
...
This dot represents the vertex of the angle
...
• Find 300° on the scale and mark a small dot at the edge of the protractor
...
• Mark the angle with a small arc as shown below
...
1
...
Using a protractor, draw each of the following angles
...
Therefore, Angle can also be written at as = 430° – 360° = 70°
• Draw a straight line AB
...
This dot represents the vertex of the angle
...
• Find 70° on the scale and mark a small dot at the edge of the protractor
...
• Mark the angle with a small arc as shown below
...
1
...
Using a protractor, draw each of the following angles
...
Therefore, Angle can also be written as=-40° + 360° = 320°
• Draw a straight line AB
...
This dot represents the vertex of the angle
...
• Find 320° on the scale and mark a small dot at the edge of the protractor
...
• Mark the angle with a small arc as shown below
...
1
...
Using a protractor, draw each of the following angles
...
1
...
Using a protractor, draw each of the following angles
...
Therefore, Angle can also be written as=-310° + 360° = 50°
• Draw a straight line AB
...
This dot represents the vertex of the angle
...
• Find 50° on the scale and mark a small dot at the edge of the protractor
...
• Mark the angle with a small arc as shown below
...
1
...
Using a protractor, draw each of the following angles
...
Therefore, Angle can also be written as=-400° + 360° = -40°
The angle is still negative, so we will further add 360° to it
...
• Place a dot at B
...
• Place the centre of the protractor at B and the baseline of the protractor along the arm
BA
...
• Join the vertex B to the small dot with a ruler to form the second arm, BC, of the angle
...
Q
...
Express each of the following angles in radians
36°
Answer : Formula : Angle in radians = Angle in degrees ×
𝜋
180
Q
...
A
...
2
...
Express each of the following angles in radians
225°
Answer : Formula : Angle in radians = Angle in degrees ×
𝜋
180
Q
...
D
...
2
...
Express each of the following angles in radians
400°
Answer : Formula : Angle in radians = Angle in degrees ×
𝜋
180
Q
...
F
...
’
Answer : Formula : Angle in radians = Angle in degrees ×
𝜋
180
Q
...
G
...
2
...
Express each of the following angles in radians
-22°30’
Answer : Formula : Angle in radians = Angle in degrees ×
𝜋
180
Q
...
Express each of the following angles in degrees
...
’
The angle in seconds = Decimal of angle in minutes x 60
...
’
The angle in seconds = Decimal of angle in minutes x 60
...
4
...
Find all the angles in degrees and radians
...
Given that the greatest angle is double the least
...
Therefore the required angles are 40° 60° 80°
Q
...
The difference between the two acute angles of a right triangle is
Answer : The angle in degree =
= 36°
Let, two acute angles are x and y
So,
...
(1)
x+ y= 90°
...
6
...
(Take
Answer :
Therefore radius is 42 cm
)
Q
...
Find the length of an arc of a circle of radius 14 cm which subtends an angle
of 360 at the centre
Answer : Angle in radians = Angle in degrees ×
𝜋
180
Therefore the length of the arc is 8
...
8
...
9
...
5 cm
...
5 x 60 = 30’
Therefore angle subtended at the center is 31° 30’
Q
...
In a circle of diameter 30 cm, the length of a chord is 15 cm
...
Answer : Diameter = 30 cm
Length of chord = 15 cm
Radius = 15 cm [r = 0
...
θ = 60°
We know that l = r × θ
Therefore, the length of the minor arc is 15
...
11
...
12
...
How many centimetres does its
extremity move in 20 minutes?
Answer : For 20 minutes = θ = 4 x 30° = 120°
We know that l = r × θ
l=
Therefore, the length is equal to 88 cm
...
13
...
Through how many radians
does it turn in 1 second?
Answer : Given that Number of revolutions per minute = 180
Then per second, it will be = 180/60 =3
We know that In one complete revolution, the wheel turns at an angle of 2
rad
...
Q
...
A train is moving on a circular curve of radius 1500 m at the rate of 66 km
per hour
...
Train speed at rate of 66km/hr = 18
...
33 m
Distance covered in 10 second = 18
...
33m
We know that θ = Distance / radius
θ = 183
...
122 radian
Q
...
A wire of length 121 cm is bent so as to lie along the arc of a circle of
radius 180 cm
...
Answer : θ will be in degrees
...
⇒ 121 = θ / 360° × 2πr
= 121 = θ / 360° × 2 × 22 / 7 × 180
= 121 = θ / 360° × 360 × 22 / 7
= 121 = θ × 22 / 7
⇒ θ = 121 × 7 / 22
= 38
...
5°
Which can also be written as 38° 30
...
16
...
Express the least angle in radians
...
[We know that → a+(n-1) d= last term= 2x]
⇒ 180°= 3x
⇒ x= 60°
Now, 60° is least angle
Title: Mathematics class 11th R.S Aggarwal solutions of chapter 14.
Description: Mathematics class 11th R.s Aggarwal solutions of chapter 14.
Description: Mathematics class 11th R.s Aggarwal solutions of chapter 14.