R S Aggarwal Solutions given in this section contain detailed step-by-step explanations to all Chapter 14 exercise questions. This chapter in RS Aggarwal Solutions deals with measurement of angles of triangles and the problems related to these angles. Students will also learn about the relationship between degree and radian. This R S Aggarwal study material for Class 11 Maths helps students in mastering all the concepts effectively. Check the detailed R S Aggarwal Solutions for Class 11 chapters and start practicing to score good marks.

## Download PDF of R S Aggarwal Solutions for Class 11 Maths Chapter 14 Measurement of Angles

### Access Answers to Maths R S Aggarwal Chapter 14 Measurement of Angles

Exercise 14 Page No: 516

**Question 1: Using a protector, draw each of the following angles.**

**(i) 60 ^{0} (ii) 130^{0} (iii) 300^{0} (iv) 430^{0}**

**Solution:**

**(i)**

Step 1: Draw a line AB.

Step 2: Place the baseline of the protractor along BA and make sure centre of the protractor lie at point B.

Step 3: Find 60^{0} on the scale of the protractor and mark a small dot at the edge and named as P as shown below:

Step 4: Join P to B with a ruler to form the second arm, BP, of the angle.

Mark the angle with a small arc as shown below:

**(ii) 130 ^{0} **

Step 1: Draw a line AB.

Step 2: Place the baseline of the protractor along BA and make sure centre of the protractor lie at point B.

Step 3: Find 130^{0} on the scale of the protractor and mark a small dot at the edge and named as C as shown below:

Step 4: Join C to B with a ruler to form the second arm, BC, of the angle.

Mark the angle with a small arc as shown below:

**(iii) 300 ^{0} **

Step 1: Draw a line AB.

Step 2: Place the baseline of the protractor along BA and make sure centre of the protractor lie at point B.

Step 3: Find 300^{0} on the scale of the protractor and mark a small dot at the edge and named as C as shown below:

Step 4: Join C to B with a ruler to form the second arm, BC, of the angle.

Mark the angle with a small arc as shown below:

**(iv) 430 ^{0}**

We know, adding or subtracting 360^{0} from a particular angle does not changes its position.

Therefore, given angle can also be written as:

430^{0} – 360^{0} = 70^{0}

Now, we have to draw an angle for 70^{0}

Step 1: Draw a line AB.

Step 3: Find 430^{0} on the scale of the protractor and mark a small dot at the edge and named as C as shown below:

Step 4: Join C to B with a ruler to form the second arm, BC, of the angle.

Mark the angle with a small arc as shown below:

**Question 2: Express each of the following angles in radians.**

**(i) 36 ^{0} (ii) 120^{0} (iii) 225^{0} (iv) 330^{0 }**

**(v) 400 ^{0} (vi) 7^{0} 30’ (vii) -270^{0} (viii) –(22^{0} 30’)**

**Solution:**

We know, Angle in radians = Angle in degrees x π/180^{0}

**(i) 36 ^{0} **

Angle in radians = 36^{0 }x π/180^{0
}

= π/5

**(ii) 120 ^{0} **

Angle in radians = 120^{0 }x π/180^{0
}

= 2π/3

** (iii) 225 ^{0} **

Angle in radians = 225^{0 }x π/180^{0
}

= 5π/4

**(iv) 330 ^{0}**

Angle in radians = 330^{0 }x π/180^{0
}

= 11π/6

**(v) 400 ^{0} **

Angle in radians = 400^{0 }x π/180^{0
}

= 20π/9

**(vi) 7 ^{0} 30’ **

Convert 30′ into degrees = (angle in minutes)/60 = (30/60) degrees = 0.5 degrees

Total angle = (7 + 0.5) degrees = 7.5 degrees or 7.5^{0}

Angle in radians = 7.5^{0 }x π/180^{0
}

= π/24

**(vii) -270 ^{0} **

Angle in radians = -270^{0 }x π/180^{0
}

= -3π/2

** (viii) –(22 ^{0} 30’)**

Convert 30′ into degrees = (angle in minutes)/60 = (30/60) degrees = 0.5 degrees

Total angle = (22 + 0.5) degrees = 22.5 degrees or 22.5^{0}

Angle in radians = -22.5^{0 }x π/180^{0
}

= -π/8

**Question 3: Express each of the following angles in degrees.**

**Solution:**

We know that.

**(i)** Angle in degrees = 5π/ 12 x 180/π = 75

**(ii)** Angle in degrees = -18π/5 x 180/π = -648

**(iii)** Angle in degrees = 5/6 x 180/π = 47.7272^{0}

Write Angle in degrees, minutes and second:

We know,

The angle in minutes = Decimal of angle in radian x 60’

The angle in seconds = Decimal of angle in minutes x 60’’

Therefore, 0.7272^{0} = 0.7272 x 60′ = 43.632′

Angle in seconds = 0.632 x 60” = 37.92” or 38”

Final angle = 47^{0} 43′ 38”

**(iv)** Angle in degrees = -4 x 180/π = = -229.0909^{0}

Write Angle in minutes:

We know,

The angle in minutes = Decimal of angle in radian x 60’

The angle in seconds = Decimal of angle in minutes x 60’’

Therefore, 0.0909^{0} = 0.0909 x 60′ = 5.4545′

Angle in seconds = 0.4545 x 60” = 27.27”

Final angle = -(229^{0} 5′ 27”)

**Question 4: The angles of a triangle are in AP, and the greatest angle is double the least. Find all the angles in degrees and radians.**

**Solution**: Let a – d, a, a + d be the three angles of the triangle that form AP.

Since greatest angle is double the least. …………… (given)

So, a + d = 2(a – d)

or a + d = 2a – 2d

or a = 3d ……(1)

Again, by angle sum property, we know

Sum of all the angles = 180 degrees

So, (a – d) + a + (a + d) = 180^{0}

or 3a = 180^{0}

or a = 60^{0} …… (2)

From (1) and (2), we get

3d = 60^{0}

or d = 20^{0}

Now, the angles are,

a – d = 60^{0}– 20^{0} = 40^{0}

a = 60^{0}

a + d = 60^{0} + 20^{0} = 80^{0}.

Therefore the required angles are 40^{0}, 60^{0} and 80^{0} .

**Question 5: The difference between the two acute angles of a right triangle is (π/5)^c. Find these angles in radians and degrees.**

**Solution**:

Angle in degree = π/5 x 180/π = 36°

Let x and y are two acute angles of a right triangle.

So, x – y= 36° ……(1)

Also we know,

x + y = 90° ……(2)

Solving (1) and (2), we get

2x= 126°

or x = 63°

Form (2), 63° + y = 90°

or y = 27°

Therefore, two acute angles are 63° and 27°.

**Represent angle into radian:**

We know, Angle in radians = Angle in degrees x π/180^{0}

Angle in radians = 63^{0 }x π/180^{0
}

= 7π/20

And Angle in radians = 27^{0 }x π/180^{0
}

= 3π/20

**Question 6: Find the radius of a circle in which a central angle of 45 ^{0} intercepts an arc of length 33 cm. (Take π = 22/7)**

**Solution: **

We know,

Central angle (θ) = (length arc)/radius ….(1)

Convert angle in radian:

Angle in radians = Angle in degrees x π/180^{0 } = 45^{0 }x π/180^{0 } = π/4

From (1),

Radius = (length arc)/Central angle

= 33/(π/4)

= 132 x 7/22 = 42

Therefore radius is 42 cm.

## R S Aggarwal Solutions for Chapter 14 Measurement of Angles

R S Aggarwal Chapter 14 – Measurement of Angles, contains the answers to all the questions listed in this chapter. Download this study material for free and get a clear idea about various concepts covered here. Let us have a look at some of the topics that are being covered here:

- Positive and negative angles
- Measuring angles
- Relationship between degree and radian

Solve and practice the R S Aggarwal Solutions for Class 11 to score high in your exams.