RD Sharma Solutions Class 11 Maths Chapter 4 – Measurement of Angles (Updated For 2021-22)

RD Sharma Solutions Class 11 Maths Chapter 4: If you are new to this chapter and found this topic tricky to understand such as angles and radians measure, their conversions, and inter-conversions then this article belongs to you. In this article, we are going to talk about the RD Sharma Solutions Class 11 Maths Chapter 4 Measurement of Angles.

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RD Sharma Solutions Class 11 Maths Chapter 4

Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 4

RD Sharma Solutions Class11 Exercise-4.1

RD Sharma Class 11 Maths Chapter 4 Exercise 4.1 Solutions

 The exercise-4.1 starts with a basic introduction of angles, radians, and the way to convert angles into radians. In this exercise, you will encounter questions like these; Convert 40 degrees 60 minutes into radians.

RD Sharma is the one-stop RD Sharma Solution to every problem we encounter in this chapter. Please go through the entire concept carefully as the concepts are going to help you in the chapters related to trigonometry.

Important Topics from RD Sharma Solutions Class 11 Maths Chapter 4

The chapter has only a few topics important being listed below:

•       Radians
•       Conversion of degree into radians
•       Solving mixed questions based on degree minutes and seconds into radians.

RD Sharma Solutions Class 11 Maths Chapter 4 – Measurement of Angles

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• Once you are done with the chapter, you can solve any questions that pop up in the examinations
• Consider this as a key to unlock your concepts when feeling stuck.
• You will get a deeper understanding of concepts by the step-by-step solutions given in the book.

Access RD Sharma Solutions Class 11 Maths Chapter 4

1. Find the degree measure corresponding to the following radian measures (Use π = 22/7)
(i) 9π/5 (ii) -5π/6 (iii) (18π/5) ^c (iv) (-3) ^c (v) 11^c (vi) 1^c

Solution:

We know that π rad = 180° ⇒ 1 rad = 180°/ π

(i)  9π/5

[(180/π) × (9π/5)] ^o

 

Substituting the value of π = 22/7

[180/22 × 7 × 9 × 22/(7×5)]

(36 × 9) °

324°

∴ Degree measure of 9π/5 is 324°

(ii) -5π/6 

[(180/π) × (-5π/6)] ^o

 

Substituting the value of π = 22/7

[180/22 × 7 × -5 × 22/(7×6) ]

(30 × -5) °

– (150) °

∴ Degree measure of -5π/6 is -150°

(iii) (18π/5)

[(180/π) × (18π/5)] ^o

 

Substituting the value of π = 22/7

[180/22 × 7 × 18 × 22/(7×5)]

(36 × 18) °

648°

∴ Degree measure of 18π/5 is 648°

(iv) (-3) ^c 

[(180/π) × (-3)] ^o

 

Substituting the value of π = 22/7

[180/22 × 7 × -3] ^o

 

(-3780/22) ^o

(-171 18/22) ^o

(-171 ^o (18/22 × 60)’)

(-171^o (49 1/11)’)

(-171^o 49′ (1/11 × 60)’)

– (171° 49′ 5.45”)

≈ – (171° 49′ 5”)

∴ Degree measure of (-3) ^c is -171° 49′ 5”

(v) 11^c

(180/ π × 11) ^o

Substituting the value of π = 22/7

(180/22 × 7 × 11) ^o

(90 × 7) °

630°

∴ Degree measure of 11^c is 630°

(vi) 1^c

(180/ π × 1) ^o

Substituting the value of π = 22/7

(180/22 × 7 × 1) ^o

(1260/22) ^o

(57 3/11) ^o

(57^o (3/11 × 60)’)

(57^o (16 4/11)’)

(57^o 16′ (4/11 × 60)’)

(57^o 16′ 21.81”)

≈ (57^o 16′ 21”)

∴ Degree measure of 1^c is 57^o 16′ 21”

2. Find the radian measure corresponding to the following degree measures:
(i) 300^o (ii) 35^o (iii) -56^o (iv)135^o (v) -300^o
(vi) 7^o 30′ (vii) 125^o 30’ (viii) -47^o 30′

Solution:

We know that 180° = π rad ⇒ 1° = π/ 180 rad

(i) 300°

(300 × π/180) rad

5π/3

∴ Radian measure of 300^o is 5π/3

(ii) 35°

(35 × π/180) rad

7π/36

∴ Radian measure of 35^o is 7π/36

(iii) -56°

(-56 × π/180) rad

-14π/45

∴ Radian measure of -56° is -14π/45

(iv) 135°

(135 × π/180) rad

3π/4

∴ Radian measure of 135° is 3π/4

(v) -300°

(-300 × π/180) rad

-5π/3

∴ Radian measure of -300° is -5π/3

(vi) 7° 30′

We know that, 30′ = (1/2) °

7° 30′ = (7 1/2) °

= (15/2) ^o

= (15/2 × π/180) rad

= π/24

∴ Radian measure of 7° 30′ is π/24

(vii) 125° 30′

We know that, 30′ = (1/2) °

125° 30’ = (125 1/2) °

= (251/2) ^o

= (251/2 × π/180) rad

= 251π/360

∴ Radian measure of 125° 30′ is 251π/360

(viii) -47° 30′

We know that, 30′ = (1/2) °

-47° 30’ = – (47 1/2) °

= – (95/2) ^o

= – (95/2 × π/180) rad

= – 19π/72

∴ Radian measure of -47° 30′ is – 19π/72

3. The difference between the two acute angles of a right-angled triangle is 2π/5 radians. Express the angles in degrees.

Solution:

Given the difference between the two acute angles of a right-angled triangle is 2π/5 radians.

We know that π rad = 180° ⇒ 1 rad = 180°/ π

Given:

2π/5

(2π/5 × 180/ π) ^o

Substituting the value of π = 22/7

(2×22/(7×5) × 180/22 × 7)

(2/5 × 180) °

72°

Let one acute angle be x° and the other acute angle be 90° – x°.

Then,

x° – (90° – x°) = 72°

2x° – 90° = 72°

2x° = 72° + 90°

2x° = 162°

x° = 162°/ 2

x° = 81° and

90° – x° = 90° – 81°

= 9°

∴ The angles are 81^o and 9^o

4. One angle of a triangle is 2/3x grades, and another is 3/2x degrees while the third is πx/75 radians. Express all the angles in degrees.

Solution:

Given:

One angle of a triangle is 2x/3 grades and another is 3x/2 degree while the third is πx/75 radians.

We know that, 1 grad = (9/10) ^o

2/3x grad = (9/10) (2/3x) ^o

= 3/5x^o

We know that, π rad = 180° ⇒ 1 rad = 180°/ π

Given: πx/75

(πx/75 × 180/π) ^o

(12/5x) ^o

We know that, the sum of the angles of a triangle is 180°.

3/5x^o + 3/2x^o + 12/5x^o = 180^o

(6+15+24)/10x^o = 180^o

Upon cross-multiplication we get,

45x^o = 180^o × 10^o

= 1800^o

x^o = 1800^o/45^o

= 40^o

∴ The angles of the triangle are:

3/5x^o = 3/5 × 40^o = 24^o

3/2x^o = 3/2 × 40^o = 60^o

12/5 x^o = 12/5 × 40^o = 96^o

5. Find the magnitude, in radians and degrees, of the interior angle of a regular:
(i) Pentagon (ii) Octagon (iii) Heptagon (iv) Duodecagon.

Solution:

We know that the sum of the interior angles of a polygon = (n – 2) π

And each angle of polygon = sum of interior angles of polygon / number of sides

Now, let us calculate the magnitude of

(i) Pentagon

Number of sides in pentagon = 5

Sum of interior angles of pentagon = (5 – 2) π = 3π

∴ Each angle of pentagon = 3π/5 × 180^o/ π = 108^o

(ii) Octagon

Number of sides in octagon = 8

Sum of interior angles of octagon = (8 – 2) π = 6π

∴ Each angle of octagon = 6π/8 × 180^o/ π = 135^o

(iii) Heptagon

Number of sides in heptagon = 7

Sum of interior angles of heptagon = (7 – 2) π = 5π

∴ Each angle of heptagon = 5π/7 × 180^o/ π = 900^o/7 = 128^o 34′ 17”

(iv) Duodecagon

Number of sides in duodecagon = 12

Sum of interior angles of duodecagon = (12 – 2) π = 10π

∴ Each angle of duodecagon = 10π/12 × 180^o/ π = 150^o

6. The angles of a quadrilateral are in A.P., and the greatest angle is 120^o. Express the angles in radians.

Solution:

Let the angles of quadrilateral be (a – 3d) °, (a – d) °, (a + d) ° and (a + 3d) °.

We know that, the sum of angles of a quadrilateral is 360°.

a – 3d + a – d + a + d + a + 3d = 360°

4a = 360°

a = 360/4

= 90°

Given:

The greatest angle = 120°

a + 3d = 120°

90° + 3d = 120°

3d = 120° – 90°

3d = 30°

d = 30°/3

= 10^o

∴ The angles are:

(a – 3d) ° = 90° – 30° = 60°

(a – d) ° = 90° – 10° = 80°

(a + d) ° = 90° + 10° = 100°

(a + 3d) ° = 120°

Angles of quadrilateral in radians:

(60 × π/180) rad = π/3

(80 × π/180) rad = 4π/9

(100 × π/180) rad = 5π/9

(120 × π/180) rad = 2π/3

7. The angles of a triangle are in A.P., and the number of degrees in the least angle is to the number of degrees in the mean angle as 1:120. Find the angle in radians.

Solution:

Let the angles of the triangle be (a – d) °, a° and (a + d) °.

We know that, the sum of the angles of a triangle is 180°.

a – d + a + a + d = 180°

3a = 180°

a = 60°

Given:

Number of degrees in the least angle / Number of degrees in the mean angle = 1/120

(a-d)/a = 1/120

(60-d)/60 = 1/120

(60-d)/1 = 1/2

120-2d = 1

2d = 119

d = 119/2

= 59.5

∴ The angles are:

(a – d) ° = 60° – 59.5° = 0.5°

a° = 60°

(a + d) ° = 60° + 59.5° = 119.5°

Angles of triangle in radians:

(0.5 × π/180) rad = π/360

(60 × π/180) rad = π/3

(119.5 × π/180) rad = 239π/360

8. The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

Solution:

Let the number of sides in the first polygon be 2x and

The number of sides in the second polygon be x.

We know that, angle of an n-sided regular polygon = [(n-2)/n] π radian

The angle of the first polygon = [(2x-2)/2x] π = [(x-1)/x] π radian

The angle of the second polygon = [(x-2)/x] π radian

Thus,

[(x-1)/x] π / [(x-2)/x] π = 3/2

 

(x-1)/(x-2) = 3/2

Upon cross-multiplication we get,

2x – 2 = 3x – 6

3x-2x = 6-2

x = 4

∴ Number of sides in the first polygon = 2x = 2(4) = 8

Number of sides in the second polygon = x = 4

9. The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Solution:

Let the angles of the triangle be (a – d) ^o, a^o and (a + d) ^o.

We know that, the sum of angles of triangle is 180°.

a – d + a + a + d = 180°

3a = 180°

a = 180°/3

= 60^o

Given:

Greatest angle = 5 × least angle

Upon cross-multiplication,

Greatest angle / least angle = 5

(a+d)/(a-d) = 5

(60+d)/(60-d) = 5

By cross-multiplying we get,

60 + d = 300 – 5d

6d = 240

d = 240/6

= 40

Hence, angles are:

(a – d) ° = 60° – 40° = 20°

a° = 60°

(a + d) ° = 60° + 40° = 100°

∴ Angles of triangle in radians:

(20 × π/180) rad = π/9

(60 × π/180) rad = π/3

(100 × π/180) rad = 5π/9

10. The number of sides of two regular polygons is 5:4 and the difference between their angles is 9^o. Find the number of sides of the polygons.

Solution:

Let the number of sides in the first polygon be 5x and

The number of sides in the second polygon be 4x.

We know that, angle of an n-sided regular polygon = [(n-2)/n] π radian

The angle of the first polygon = [(5x-2)/5x] 180^o

The angle of the second polygon = [(4x-1)/4x] 180^o

Thus,

[(5x-2)/5x] 180^o – [(4x-1)/4x] 180^o = 9

 

180^o [(4(5x-2) – 5(4x-2))/20x] = 9

Upon cross-multiplication we get,

(20x – 8 – 20x + 10)/20x = 9/180

2/20x = 1/20

2/x = 1

x = 2

∴Number of sides in the first polygon = 5x = 5(2) = 10

Number of sides in the second polygon = 4x = 4(2) = 8

Key Take Away from Class 11 Maths RD Sharma Solutions

•       Understanding is the key.
•       Do not learn answers or the questions for the ease of solving them.
•       Analyze every problem you did wrong.
•       Do not leave any concept in between. Go through again because everything you leave will create a void for further levels.

Access Other Important Chapters of RD Sharma Solutions Class 11 Maths Chapter 4

• Chapter 1 – Sets
• Chapter 2 – Relations
• Chapter 3 – Functions
• Chapter 5 – Trigonometric Functions
• Chapter 6 – Graphs of Trigonometric Functions
• Chapter 7 – Trigonometric Ratios of Compound Angles
• Chapter 8 – Transformation Formulae
• Chapter 9 – Trigonometric Ratios of Multiple and Sub Multiple Angles
• Chapter 10 – Sine and Cosine Formulae and Their Applications
• Chapter 11 – Trigonometric Equations
• Chapter 12 – Mathematical Induction
• Chapter 13 – Complex Numbers
• Chapter 14 – Quadratic Equations
• Chapter 15 – Linear Inequations

Lastly, all the best for your CBSE class11 examinations!

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