Class 11 RD Sharma Solutions - Chapter 4 Measurement of Angles - Exercise 4.1 | Set 1

Measurement of angles is a fundamental concept in geometry that deals with quantifying the amount of rotation between two intersecting lines or rays. In Chapter 4 of RD Sharma's Class 11 mathematics textbook, students are introduced to various aspects of angle measurement, including different units of measurement (degrees, radians, and gradians), the relationships between these units, and methods for converting between them.

Understanding angle measurement is crucial for further studies in trigonometry, calculus, and higher mathematics, as well as for practical applications in fields like engineering, architecture, and navigation.

Question 1. Find the degree measure corresponding to the following radian measures using π = 22/7:

(i) 9π/5

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)^o 

Hence, (9π/5)^c = (9π/5 × 180/π)^o = 324^o

Thus, (9π/5)^c = 324^o

(ii) −5π/6

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)^o

Hence, (−5π/6)^c = (−5π/6 × 180/π)^o = −150^o

Thus, (9π/5)c = −150^o

(iii) 18π/5

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)^o

Hence, (18π/5)^c = (18π/5 × 180/π)^o = 648^o

Thus, (18π/5)^c = 648^o

(iv) −3

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)^o

Hence, (−3)^c = (−3 × 180/π)^o = (180 × 7 × −3/22)^o = (−171⁹/₁₁) = −171^o(9 × 60/11)' =−171^o49'5''

Thus, (−3)^c = −171^o49'5''

(v) 11

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)⁰

Hence, (11)^c = (11 × 180/π)^o = (11 × 180 × 7/22) = 630^o

Thus, (11)^c =630^o

(vi) 1

Solution:

We know that π radians = 180^o or 1 radian = 1^c = (180/π)⁰

Hence, (1)^c = (1 × 180/π)^o = (180 × 7/22) = 57^o(3 × 60/11) = 57^o16¹(4 × 60/11)¹¹ = 57^o16'21''

Thus, (1)^c = 57^o16'21''

Question 2. Find the radian measure corresponding to the following degree measures using:

(i) 300^o

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 300⁰ = 300 × π/180 = 5π/3

Thus, 300^o = 5π/3 radians

(ii) 35^o

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 35^o = 35 × π/180 = 7π/36

Thus, 35^o = 7π/36 radians

(iii)−56^o

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, −56^o = −56^o × π/180 = −14π/45

Thus, −56^o = −14π/45 radians

(iv) 135^o

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 135^o = 135 × π/180 = 3π/4

Thus, 135^o = 3π/4 radians

(v) −300^o

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, −300⁰ = −300 × π/180 = −5π/3

Thus, −300^o = −5π/3 radians

(vi) 7^o30'

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 7^o30' = (7 × π/180)^C × (30/60)^o = (7'/₂)^o ×(π/180)^C = (15π/360)^c = π/24

Thus, 7^o30' = π/24 radians

(vii) 125^o30'

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 125^o30' = 125^o(30/60)^o = (125'/₂)^o = 251π/360 

Thus, 125^o30' = 251π/360 radians

(viii) −47^o30'

Solution:

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, −47^o30' = −47^o(30/60)^o = (−47'/₂)^o = (−95/2)^o = (−95/2 × π/180)^o = −19π/72

Thus, −47^o30' = −19π/72 radians

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

Solution:

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

Hence, 2π/5 radians = (2π/5 × 180/ π)^o.Substituting the value of π = 22/7, we get

2π/5 radians = (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° = 162° ⇒ x° = 81° and

Now, 90° – x° = 90° – 81° = 9°

∴ The angles are 81^o and 9^o.

Question 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 grade = (9/10)^o ⇒ 2/3x grade = (9/10 × 2/3x)^o = 3/5x^o

Also since, π radians = 180° ⇒ 1 radian = 180°/π ⇒ πx/75 radians= (πx/75 × 180/π)^o = (12/5x)^o

Since, 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 = 180^o ⇒ x^o = 180^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

Question 5.Find the magnitude, in radians and degrees, of the interior angle of a regular:

(i) Pentagon

Solution:

Since 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 

Using this rationale,

Number of sides in pentagon = 5

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

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 3π radians = 3π × 180^o/π = 540^o

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

(ii) Octagon 

Solution:

Since 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 

Number of sides in octagon = 8

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

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 6π radians = 6π × 180o/π = 1080^o

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

(iii) Heptagon 

Solution:

Since 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

Number of sides in heptagon = 7

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

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 5π radians = 5π × 180^o/π = 900^o

∴ Each angle of heptagon = 5π/7 × 180^o/ π = 900^o/7 = 128^o34′17”

(iv) Duo decagon

Solution:

Since 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

Number of sides in duo decagon = 12

Sum of interior angles of duo decagon = (12 – 2)π = 10π radians

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 5π radians = 10π × 180^o/π = 1800^o

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

Question 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= 90°

Given:

The greatest angle = 120° ⇒ a + 3d = 120° ⇒ 90° + 3d = 120° ⇒ d = 30°/3 = 10^o

∴ The angles are:

(a – 3d)° = 90° – 30° = 60°, (a – d)° = 90° – 10° = 80°, (a + d)° = 90° + 10° = 100° and (a + 3d)° = 120°

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Using the above rationale, angles of quadrilateral in radians are as follows:

(60 × π/180) radians = π/3, (80 × π/180) radians = 4π/9, (100 × π/180) radians= 5π/9 and (120 × π/180) radians = 2π/3.

Thus, the angles of quadrilateral in radians are π/3, 4π/9, 5π/9 and 2π/3.

Question 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°

It is give that, 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 ⇒ 120-2d = 1⇒ 2d = 119 ⇒ d = 119/2 = 59.5

∴ The angles (in degrees) are:

(a – d)° = 60° – 59.5° = 0.5°, a° = 60° and (a + d)° = 60° + 59.5° = 119.5°

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Using the above rationale, angles of quadrilateral in radians are as follows:

(0.5 × π/180) radians = π/360, (60 × π/180) radians = π/3and (119.5 × π/180) radians = 239π/360

Thus, the angles of triangle in radians are π/360, π/3 and 239π/360.

Question 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 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

Cross multiplying the above 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

Question 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

We are given that the greatest angle = 5 × least angle

Hence, 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° and (a + d)° = 60° + 40° = 100°

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Using the above rationale, angles of quadrilateral in radians are as follows:

(20 × π/180) radians = π/9, (60 × π/180) radians = π/3 and (100 × π/180) radians = 5π/9

Hence, the angles of the triangle in radians are π/9, π/3 and 5π/9.

Question 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 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

Summary

Chapter 4 of RD Sharma's Class 11 mathematics textbook focuses on the measurement of angles, providing students with a comprehensive understanding of different angle measurement systems and their interrelationships. The chapter covers the three main units of angle measurement - degrees, radians, and gradians - and explores techniques for converting between these units. It also introduces concepts such as complementary and supplementary angles, coterminal angles, and the representation of angles in standard position. By mastering these concepts, students develop a strong foundation for more advanced topics in trigonometry and calculus, while also gaining practical skills applicable to various real-world scenarios involving angular measurements and rotations.