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

• Last Updated : 28 Apr, 2021

### (i) 9π/5

Solution:

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

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

Thus, (9π/5)c = 324o

### (ii) −5π/6

Solution:

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

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

Thus, (9π/5)c = −150o

### (iii) 18π/5

Solution:

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

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

Thus, (18π/5)c = 648o

### (iv) −3

Solution:

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

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

Thus, (−3)c = −171o49’5”

### (v) 11

Solution:

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

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

Thus, (11)c =630o

### (vi) 1

Solution:

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

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

Thus, (1)c = 57o16’21”

### (i) 300o

Solution:

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

Hence, 3000 = 300 × π/180 = 5π/3

### (ii) 35o

Solution:

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

Hence, 35o = 35 × π/180 = 7π/36

### (iii)−56o

Solution:

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

Hence, −56o = −56o × π/180 = −14π/45

### (iv) 135o

Solution:

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

Hence, 135o = 135 × π/180 = 3π/4

### (v) −300o

Solution:

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

Hence, −3000 = −300 × π/180 = −5π/3

### (vi) 7o30′

Solution:

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

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

### (vii) 125o30′

Solution:

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

Hence, 125o30′ = 125o(30/60)o = (125’/2)o = 251π/360

### (viii) −47o30′

Solution:

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

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

### 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 81o and 9o.

### 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 = 3/5xo

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°.

⇒ 3/5xo + 3/2xo + 12/5xo = 180o ⇒ (6+15+24)/10xo = 180o

Upon cross-multiplication we get, 45xo = 180o × 10o = 180o ⇒ xo = 180o/45o = 40o

∴ The angles of the triangle are:

3/5xo = 3/5 × 40o = 24o

3/2xo = 3/2 × 40o = 60o

12/5 xo = 12/5 × 40o = 96o

### (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 × 180o/π = 540o

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

### (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 = 1080o

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

### (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 × 180o/π = 900o

∴ Each angle of heptagon = 5π/7 × 180o/ π = 900o/7 = 128o34′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 × 180o/π = 1800o

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

### Question 6. The angles of a quadrilateral are in A.P., and the greatest angle is 120o. 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 = 10o

∴ The angles are:

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

We know 180o = π radians = πc or 1o = (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 180o = π radians = πc or 1o = (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, ao 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 = 60o

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 180o = π radians = πc or 1o = (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 9o. 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] 180o

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

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

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