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

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

**(i) 9**π**/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) −5**π**/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) 18**π**/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}= (−171^{9}/_{11}) = −171^{o}(9 × 60/11)’ =^{ }−171^{o}49’5”

Thus, (−3)^{c}= −171^{o}49’5”

**(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^{o}16^{1}(4 × 60/11)^{11}= 57^{o}16’21”

Thus, (1)^{c}= 57^{o}16’21”

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

**(i) 300**^{o}

^{o}

**Solution:**

We know 180

^{o }= π radians = π^{c }or 1^{o }= (π/180)^{c}Hence, 300

^{0 }= 300 × π/180 = 5π/3

Thus, 300^{o}= 5π/3 radians

**(ii) 35**^{o}

^{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}

^{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}

^{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}

^{o}

**Solution:**

We know 180

^{o }= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, −300

^{0}= −300 × π/180 = −5π/3

Thus, −300^{o}= −5π/3 radians

**(vi) 7**^{o}30′

^{o}30′

**Solution:**

We know 180

^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, 7

^{o}30′ = (7 × π/180)^{C}× (30/60)^{o}= (7’/_{2})^{o }×^{ }(π/180)^{C }= (15π/360)^{c}= π/24

Thus, 7^{o}30′ = π/24 radians

**(vii) 125**^{o}30′

^{o}30′

**Solution:**

We know 180

^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, 125

^{o}30′ = 125^{o}(30/60)^{o}= (125’/_{2})^{o}= 251π/360

Thus, 125^{o}30′ = 251π/360 radians

**(viii) −47**^{o}30′

^{o}30′

**Solution:**

We know 180

^{o}= π radians = π^{c}or 1^{o}= (π/180)^{c}Hence, −47

^{o}30′ = −47^{o}(30/60)^{o}= (−47’/_{2})^{o}= (−95/2)^{o}= (−95/2 × π/180)^{o}= −19π/72

Thus, −47^{o}30′ = −19π/72 radians

**Question 3. The difference between the acute angles of a **right-angled** triangle is 2**π** 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 get2π/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 **π**x/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^{o}34′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.

^{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.

^{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] = 9Upon 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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