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RD Chapter 4- Measurement of Angles Interview Questions Answers

Question 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) 11c (vi) 1c

Answer 1 :

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 thevalue of π = 22/7

[180/22 × 7× -3] o

(-3780/22) o

(-171 18/22) o

(-171 o (18/22 × 60)’)

(-171o (49 1/11)’)

(-171o 49′ (1/11 × 60)’)

– (171° 49′ 5.45”)

≈ – (171° 49′ 5”)

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

(v) 11c

(180/ π × 11) o

Substituting the value of π = 22/7

(180/22 × 7 × 11) o

(90 × 7) °

630°

Degree measure of 11c is 630°

(vi) 1c

(180/ π × 1) o

Substituting thevalue of π = 22/7

(180/22 × 7 × 1) o

(1260/22) o

(57 3/11) o

(57o (3/11 × 60)’)

(57o (16 4/11)’)

(57o 16′ (4/11 × 60)’)

(57o 16′ 21.81”)

≈ (57o 16′ 21”)

Degree measure of 1c is 57o 16′ 21”

Question 2 :

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

Answer 2 :

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

(i) 300°

(300 × π/180) rad

5π/3

Radian measure of 300o is 5π/3

(ii) 35°

(35 × π/180) rad

7π/36

Radian measure of 35o 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’ = – (471/2) °

= – (95/2) o

= – (95/2 × π/180)rad

= – 19π/72

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

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

Answer 3 :

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

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

Given:

2π/5

(2π/5 × 180/ π) o

Substituting thevalue of π = 22/7

(2×22/(7×5) × 180/22× 7)

(2/5 × 180) °

72°

Let one acute anglebe 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 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.

Answer 4 :

Given:

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

We know that, 1 grad= (9/10) o 

2/3x grad = (9/10)(2/3x) o

= 3/5xo

We know that, π rad= 180°  1rad = 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°.

3/5xo + 3/2xo +12/5xo = 180o

(6+15+24)/10xo = 180o

Uponcross-multiplication we get,

45xo = 180o ×10o

= 1800o

xo = 1800o/45o

= 40o

 Theangles of the triangle are:

3/5xo = 3/5 × 40o =24o

3/2xo = 3/2 × 40o =60o

12/5 xo = 12/5 × 40o =96o

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

Answer 5 :

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

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

Now, let uscalculate the magnitude of

(i) Pentagon

Number of sides inpentagon = 5

Sum of interiorangles of pentagon = (5 – 2) π = 3π

 Eachangle of pentagon = 3π/5 × 180o/ π =108o


(ii) Octagon

Number of sides inoctagon = 8

Sum of interiorangles of octagon = (8 – 2) π = 6π

 Eachangle of octagon = 6π/8 × 180o/ π =135o 


(iii) Heptagon

Number of sides inheptagon = 7

Sum of interiorangles of heptagon = (7 – 2) π = 5π

 Eachangle of heptagon = 5π/7 × 180o/ π =900o/7 = 128o 34′17”  


(iv) Duodecagon

Number of sides induodecagon = 12

Sum of interiorangles of duodecagon = (12 – 2) π = 10π

 Eachangle of duodecagon = 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.

Answer 6 :

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

We know that, thesum 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

= 10o


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

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.

Answer 7 :

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

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

a – d + a + a + d =180°

3a = 180°

a = 60°

Given:

Number of degrees inthe 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

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.

Answer 8 :

Let the number of sides in the firstpolygon be 2x and

The number of sidesin the second polygon be x.

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

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

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

Thus,

[(x-1)/x] π/ [(x-2)/x] π = 3/2

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

Uponcross-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 inthe 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.

Answer 9 :

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

We know that, thesum of angles of triangle is 180°.

a – d + a + a + d =180°

3a = 180°

a = 180°/3

= 60o

Given:

Greatest angle = 5 ×least angle

Uponcross-multiplication,

Greatest angle /least angle = 5

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

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

By cross-multiplyingwe 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°


 Anglesof triangle in radians:

(20 × π/180) rad =π/9

(60 × π/180) rad = π/3

(100 × π/180) rad =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.

Answer 10 :

Let the number of sides in the firstpolygon be 5x and

The number of sidesin the second polygon be 4x.

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

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

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

Thus,

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

180o [(4(5x-2) – 5(4x-2))/20x] =9

Uponcross-multiplication we get,

(20x – 8 – 20x +10)/20x = 9/180

2/20x = 1/20

2/x = 1

x = 2

Numberof sides in the first polygon = 5x = 5(2) = 10

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


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RD Chapter 4- Measurement of Angles Contributors

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