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# RD Chapter 4- Measurement of Angles Interview Questions Answers

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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

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′

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

(i) 300°

5π/3

Radian measure of 300o is 5π/3

(ii) 35°

7π/36

Radian measure of 35o is 7π/36

(iii) -56°

-14π/45

Radian measure of -56° is -14π/45

(iv) 135°

3π/4

Radian measure of 135° is 3π/4

(v) -300°

-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

= 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

= 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

= – 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.

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

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.

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

= 3/5xo

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.

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.

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°

(80 × π/180) rad = 4π/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.

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°

(60 × π/180) rad = π/3

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.

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.

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°

(60 × π/180) rad = π/3

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.

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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