RD Chapter 4 Measurement of Angles Solutions
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