# Angle Measurement – 01

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## Minutes x 60 = Seconds

• Express following angles in degrees, minutes, and seconds
• 74.87°

74.87° = 74° + 0.87°

∴  74.87° = 74° + 0.87 x 60′

∴  74.87° = 74° + 52.2′

∴  74.87° = 74° + 52′ + 0.2′

∴  74.87° = 74° + 52′ + 0.2 x 60”

∴  74.87° = 74° + 52′ + 12”

∴  74.87° = 74°,52′,12”

• – 30.6947°

– 30.6947° = – (30° + 0.6947°)

∴  – 30.6947° = – (30° + 0.6947 x 60′)

∴  – 30.6947° = – (30° + 41.682′)

∴  – 30.6947° = – (30° + 41′ + 0.682′)

∴  – 30.6947° = – (30° + 41′ + 0.682 x 60”)

∴  – 30.6947° = – (30° + 41′ + 40.92”)

∴  – 30.6947° = – (30° + 41′ + 41”)

∴  – 30.6947° = – 30°,41′,41” approx.

• 321.9°

321.9°  = 321° + 0.9°

∴  321.9° = 321° + 0.9 x 60′

∴  321.9° = 321° + 54′

∴  321.9° = 321°,54′,0”

• 200.6°
• 200.6°  = 200° + 0.6°

∴  200.6° = 200° + 0.6 x 60′

∴  200.6° = 200° + 36′

∴  200.6° = 200°,36′,0”

• 11.0133°

11.0133° = 11° + 0.0133°

• ∴  11.0133° =  11° + 0.0133 x 60′

∴ 11.0133° =  11° + 0.798′

∴  11.0133° =  11° + 0′ + 0.798′

∴  11.0133° =  11° + 0′ + 0.798 x 60”

∴  11.0133° =  11° + 0′ +  47.88”

∴  11.0133° =  11° + 0′ + 48”

∴ 11.0133° = 11°,48” approx.

• 94.3366°

94.3366° = 94° + 0.3366°

• ∴  94.3366° =  94° + 0.3366 x 60′

∴ 94.3366° =  94° + 20.196′

∴  94.3366° =  94° + 20′ + 0.196′

∴  94.3366° =  94° + 20′ + 0.196 x 60”

∴  94.3366° =  94° + 20′ +  11.76”

∴  94.3366° =  94° + 20′ + 12”

∴ 11.0133° = 94°,20′,12”  approx.

## Degrees x π/180 = Radians

 Sr. No. Angle in degrees Conversion Angle in radians 1 30° 30 x π/180 (π/6)c 2 45° 45 x π/180 (π/4)c 3 60° 60 x π/180 (π/3)c 4 90° 90 x π/180 (π/2)c 5 120° 120 x π/180 (2π/3)c 6 135° 135 x π/180 (3π/4)c 7 180° 180 x π/180 (π)c 8 75° 75 x π/180 (5π/12)c 9 -270° – 270 x π/180 – (3π/2)c 10 – (1/3)° – (1/3) x π/180 – (π/540)c 11 225° 225 x π/180 (5π/4)c 12 945° 945 x π/180 (21π/4)c 13 – 600° – 600 x π/180 – (10π/3)c 14 – (1/5)° – (1/5) x π/180 (π/900)c 15 -108° – 108 x π/180 – (3π/5)c 16 – 144° -144 x π/180 -(4π/5)c

## Radians  x 180/π = Degrees

 Sr. No. Angle in radians Conversion Angle in degrees 1 (π/6)c (π/6) x (180/π) 30° 2 (π/4)c (π/4) x (180/π) 45° 3 (π/3)c (π/3) x (180/π) 60° 4 (π/2)c (π/2) x (180/π) 90° 5 (3π/2)c° (3π/2) x (180/π) 120° 6 (3π/4)c (3π/4) x (180/π) 135° 7 (π)c (π) x (180/π) 180° 8 (7π/8)c (7π/8)x (180/π) 157.5° 9 – (9π/2)c – (9π/2) x (180/π) – 810° 10 (4.4)c (4.4) x (180/π) 252° 11 (5π/12)c (5π/12) x (180/π) 75° 12 – (7π/12)c -(7π/12) x (180/π) -105° 13 8c 8 x (180/π) (1440/π)° 14 (1/3)c (1/3) x (180/π) (60/π)° 15 (5π/7)c (5π/7) x (180/π) (900/7π)° 16 -(2π/9)c -(2π/9) x (180/π) – 40° 17 – (7π/24)c – (7π/24) x (180/π) – 52.5°

#### Example – 01:

• The difference between the two acute angles of right angle triangle is (2π/5)c.  Find the angles in degrees.
• Solution:

Let the two acute angles be x and y in degrees

Given their difference is  (2π/5)c.=  (2π/5) x  (180/π) = 72°

∴  x – y = 72° ………   (1)

Now the sum of acute angles of triangle is always 90°

∴  x + y = 90° ………   (2)

Solving equation (1) and (2) we get

x = 81° and y = 9°

Ans: The acute angles of triangle are 81° and 9°

#### Example – 02:

• The difference between the two acute angles of right angle triangle is (3π/10)c.  Find the angles in degrees.
• Solution:

Let the two acute angles be x and y in degrees

Given their difference is  (3π/10)c.=  (3π/10) x  (180/π) = 54°

∴  x – y = 54° ………   (1)

Now the sum of acute angles of triangle is always 90°

∴  x + y = 90° ………   (2)

Solving equation (1) and (2) we get

x = 72° and y = 18°

Ans: The acute angles of triangle are 72° and 18°

#### Example – 03:

• The sum of two angles is 5πc and their difference is 60°. Find the angles in degrees.
• Solution:

Let the two acute angles be x and y in degrees

Given their sum is 5πc.=  (5π) x  (180/π) = 900°

∴  x + y = 900° ………   (1)

Given their differene is 60°

∴  x – y = 60° ………   (2)

Solving equation (1) and (2) we get

x = 480° and y = 420°

Ans: The acute angles of triangle are 480° and 420°

#### Example – 04:

• The measures of angle of triangle are in the ratio 2:3:5. Find their measures in radians.
• Solution:

The  angles.of triangle are in the ratio 2:3:5

Let the three angles be 2k, 3k, and 5k.

Now the sum of all angles of triangle is 180°

∴  2k + 3k + 5k = 180°

∴  10k   = 180°

∴  k   = 18°

∴ The three angles are ( 2 x 18° = 36°), (3 x 18° = 54°), and (5 x 18°) = 90°

36° = 36 x π/180 = (π/5)c

54° = 54 x π/180 = (3π/10)c

90° = 90 x π/180 = (π/2)c

Ans: The angles of quadrilateral are (π/5)c, (3π/10)c, and (π/2)c,

#### Example – 05:

• One angle of triangle is (2π/9)c and the measures of other two angles.are in the ratio 4:3. Find their measures in degrees and radians.
• Solution:

one of the angle of triangle is of measure (2π/9).=  (2π/9) x  (180/π) = 40°

other two angles.are in the ratio 4:3

Let the two angles be 4k, and 3k.

Now the sum of all angles of triangle is 180°

∴  4k + 3k + 40°  = 180°

∴  7k   = 140°

∴  k   = 20°

∴ The two angles are ( 4 x 20° = 80°) and  (3 x 20° = 60°)

80° = 80 x π/180 = (4π/9)c

60° = 60 x π/180 = (π/3)c

Ans: The  two angles of triangle are 80° and 60° or (4π/9)and (π/3)c,

#### Example – 08:

• In ΔABC, m∠A = (2π/3)c and m∠B = 45°. Find m∠C in both the system.
• Solution:

m∠A = (2π/3)c =  (2π/3) x  (180/π) = 120°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  120° +  45° + m∠C = 180°

∴  m∠C = 180° – 165°

∴  m∠C = 15° = 15 x (π /180) = (π/12)c

Ans:  m∠C = 15° or (π/12)c

#### Example – 07:

• If the radian measures of two angles of triangle are (5π/9)c and (5π/18)c . Find the measure of the third angle in radians and degrees.
• Solution:

Let m∠A = (5π/9)c =  (5π/9) x  (180/π) = 100°

Let m∠B = (5π/18)c =  (5π/18) x  (180/π) = 50°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  100° +  50° + m∠C = 180°

∴  m∠C = 180° – 150°

∴  m∠C = 30° = 30 x (π /180) = (π/6)c

Ans: Measure of third angle is (π/6)or  30°

#### Example – 08:

• If the radian measures of two angles of triangle are (3π/5)c and (4π/15)c . Find the measure of the third angle in radians and degrees.
• Solution:

Let m∠A = (3π/5)c =  (3π/5) x  (180/π) = 108°

Let m∠B = (4π/15)c =  (4π/15) x  (180/π) = 48°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  108° +  48° + m∠C = 180°

∴  m∠C = 180° – 156°

∴  m∠C = 24° = 24 x (π /180) = (2π/15)c

Ans: Measure of third angle is (2π/15)or  24°

#### Example – 09:

• In ΔLMN, m∠L = (3π/4)c and m∠N = 30°. Find m∠M in both the system.
• Solution:

m∠L = (3π/4)c =  (3π/4) x  (180/π) = 135°

Now the sum of all angles of triangle is 180°

m∠L + m∠M + m∠N = 180°

∴  135° +  m∠M + 30° = 180°

∴  m∠M = 180° – 165°

∴  m∠M = 15° = 15 x (π /180) = (π/12)c

Ans:  m∠M = 15° or  (π/12)c

#### Example – 10:

• One angle of quadrilateral is (2π/9)c and the measures of other three angles.are in the ratio 3:5:8. Find their measures in radians.
• Solution:

one of the angle of quadrilateral is of measure (2π/9).=  (2π/9) x  (180/π) = 40°

other three angles.are in the ratio 3:5:8

Let the three angles be 3k, 5k, and 8k.

Now the sum of all angles of quadrilateral is 360°

∴  3k + 5k + 8k + 40°  = 360°

∴  16k   = 320°

∴  k   = 20°

∴ The three angles are ( 3 x 20° = 60°), (5 x 20° = 100°), and (8 x 20°) = 160°

60° = 60 x π/180 = (π/3)c

100° = 100 x π/180 = (5π/9)c

160° = 160 x π/180 = (8π/9)c

Ans: The angles of quadrilateral are (2π/9)c, (π/3)c, (5π/9)c, and (8π/9)c,

#### Example – 11:

• One angle of quadrilateral is (2π/5)c and the measures of other three angles.are in the ratio 2:3:4. Find their measures in degrees and radians.
• Solution:

one of the angle of quadrilateral is of measure (2π/5).=  (2π/5) x  (180/π) = 72°

other three angles.are in the ratio 2:3:4

Let the three angles be 2k, 3k, and 4k.

Now the sum of all angles of quadrilateral is 360°

∴  2k + 3k + 4k + 72°  = 360°

∴  9k   = 288°

∴  k   = 32°

∴ The three angles are ( 2 x 32° = 64°), (3 x 32° = 96°), and (4 x 32°) = 128°

64° = 64 x π/180 = (16π/45)c

96° = 96 x π/180 = (24π/45)c

128° = 108 x π/180 = (32π/4)c

Ans: The angles of quadrilateral are (2π/9)c, (π/3)c, (5π/9)c, and (8π/9)c,

#### Example – 12:

• The measures of angles of quadrilateral are in the ratio 2:3:6:7. Find their measures in degrees and radians.
• Solution:

The measures of angles are in the ratio 2:3:6:7

Let the measures of angles be 2k, 3k, 6k  and 7k.

Now the sum of all angles of quadrilateral is 360°

∴  2k + 3k + 6k + 7k  = 360°

∴  18k   = 360°

∴  k   = 20

∴ The measures of angles are ( 2 x 20° = 40°), (3 x 20° = 60°), (6 x 20° = 120°), and (7 x 20°) = 140°

40° = 40 x π/180 = (2π/9)c

60° = 60 x π/180 = (π/3)c

120° = 120 x π/180 = (2π/3)c

140° = 140 x π/180 = (7π/9)c

Ans: The measures of angles of quadrilateral are 40°, 60°, 120°, and 140°

or (2π/9)c, (π/3)c, (2π/3)c, and (7π/9)c,

#### Example – 13:

• The measures of angles of quadrilateral are in the ratio 3:4:5:6. Find their measures in degrees and radians.
• Solution:

The measures of angles are in the ratio 3:4:5:

Let the measures of angles be 3k, 4k, 5k  and 6k.

Now the sum of all angles of quadrilateral is 360°

∴  3k + 4k + 5k + 6k  = 360°

∴  18k   = 360°

∴  k   = 20

∴ The measures of angles are ( 3 x 20° = 60°), (4 x 20° = 80°), (5 x 20° = 100°), and (6 x 20°) = 120°

60° = 60 x π/180 = (π/3)c

80° = 80 x π/180 = (2π/9)c

100° = 100 x π/180 = (5π/9)c

120° = 120 x π/180 = (2π/3)c

Ans: The measures of angles of quadrilateral are 60°, 80°, 100°, and 120°

or (π/3)c, (2π/9)c, (5π/9)c, and (2π/3)c,

#### Example – 14:

• The angles of triangle are in A.P. and the greatest angle is 84°. Find all the three angles in radians.
• Solution:

Let the three angles of a triangle in A.P. be (a – d), a, (a + d) in degrees

Now the sum of all angles of a triangle is 180°

(a – d) + a + (a + d) = 180°

∴ 3a = 180°

∴ a = 60° = 60 x π/180 = (π/3)c

Now the greatest angle is 84°

∴ a + d = 84°

∴ 60° + d = 84° = 84 x π/180 = (7π/15)c

∴ d = 24°

∴ a – d = 60° – 24° = 36° = 36 x π/180 = (π/5)c

Ans: Measure of the angleare of triangle are (π/5)c, (π/3)c, (7π/15)c,

#### Example – 15:

• The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Express the least angle in radians.
• Solution:

Let the four angles of quadrilateral in A.P. be (a – 3d), (a -d), (a +d), and (a + 3d) in degrees

Now the sum of all angles of quadrilateral is 360°

(a- 3d) + (a – d) + (a + d) + (a + 3d) = 360°

∴ 4a = 360°

∴ a = 90°

Now the greatest angle is double the least

∴ a + 3d = 2(a – 3d)

∴ 90 + 3d = 2(90 – 3d)

∴ 90 + 3d = 180 – 6d

∴ 9d = 90

∴ d = 10°

Least angle = a – 3d = 90° – 3 x 10° = 90° – 30° = 60° = 60 x (π /180) = (π/3)c

Ans: The least angle is (π/3)c.

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