RS Aggarwal And Veena Aggarwal Class 9 Math Seventh Chapter Lines and Angles Exercise 7B Solution
EXERCISE 7B
(1) ∠BOC + ∠AOC = 180
⇒ x + 62 = 180
⇒ x = 180 – 62 = 118
(2) ∠AOC + ∠COD + ∠DOB = 180
⇒ 3x – 7 + 55 + x + 20 = 180
⇒ 4x + 75 – 7 = 180
⇒ 4x = 180 – 68 = 112
⇒ x = 28
Hence, ∠AOC = (3×28 – 7)o = (84 – 7)o = 77o
And ∠BOD = 28 + 20 = 48o
(3) ∠AOC + ∠COD + ∠DOB = 180
⇒ 3x +7 + 2x – 19 + x = 180
⇒ 6x – 12 = 180
⇒ 6x = 180 + 12 = 192
⇒ x = 32
Hence, ∠AOC = (3×32 + 7)o = (96 + 7)o = 103o
∠COD = (2×32 – 19)o = (64 – 19)o = 45o
∠BOD = 32o
(4) Let the three angles are 5m, 4m and 6m.
∴ 5m + 4m + 6m = 180
⇒ 15m = 180
⇒ m = 12
Therefore the value of x = (5 × 12)o = 60o
The value of y = (4 × 12)o = 48o
The value of z = (6 × 12)o = 72o(5) 3x + 20 + 4x – 36 = 180
⇒ 7x – 16 = 180
⇒ 7x = 180 + 16 = 196
⇒ x = 28
(6) Since ∠AOC + ∠AOD = 180
⇒ 50 + ∠AOD = 180
⇒ ∠AOD = 180 – 50 = 130
∠AOD and ∠BOC are vertically opposite angles.
So, ∠AOD = ∠BOC
⇒ ∠BOC = 130o
∠BOD and ∠AOC are vertically opposite angles.
So, ∠BOD = ∠AOC
⇒ ∠BOD = 50o
(7) Since, ∠COE and ∠DOF are vertically opposite angles, we have,
∠COE = ∠DOF
⇒ ∠z = 50
Also ∠BOD and ∠COA are vertically opposite angles.
So, ∠BOD = ∠COA
⇒ ∠t = 90o
As ∠COA and ∠AOD from a linear pair,
∠COA + ∠AOD = 180
⇒ ∠COA + ∠AOF + ∠FOD = 180
⇒ 90 + x + 50 = 180
⇒ x + 140 = 180
⇒ x = 180 – 140 = 40
Since ∠EOB and ∠AOF are vertically opposite angles
So, ∠EOB = ∠AOF
⇒ y = x = 40
Thus, x = 40 = y = 40, z = 50 and t = 90
(8) ∠AOD and ∠COB are vertically opposite.
So, 2x = ∠COB
∴ ∠EOC + ∠COB + ∠BOF = 180
⇒ 5x + 2x + 3x = 180
⇒ 10x = 180
⇒ x = 18
Therefore, ∠EOC =(5 × 18)o = 90o
∠COB = (2 × 18)o = 36o = ∠AOD
∠BOF = (3 × 18)o = 54o = ∠AOE
(9) Let the two adjacent angles be 5x aqnd 4x.
Now since these angles form a linear pair.
∴ 5x + 4x = 180
⇒ 9x = 180
⇒ x = 20
Therefore the required angles are (5×20) = 100o and (4×20) = 80o.
(10) Let two straight lines AB and CD intersect at O and ∠AOC = 90o.
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Now, ∠AOC and ∠BOD are vertically opposite.
SO, ∠AOC = ∠BOD
⇒ ∠BOD = 90o
Also, as ∠AOC and ∠AOD form a linear pair.
⇒ 90o + ∠AOD = 180o
⇒ ∠AOD = 180 – 90 = 90o
Since, ∠BOC = ∠AOD vertically opposite
∠BOC = 90o
(11) ∠AOD and ∠BOC are vertically opposite angles.
∴ ∠AOD = ∠BOC
Now, ∠AOD + ∠BOC = 280o
⇒∠AOD + ∠AOD = 280
⇒ 2∠AOD = 280
⇒ ∠AOD = 140o
⇒ ∠BOC = ∠AOD = 140o
As, ∠AOC and ∠AOD are from a linear pair.
So, ∠AOC + ∠AOD = 180o
⇒ ∠AOC + 140 = 180
⇒ ∠AOC = 180 – 140 = 40o
Since, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠BOD = 40o
∴ ∠BOC = 140o, ∠AOC = 40o, ∠AOD = 140o and ∠BOD = 40o.
(12) Let ∠AOC = 5x and ∠AOD = 7x
Then, ∠AOC + ∠AOD = 180o (linear pair of angles)
⇒ 5x + 7x = 180
⇒ 12x = 180
⇒ x = 15
⇒ ∠AOC = 5 × 15 = 75o and ∠AOD = 7 × 15 = 105
Now, ∠AOC = ∠BOD (vertically opposite angles)
⇒ ∠BOD = 75o
Also, ∠AOD = ∠BOC (vertically opposite angles)
⇒ ∠BOC = 105o
(13) ∠BOD = 40o
⇒ ∠AOC = ∠BOD = 40o (vertically opposite angles)
∠AOE = 35o
⇒ ∠BOF = ∠AOE = 35o (vertically opposite angles)
∠AOB is a straight angle.
⇒ ∠AOB = 180o
⇒ ∠AOE + ∠EOD + ∠BOD = 180
⇒ 35 + ∠EOD + 40 = 180
⇒ ∠EOD = 180 – 75 = 105o
Now, ∠COF = ∠EOD = 105o (vertically opposite angles)
(14) ∠AOC + ∠BOC = 180o (linear pair of angles)
⇒ x + 125 = 180
⇒ x = 180 – 125 = 55o
Now, ∠AOD = ∠BOC (vertically opposite angles)
⇒ y = 125o
Also, ∠BOD = ∠AOC (vertically opposite angles)
⇒ z = 55o
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