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Answers
Step-by-step explanation:
In fig.(i)
AOB is a straight line.
angle AOC and angle BOC are linear pairs on the same line.
SO AOC + BOC = 180°
= 62° + X = 180°
= x = 180° - 62°
= x = 118°
Hance, angle BOC = X = 118°
In fig.(ii)
Given, AOC = (3x-5)°
COD = 55°
BOD = (x+20)°
AOC + COD + BOD = 180°. [ angle AOC , COD and BOD form a linear pair on the same line]
= (3x+5)° + 55° + (x+20)°= 180°
= 3x+ 5 + 55 +x + 20 = 180°
= 3x+x +55+20+5 = 180°
= 4x + 80 = 180°
= 4x = 180- 80
= 4x = 100
= x = 100/4
= x = 25
Therefore,
angle AOC = (3x+5)° = (3×25+5)°= 80°
angle BOD = (x+20)° = (25+20)° = 45°
angle COD = 55°
In fig.(iii)
Given, AOC = (3x+7)°
COD = (2x-19)°
BOD = x°
AOC+COD+BOD = 180°
= (3x+ 7 + 2x-19 + x = 180°
= 3x+x+2x+7-19 = 180°
= 6x - 12 = 180°
= 6x = 180+ 12
= x = 192/6
= x = 32
Therefore,
AOC=(3x+7)° = (3×32+7)°= 103°
COD=(2x-19)°= (2×32-19)°= 45°
BOD=X° = 32°
In fig.(v)
GIVEN, AOC= (3x+20)°
BOC= (4x-36)°
AOC + BOC = 180°. [ they form linear pairs on the same line]
= 3x+20 + 4x-36 = 180°
= 3x+4x+20-36 = 180°
= 7x-16= 180°
= 7x = 180°+16
= x = 196/7
= x = 28
Therefore,
AOC= (3x+20)°= (3×28+20)°= 104°
BOC = (4x-36)°= (4×28-36)°=76°
In fig.(iii)
I don't know if x=y=z then
XOP + POQ + YOQ = 180°
x+y+z = 180° [They form a linear pairs on the same line]
= x+x+x = 180° [x=y=z]
= 3x = 180°
= x = 180°/3
= x = 60°
Therefore, x=y=z=60°