three lines AB
In the given figure
8
intersecting at o find the
vakees of Ч and it is being given
x: 4:2=
and
than
2!315
Answers
Answer:
\red { Value \: of \:x } = 36\degreeValueofx=36°
\red { Value \: of \:y } = 54\degreeValueofy=54°
\red { Value \: of \:z } = 90\degreeValueofz=90°
Step-by-step explanation:
Given:
Three lines AB,CD and EF intersecting at O. x:y:z=2:3:5.
Solution:
\angle AOE = \angle FOB∠AOE=∠FOB
\blue { ( Vertically \: opposite \: angles)}(Verticallyoppositeangles)
Similarly.,
\angle EOD = \angle FOC = y∠EOD=∠FOC=y
\angle BOD = \angle AOC = z∠BOD=∠AOC=z
\angle AOE + \angle EOD + \angle DOB + \angle BOF + \angle FOC + \angle COA = 360\degree∠AOE+∠EOD+∠DOB+∠BOF+∠FOC+∠COA=360°
\orange { ( Sum \: the \: angles \:around \: a \:point )}(Sumtheanglesaroundapoint)
\implies x + y + z + x + y + z = 360\degree⟹x+y+z+x+y+z=360°
\implies 2x + 2y + 2z = 360\degree⟹2x+2y+2z=360°
\implies 2(x+y+z) = 360\degree⟹2(x+y+z)=360°
\implies x + y + z = \frac{360}{2} = 180\degree⟹x+y+z=
2
360
=180°
x:y:z = 2:3:5 \:(given)x:y:z=2:3:5(given)
Let \: x = 2mLetx=2m
y = 3my=3m
z = 5mz=5m
x + y + z = 180\degreex+y+z=180°
\implies 2m+3m+5m = 180⟹2m+3m+5m=180
\implies 10m = 180\degree⟹10m=180°
\implies m = \frac{180}{10} = 18⟹m=
10
180
=18
Therefore.,
\red { Value \: of \:x } = 2m = 2 \times 18 = 36\degreeValueofx=2m=2×18=36°
\red { Value \: of \:y } = 3m = 3\times 18 = 54\degreeValueofy=3m=3×18=54°
\red { Value \: of \:z } = 5m = 5\times 18 = 90\degreeValueofz=5m=5×18=90°
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