Question

In the given figure three lines AB,CD and EF intersecting at O.Find the values of x,y and z it is being given that x:y:z=2:3:5.

Answer 1

Answer:

$\red { Value \: of \:x } = 36\degree$

$\red { Value \: of \:y } = 54\degree$

$\red { Value \: of \:z } = 90\degree$

Step-by-step explanation:

Given:

Three lines AB,CD and EF intersecting at O. x:y:z=2:3:5.

Solution:

$\angle AOE = \angle FOB$

$\blue { ( Vertically \: opposite \: angles)}$

Similarly.,

$\angle EOD = \angle FOC = y$

$\angle BOD = \angle AOC = z$

$\angle AOE + \angle EOD + \angle DOB + \angle BOF + \angle FOC + \angle COA = 360\degree$

$\orange { ( Sum \: the \: angles \:around \: a \:point )}$

$\implies x + y + z + x + y + z = 360\degree$

$\implies 2x + 2y + 2z = 360\degree$

$\implies 2(x+y+z) = 360\degree$

$\implies x + y + z = \frac{360}{2} = 180\degree$

$x:y:z = 2:3:5 \:(given)$

$Let \: x = 2m$

$y = 3m$

$z = 5m$

$x + y + z = 180\degree$

$\implies 2m+3m+5m = 180$

$\implies 10m = 180\degree$

$\implies m = \frac{180}{10} = 18$

Therefore.,

$\red { Value \: of \:x } = 2m = 2 \times 18 = 36\degree$

$\red { Value \: of \:y } = 3m = 3\times 18 = 54\degree$

$\red { Value \: of \:z } = 5m = 5\times 18 = 90\degree$

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Answer 2

Answer:

answer is

x=36 Y = 54 Z = 90