Question

In fig. given Line AB and CD intersect at O. If ∠AOC + ∠BOE =70˚ and ∠BOD =40˚, Find ∠COE .​

Answer 1

Answer:

<AOC=<BOE     (V.O.A)

<AOC=40

<AOC+<BOE=70

40+<BOE=70

<BOE=30

<BOE+<AOC+<COE=180

70+<COE=180

<COE=110

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Answer 2

Given :

• ∠AOC + ∠BOE = 70°
• ∠BOD = 40°

To Find :

• ∠COE

Solution :

$\longmapsto\tt{\angle{AOC}=\angle{BOD}\:(V.O.A)}$

$\longmapsto\tt\bf{\angle{AOC}=40\degree}$

Also ,

$\longmapsto\tt{\angle{AOC}+\angle{BOE}=70\degree\:(Given)}$

$\longmapsto\tt{40\degree+\angle{BOE}=70\degree}$

$\longmapsto\tt{\angle{BOE}=70\degree-40\degree}$

$\longmapsto\tt\bf{\angle{BOE}=30\degree}$

Now ,

$\longmapsto\tt{\angle{AOC}+\angle{COE}+\angle{BOE}=180\degree\:(Angles\:on\:one\:line)}$

$\longmapsto\tt{40\degree+\angle{COE}+30\degree=180\degree}$

$\longmapsto\tt{70\degree+\angle{COE}=180\degree}$

$\longmapsto\tt{\angle{COE}=180\degree-70\degree}$

$\longmapsto\tt\bf{\angle{COE}=110\degree}$