Question

In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

answer this question please ​.

Answer 1

Given :

• ∠AOC + ∠BOE = 70°
• ∠BOD = 40°

To Find :

• ∠BOE and reflex of ∠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 ,

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

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

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

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

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

Reflex of ∠COE :

$\longmapsto\tt{360\degree-110\degree}$

$\longmapsto\tt\bf{250\degree}$

Answer 2

$\huge{\underline{\mathcal{\red{A}\green{n}\pink{s}\orange{w}\blue{e}\pink{R:-}}}}$

$\large\underline\bold\red{Given:-}$

ㄥBOD = 40°

$\large\underline\bold\red{To\:Find:-}$

• ㄥBOE and

• reflex ㄥCOE.

$\large\underline\bold\red{Solution:-}$

Since , AB and CD. intersects ,

ㄥAOC = ㄥBOD ( Vertically opp. angles)

ㄥAOC = 40°

Now ,

ㄥAOC + BOE = 70°

➦ ㄥBOE = 70° - 40° = 30°

We Need To Find reflex ㄥCOE

Reflex ㄥCOE = 360° - ㄥCOE

Now ,

ㄥAOC + ㄥCOE + ㄥBOE = 180°

➦ ㄥCOE + (ㄥAOC + ㄥBOE ) = 180°

➦ㄥCOE (40° + 30°) = 180°

➦ㄥCOE = 180° - 70° = 110°

Reflex ㄥCOE = 360° - 110° = 250°

Hence, ReflexㄥCOE = 250°

$\rule{200px}{.3ex}$