Question

In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Anyone plz help with this que i am stuck with this que. for an hour .... plz help ​

Answer 1

SOLUTION :

∠AOC + ∠BOE = 70°

∠AOC + ∠COE + ∠BOE = 180°

[ linear pair ]

So,

if ∠AOC + ∠BOE = 70°

so,

→ 70° + ∠COE = 180°

→ ∠COE = 180 - 70

→ ∠COE = 110°

.

∠BOD = ∠AOC [ Vertically Opposite Angles ]

.

Now,

→ ∠AOC + ∠COE + ∠BOE = 180°

→ 40° + 110° + ∠BOE = 180°

→ 150° + ∠BOE = 180°

→ ∠BOE = 180° - 150°

→ ∠BOE = 30°

.

∠BOD + ∠DOA = 180° [Liner Pair]

→ 40° + ∠DOA = 180°

→ ∠DOA = 180° - 40°

→ ∠DOA = 140°

Hence,

reflex angle ( ∠COE ) = ∠AOC + ∠DOE + ∠BOD + ∠BOE

reflex angle ( ∠COE ) = 40° + 140° + 40° + 30°

reflex angle ( ∠COE ) = 250°

Answer 2

Given:∠BOD=40⁰

Since AB and CD intersects, ∠AOC=∠BOD(vertically opposite angles)

∠AOC=40⁰

Also,∠AOC+∠BOE=70⁰

⇒∠BOE=70 −∠AOC=70 − 40 = 30⁰

We need to find reflex∠COE

Reflex∠COE=360⁰ −∠COE

Now, ∠AOC+∠COE+∠BOE=180⁰

⇒∠COE+(∠AOC+∠BOE)=180⁰

⇒∠COE+(40⁰ + 30⁰ )=180⁰

⇒∠COE = 180⁰ −70⁰ = 110⁰

Reflex∠COE=360⁰ − 110⁰ = 250⁰

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