Question

Question for Brainly Stars and Mods

Mathematics

Class 9th
Lines and Angles

Question- In fig 6.13 , lines AB and CD intersect at O . if angle AOC + angle BOE = 70° and angle BOD = 40° , then find angle BOE and reflex angle COE .

Spammed answers will be reported..​

Answer 1

Step-by-step explanation:

∠AOC + ∠BOE = 70⁰

∠AOC + ∠COE + ∠BOE = 180°

[linear pair]

So,

if ∠AOC + ∠BOE = 70°

→ 70° + ∠COE = 180°

∠COE = 180-70

∠COE = 110⁰

∠BOD = ∠AOC [ Vertically Opposite Angles]

Now,

→ ∠AOC + ∠COE + ∠BOE = 180°

→40° +110° + BOE = 180°

→150° + BOE = 180°

→∠BOE = 180° - 150⁰

→∠BOE = 30°

∠BOD + ∠DOA = 180° [Liner Pair]

→40° + DOA = 180°

→∠DOA = 180° - 40°

→∠DOA = 140°

Hence,

reflex angle (COE) = <AOC + <DOE + ZBOD + ZBOE

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

reflex angle (∠COE) = 250⁰

$\huge{ son \: prodigal}$

Answer 2

Solution :

➣ To Find :

• ∠BOE

Since,

∠ AOC + ∠ BOE = 70°

We know,

• Angles along straight line = 180°

So,

∠AOC + ∠BOE + ∠COE = 180°

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180° - 70°

∴ ∠COE = 110°

➣ Given :

• ∠ BOD = 40°

Linear Pair = 180° ( COD )

∠ COE = 110°

So,

∠COE + ∠BOE + ∠BOD = 180°

⇒ 110° + ∠BOE + 40° = 180°

⇒ 150° + ∠BOE = 180°

⇒ ∠BOE = 180° - 150°

∴ ∠BOE = 30°

➣ Reflex ∠COE :

Here,

• Reflex ∠COE =

∠COA + ∠AOD + ∠BOD + ∠BOE

Since,

∠BOE = 30° and

∠AOC + ∠BOE = 70°

⇒ ∠AOC = 70° - 30°

∴ ∠AOC = 40°

➣ Finding ∠AOD :

∠BOD + ∠AOD = 180° (linear pair)

⇒ 40° + ∠AOD = 180°

⇒ ∠AOD = 180° - 40°

∴ ∠AOD = 140°

☆ Putting Values :

⇒ 40° + 140° + 40° + 30° = Reflex ∠COE

∴ Reflex ∠COE = 250°

• Question :

Find

• ∠BOE

• Reflex ∠COE

➩ ∠BOE = 30°

➩ Reflex ∠COE = 250°