in figure lines ab and CD intersect at 0 if angle AOC + angle BOC is equal to 70 degree and Angle Bod is equal to 40 degree find angle BOC and reflex angle Coe
Answers
Answer:
∠BOC = 140°
∠COE = 250°
Step-by-step explanation:
Given
∠AOC + ∠BOE = 70°
∠BOD = 40°
Lines AB and CD intersect at O
∴ ∠AOC = ∠BOD (Opposite angles)
∴ ∠AOC = 40°
∵ ∠AOC + ∠BOE = 70°
∴ ∠BOE = 70° - ∠AOC
= 70° - 40°
= 30°
∵ AB is a line and OE, OC are rays on it
∴ From linear pair axiom
∠AOC + ∠BOC = 180° .............. (1)
or, ∠BOC = 180° - 40° = 140°
Also, from (i)
∠AOC + (∠BOE + ∠COE) = 180°
or, 40° + 30° + ∠COE = 180°
or, ∠COE = 180° - 70° = 110°
Reflex angle COE = 360° - 110° = 250°
Hope this is helpful.
Answer:
Given:∠BOD=40
∘
Since AB and CD intersects, ∠AOC=∠BOD(vertically opposite angles)
∠AOC=40
∘
Also,∠AOC+∠BOE=70
∘
⇒∠BOE=70
∘
−∠AOC=70
∘
−40
∘
=30
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We need to find reflex∠COE
Reflex∠COE=360
∘
−∠COE
Now, ∠AOC+∠COE+∠BOE=180
∘
⇒∠COE+(∠AOC+∠BOE)=180
∘
⇒∠COE+(40
∘
+30
∘
)=180
∘
⇒∠COE=180
∘
−70
∘
=110
∘
Reflex∠COE=360
∘
−110
∘
=250
∘