1. In Fig. 6.13. lines AB and CD intersect at O. If
<AOC + <BOE = 70° and < BOD = 40°, find
<BOE and reflex < COE.
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Given:∠BOD=40
∘
Since AB and CD intersects, ∠AOC=∠BOD(vertically opposite angles)
∠AOC=40
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Also,∠AOC+∠BOE=70
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⇒∠BOE=70
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−∠AOC=70
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−40
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=30
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We need to find reflex∠COE
Reflex∠COE=360
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−∠COE
Now, ∠AOC+∠COE+∠BOE=180
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⇒∠COE+(∠AOC+∠BOE)=180
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⇒∠COE+(40
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+30
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)=180
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⇒∠COE=180
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−70
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=110
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Reflex∠COE=360
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−110
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=250
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Answered by
2
❤️ Solution ❤️
Since AB is a straight line ,
:. ∠AOC + ∠COE + ∠EOB = 180°
______⭐OR⭐_______
=> (∠AOE + ∠BOE) + ∠COE = 180° (OR) 70° + ∠COE = 180°
∴ ∠AOC + ∠BOE = 70° [ ]
______⭐OR⭐________
=> ∠COE = 180° - 70° = 110°
=> Reflex ∠COE = 360° - 110° = 250°
Also , AB and CD intersect at 0
∴ ∠COA = ∠BOD [ ]
But , ∠BOD = 40° [ ]
- ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
- 40° + ∠BOE = 70°
______⭐OR⭐________
- ∠BOE = 70° - 40° = 30°
,
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