EXERCISE 3.1Fig. 3.13, lines AB and CD intersect at O. IfA0C < BOE = 7

1.	EXERCISE 3.1Fig. 3.13, lines AB and CD intersect at O. IfA0C < BOE = 70Â° and BOD-40", findOE and reflex Z COEFig, 3.13
Answer» ∠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° Like my answer if you find it useful!

EXERCISE 3.1Fig. 3.13, lines AB and CD intersect at O. IfA0C < BOE = 70Â° and BOD-40", findOE and reflex Z COEFig, 3.13

Answer»

∠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°

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