In Fig. 6.13, lines AB and CD intersect at O. If _ AOC + < BOE = 70° and < BOD = 40°
find 2 BOE and reflex 2 COE.
B
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∠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°
Step-by-step explanation:
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