Question

Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE...???​

Answer 1

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°

pls follow me

Answer 2

Answer:

Given

•‹AOC + ‹BOE = 70°

•‹BOD = 40°

‹BOD = ‹AOC (Vertically opposite angles)

→ ‹AOC = 40°

‹AOC + ‹BOE = 70°

→ 40° + ‹BOE = 70°

→ ‹BOE = 70° - 40°

→ ‹BOE = 30°

‹AOC + ‹BOE + ‹COE = 180° (Linear angles)

→ 40° + 30° + ‹COE = 180°

→ 70° + ‹COE = 180°

→ ‹COE = 180° - 70°

→ ‹COE = 110°

Reflex COE = 360° - 110°

→ 250°

Therefore

‹BOE = 30°

Reflex ‹COE = 250°

Hope this helps you...

Borahae!