Question

in fig 3.13 lines AB and CD intersect at o. if angle AOC+angleBOC=70 and angle BOD=40find angleBOE and reflex angle COE​

Answer 1

Answer:

aoc + boe = 70 given

bod = 40

aoc = bod =40 (vertically opposite angle)

so

aoc + boe = 70

boe = 70-40=30

aob = 180 ( linear pair)

aoc+ coe+ boe = 180

40 + coe +30 =180

coe=180-70=110

now reflex coe

aoc+ aod+bod+boe=360

aod= coe( vertically opposite angle)

reflex coe= 360

coe= 110

reflex coe = 360-coe

360-110=250 .....!!!!!!

Answer 2

Step-by-step explanation:

∠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°

Hope it helps

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