Question

in fig 6.13 lines AB and CD intersect at O.if Angle AOC+AngleBOE =70° and Angle BOD =40° ,find ANGLE BOE and Reflex Angle COE.

Answer 1

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 .....!!!!!!
hope it help you
this is ncert class 9 question in chapter 6
exercise 6.1 question no.1

Answer 2

$\mathbb{SOLUTION}$:

∠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°