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

It is given: ∠AOC + ∠BOE = 70^o

And, ∠BOD = 40^o

Now, according to the question,

∠AOC + ∠BOE + ∠COE = 180^o (Sum of linear pair is always 180^o )

⇒ 70^o + ∠COE = 180^o

⇒ ∠COE = 110^o     ...........(1)

And, 

∠COE + ∠BOD + ∠BOE = 180^o  (Sum of linear pair is always 180^o )

putting the value of ∠COE, we get, 

110^o + 40^o + ∠BOE = 180^o

150^o + ∠BOE = 180^o

∴ ∠BOE = 30^o

Now, reflex ∠COE = 360^o - ∠COE

⇒ reflex ∠COE = 360^o - 110^o

∴ reflex ∠COE = 250^o