If the given figure, ∠ACB + ∠BAC = 80°;∠BDE = 35°; ∠BCE = 45°, then t…

We have,

∠ACB + ∠BAC = 80^o

∠BDE = 35^o

∠BCE = 45^o

Now, we know that,

∠DBC = ∠ACB + ∠BAC

∠DBC = 80^o

∠DBC + ∠BCE = ∠CPD 

∠CPD = 80^o + 45^o = 125^o

∠CED = ∠CPD + ∠BDE 

∠CED = 125^o + 35^o = 160^o