In the given figure ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40° Find ∠AED?

In ΔABC sum of all angles = 180°.

So, ∠BAC + ∠ABC + ∠ACB = 180°

30 + 50 + ∠ACB = 180

∠ACB = 100°

Since BCD represents a straight line ∠ACB + ∠ECD = 180°

So, ∠ECD = 80°

In ΔECD sum of all angles = 180°

So, ∠ECD + ∠EDC + ∠CED = 180°

60 + 40 + ∠CED = 180

∠CED = 80°

Since AEC represents a straight line, ∠CED + ∠AED = 180°

So, ∠AED = 120°