In the given figure, AB || CD, ∠BAD = 30° and ∠ECD = 50°. Find ∠CED.
A = 
D (Pair of alternate angles)
= 30O
Now, in triangle EDC we have
D = 30O and 
C = 50O
So,
CED = 180 – (
C + 
D)
= 180 – 30 – 50
=100O
10
In the given figure, AB || CD, ∠BAD = 30° and ∠ECD = 50°. Find ∠CED.
A = D (Pair of alternate angles)
= 30O
Now, in triangle EDC we have
D = 30O and C = 50O
So,
CED = 180 – (C + D)
= 180 – 30 – 50
=100O