In the figure AB || CD and ∠ECD = 100° and ∠ABE = 110°, find the value of x.
Since AB || CD
... AF || CD ….(AB was produced to F) and CF is transversal
... ∠DCF = ∠BFC = 100°
Now ∠BFC + ∠BFE = 180° ….(Linear pair)
... 100° + ∠BFE = 180°
... ∠BFE = 180° - 100° = 80°
Now in ΔBFE, we have
∠ABE = ∠BFE +∠BEF
⇒ 110° = 80° + x°
Þ x = 110° - 80° = 30°
... AF || CD ….(AB was produced to F) and CF is transversal
... ∠DCF = ∠BFC = 100°
Now ∠BFC + ∠BFE = 180° ….(Linear pair)
... 100° + ∠BFE = 180°
... ∠BFE = 180° - 100° = 80°
Now in ΔBFE, we have
∠ABE = ∠BFE +∠BEF
⇒ 110° = 80° + x°
Þ x = 110° - 80° = 30°
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