AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB =35°, then ∠CFQ will be
Given that,
AB ‖ CD and PQ cuts them
EL is bisector of ∠FEB
∠LEB = ∠FEL = 35o
∠FEB = ∠LEB + ∠FEL
= 35o + 35o
= 70o
∠FEB = ∠EFC = 70o (Alternate angles)
∠EFC + ∠CFQ = 180o (Linear pair)
70o + ∠CFQ = 180o
∠CFQ = 110o