According to question,

AB || CD

AF || CD (AB is produced to F, CF is traversal)

DCF=BFC=110^o

Now, BFC + BFO = 180^o(Sum of angles of Linear pair is 180^o)

BFO = 180^o – 110^o = 70^o

Now in triangle BOF, we have

ABO = BFO + BOF

130 = 70 + BOF

BOF = 130 – 70 =60^o

So, BOC = 60^o