The picture shows a trapezium with the given sides and angle marked according to the question.
Angle ACB=DAE=65 (alternate angles between parallel lines EA and CB).
DB=√(6^2-2^2)=√32=4√2 (Pythagoras).
DB/BC=sin65, so BC=4√2/sin65=6.24 ft approx.
This is one possible solution under the assumed pictured figure.