设船速为x千米/小时,则BE=x/3,CB=4/3x,CE=5x/3,角BAE=30,角CAE=150
在三角形CAE中,由正弦定理
sinC/AE=sin150/CE (1)
在三角形BAE中,由正弦定理
sin(120+C)/AE=sin30/BE (2)
(2)/(1)得,
sin(120+C)/sinC=5
sin120cosC+cos120sinC=5sinC
根号3/2cosC-1/2sinC=5sinC
cosC=11sinC/根号3
由(sinC)^2+(cosC)^2=1,且C