如图,∠BCD=47°+30°=77°
BC=CD=35
∴∠CBD=(180°-77°)/2=51.5°
∠ABC=135°-30°=105°
∠ABD=∠ABC-∠CBD=53.5°
△BCD中,由余弦定理：
BD²=BC²+CD²-2BC*CDcos∠BCD
=35²+35²-2×35×35×cos77°
≈1899
△ABD中,由余弦定理：
AD²=BD²+AB²-2BD*ABcos∠ABD
=1899+50²-2√1899 *50cos53.5°
=1807
由正弦定理：AD/sin∠ABD = BD/sin∠BAD
sin∠BAD= BDsin∠ABD / AD
=√1899 sin53.5° / √1807
=0.824
∠BAD=55.5°
∴∠DAE=55.5°-45°=10.5°
即目的地在A港的北偏西10.5°