正余弦定理问题为了躲避台风,两海岸巡逻船分别抛锚在两荒岛C、D处,A、B为两船舶基地,且相距√5km,观察到：∠ACB=

∠ACB=75°∠BCD=45°∠ADC=30°∠ADB=45°
设CD=X,AC=X,AD=X√3,设BC,AD交点O,∠AOC=75°,AO=AC=X
OD=X(√3-1)
BD=CD*sin45°/sin60°=X√6/3
OC=ACsin30°/sin75°
BC=Xsin75°/sin60°
X/sinB=BC/sinA,sinA/sinB=1.115
A+B=105°,A=105°-B;其中,∠ACB=B,∠BAD=A
sin(105-B)=sin105cosB-cos105sinB
=0.966cosB+0.2588sinB=1.115sinB
tanB=0.966/(1.115-0.2588)=1.1277
B=48.43°,A=26.57°
√5/sin105°=X/sin48.43°
X=1.73km=CD
两荒岛C、D间的距离为1.73公里