(1)AB为c AC为b BC为a
正弦定理S = 1/2absinC
因为sinC^2 = 1 - cosC^2
余弦定理cosC = (a^2+b^2-c^2)/2ab
由题目知 b=√2a,c=4
代入得到cosC = (3a^2-4)/2√2a^2
sinC^2 = 1 - cosC^2 = (-a^4+24a^2-16)/8a^4
S^2 = 1/2a^4 * sinC^2 = (-a^4+24a^2-16)/16
配方可知 a^2=12时 S^2最大,为8
S = 2√2
(2)过A点坐北岸的的垂线即 AC⊥BC于C点
∠CAB=15度
COS15=COS(60-45)=(√6+√2)/4
SIN15=SIN(60-45)=(√6-√2)/4
AC=AB*COS15=6/5*(√6+√2)/4 =3(√6+√2)/10
BC=AB*sin15=6/5*(√6-√2)/4 =3(√6-√2)/10
速度BC/0.1=3(√6-√2)