正弦定理和余弦定理 (26 10:24:53)

1个回答

  • (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)