三角形ABC中,BC=7,AB=3,且sinC/sinB=3/5.(1)求AC.(2)求A.

问题描述:

三角形ABC中,BC=7,AB=3,且sinC/sinB=3/5.(1)求AC.(2)求A.

正弦定理sinC/AB=sinB/AC
AC=(sinB/sinC)×AB=5/3×3=5
cosA=(AC²+AB²-BC²)/2AC×AB
=(25+9-49)/2×5×3
=-15/30
=-1/2
A=120°

c/sinC=b/sinB c=3 sinC/sinB=3/5
所以b=5 b就是ac
所以ac=5
再根据余弦定理可以求得cosA为-0.5 所以A=120

(1)AB/SINC=AC/SINBAB/AC=SIC/SINB=3/5AC=5(2)BC²=AC²+AB²-2AC×AB×COSACOSA=(AC²+AB²-BC²)/(2AC×AB)=(5²+3²-7²)/30=-15/30=-1/2因为A为三角形内角,所以A...