在△ABC中若∠C=π/3,sinA,sinB,sinC成等差数列,且CA(AB-AC)=18求c边的长是向量CA(AB-AC)
问题描述:
在△ABC中若∠C=π/3,sinA,sinB,sinC成等差数列,且CA(AB-AC)=18求c边的长
是向量CA(AB-AC)
答
应该是sinA,sinC,sinB成等差数列sinA+sinB=2sinC,∴a+b=2c.(正弦定理)∴a^2+b^2+2ab=4c^2.(1)∵向量CA(AB-AC)=18,∴向量CA·CB=18,∴|CA||CB|cosπ/3=18,即ab=36.(2)由余弦定理,c^2=a^2+b^2-ab,(3)由(1)(2)(3)...