(Ⅰ)由于S=12absinC,于是2absinC=3(a2 b2−c2)=23abcosC即:tanC=3,于是C=π3(Ⅱ)由于cos(A B)=cosAcosB−sinAsinB=−cosC=−12。由余弦定理,bccosA cacosB abcosC=bc×b2 c
(Ⅰ)由于S=12absinC,于是2absinC=3(a2 b2−c2)=23abcosC即:tanC=3,于是C=π3(Ⅱ)由于cos(A B)=cosAcosB−sinAsinB=−cosC=−12。由余弦定理,bccosA cacosB abcosC=bc×b2 c
