Question

在三角形ABC中，若cosC分之cosB＝2a+c分之负b 高二数学急需

1、求角B的大小 2、若b=跟号下13，a+c=4，求三角形ABC的面积

Answer 1

-b/(2a+c)=cosB/cosC=[(c�0�5+a�0�5-b�0�5)/(2ca)]/[(a�0�5+b�0�5-c�0�5)/(2ab)]=[b(c�0�5+a�0�5-b�0�5)]/[c(a�0�5+b�0�5-c�0�5)]c�0�6-a�0�5c-b�0�5c=2ac�0�5+2a�0�6-2ab�0�5+c�0�6+a�0�5c-b�0�5c，0=a�0�6+a�0�5c-ab�0�5+ac�0�5=a[(c�0�5+a�0�5-b�0�5)+ac]-ac=c�0�5+a�0�5-b�0�5=2accosB，cosB=-1/2=cos120°，B=120°13=b�0�5=a�0�5+ac+c�0�5=(a+c)�0�5-ac=4�0�5-ac，ac=3，S⊿=acsinB/2=3sin120°/2=3√3/4