-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�5cï¼0=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=2accosBï¼cosB=-1/2=cos120°ï¼B=120°13=b�0�5=a�0�5+ac+c�0�5=(a+c)�0�5-ac=4�0�5-acï¼ac=3ï¼Sâ¿=acsinB/2=3sin120°/2=3â3/4
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