(a+b+c)/(sinA+sinB+sinC)=2R

题目

(a+b+c)/(sinA+sinB+sinC)=2R
怎么证明?

答案

正弦定理:a/sinA=b/sinB=c/sinC=2R
(a+b+c)/(sinA+sinB+sinC)=(2R sinA+2R sinB+2RsinC)/(sinA+sinB+sinC)
=2R (sinA+sinB+sinC)/(sinA+sinB+sinC)
=2R