(a+b+c)/(sinA+sinB+sinC)=2R怎么证明?
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![(a+b+c)/(sinA+sinB+sinC)=2R怎么证明?](/uploads/image/z/5212512-0-2.jpg?t=%28a%2Bb%2Bc%29%2F%28sinA%2BsinB%2BsinC%29%3D2R%E6%80%8E%E4%B9%88%E8%AF%81%E6%98%8E%3F)
(a+b+c)/(sinA+sinB+sinC)=2R怎么证明?
(a+b+c)/(sinA+sinB+sinC)=2R
怎么证明?
(a+b+c)/(sinA+sinB+sinC)=2R怎么证明?
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