已知正弦定理a/sinA = b/sinB = c/sinC = 2R证明（a+b+c)/(sinA+sinB+sinC)=（a_数学解答

已知正弦定理a/sinA = b/sinB = c/sinC = 2R
证明（a+b+c)/(sinA+sinB+sinC)=（a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R
就是推出它们全部相等.

人气：325 ℃　时间：2020-02-05 15:11:53

解答

证明,已知
a/sinA = b/sinB = c/sinC = 2R（1）

a=2RsinA, b=2RsinB,c=2RsinC
（a+b+c)/(sinA+sinB+sinC)=2R(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R（2）
（a-b-c)/(sinA-sinB-sinC)=2R(sinA-sinB-sinC)/(sinA-sinB-sinC)=2R（3）
前2个代入后提取2R就出来了,后面3个是正弦定理已知的
所以由（1）（2）（3）得到
（a+b+c)/(sinA+sinB+sinC)=（a-b-c)/(sinA-sinB-sinC)=a/sinA = b/sinB = c/sinC = 2R