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正弦定理题目一道:A=60,a=12,则a+b+c/sinA+sinB+sinC=?
人气:350 ℃ 时间:2020-05-31 06:52:18
解答
a/sinA=b/sinB=c/sinC=2R
2R=a/sinA=12/sin60°=24/√3
(a+b+c)/(sinA+sinB+sinC)
=(2RsinA+2RsinB+2RsinC)/(sinA+sinB+sinC)
=2R
=24/√3
=8√3
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