在△ABC中,角A,B,C所对的边分别为a,b,c。且满足2acosC=2b-√3c. 题在补充。

2025-03-13 23:46:08
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回答1:

2acosC=2b-√3c.
2sinAcosC=2sinB-√3sinC
2sinAcosC=2sin(A+C)-√3sinC
2sinAcosC=2sinAcosC+2cosAsinC-√3sinC
2cosAsinC-√3sinC=0
2cosA-√3=0
cosA=√3/2
A=π/3
因,A=π/3,B=π/6,
C=π/2
所以,BC=√3AC
a=√3b
CM=1/2*BC=1/2*a=√3b/2
CM^2+AC^2=AM^2
(√3b/2)^2+b^2=(√7)^2
b^2=4
b=2 ,b=-2(舍去)
b=2
AC=2