第1个回答 2009-08-22
c2=a2+b2-2abcosC
2√3absinC=a2+b2+c2
则,2√3absinC+2abcosC=2(a2+b2)
即√3absinC+abcosC=a2+b2
得2absin(C+30度)=a2+b2
所以(a-b)^2=2ab[sin(C+30)-1]
因为(a-b)^2>=0,sin(C+30)-1<=0,则sin(C+30)-1=0
C+30=90 C=60,(a-b)^2=0,a=b
△ABC的形状为等边△
采纳哦!O(∩_∩)O谢谢!
第2个回答 2009-08-22
c^2=a^2+b^2-2abcosC
2√3absinC=a^2+b^2+c^2
则,2√3absinC+2abcosC=2(a^2+b^2)
即√3absinC+abcosC=a^2+b^2
得2sin(C+30)=a^2+b^2
所以(a-b)^2=2ab[sin(C+30)-1]
因为(a-b)^2>=0,sin(C+30)-1<=0,则sin(C+30)-1=0
C+30=90 C=60,(a-b)^2=0,a=b
△ABC的形状为等边△