在三角形ABC中，内角A,B,C对边的边长分别是a,b,c,，已知c=2，C=3分之π。
1,S=(1/2)*a*b*sinC=1/4ab=√3
ab=4√3
2,
S=acsinB*1/2=bcsinA*1/2
asinB=bsinA
a=2b
S=absinC*1/2=b*b/2
答案补充
2.后边还有,因为c^2=a^2+b62-2ab*cosC
cosC=(a*a+b*b-c*c)/2ab=(5b^2-4)/4b^2=1/2
所以5/4-1/b^2=1/2
b^2=4/3
所以S=b^2/2=4/3/2=2/3
答案补充
1,应该是这样子的
S=(1/2)*a*b*sinC=√3/4ab=√3
ab=4
c^2=a^2+b^2-2ab*cosC
4=a^2+b^2-8*1/2
a^2+b^2=4+4=8
(a+b)^2=8+8=16
(a-b)^2=8-8=0
a=b=2
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