△ABC中,a^2+b^2=c^2+a*b,且sina*sinb=3/4,试判断三角形ABC的形状
1个回答

由余弦定理a^2+b^2-c^2=2abcosC得：

a^2+b^2＝c^2+2abcosC

∵a^2+b^2=c^2+ab

∴cosC=1/2

∴C＝60°

sinAsinB

=(-1/2)[cos(A+B)-cos(A-B)]

=(-1/2)[cos(π-C)-cos(A-B)]

=(-1/2)[-cosC-cos(A-B)]

=(1/2)[cosC+cos(A-B)]

=(1/2)[(1/2)+cos(A-B)]

=(1/4)+[cos(A-B)]/2

∵sinAsinB=3/4

∴cos(A-B)＝[(3/4)-(1/4)]×2＝1

∴A＝B

∵A+B＝180-60＝120°

∴A＝B＝C＝60°

三角形ABC为正三角形(等边三角形)

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