已知三角形ABC中,a^2+b^2=c^2+ab,且sinAsinB=3/4.试判断三角形的形状解:由题意得:a^2+b^2=c^2+ab,得C=π/3∴A

解:由题意得:
a^2+b^2=c^2+ab,得C=π/3
∴A+B=2π/3.
sinAsinB=sinAsin[(2π/3)-A]
=sinA(sin2π/3cosA-cos2π/3sinA)
=根号3/4sin2A+1/4-1/4cos2A=3/4
∴sin(2A-π/6)=1.
又∵-π/6<2A-π/6<11π/6,
∴2A-π/6=π/2,
A=π/3.
∴三角形为正三角形,即锐角三角形

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