Question

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

Answer 1

a^2+b^2=c^2+ab

c^2=a^2+b^2-ab

余弦定理,c^2=a^2+b^2-2*a*b*cosC

cosC=1/2,C=60°,A+B=120°

cos(A+B)=cosAcosB-sinAsinB=-1/2,

cosAcosB=1/4,

cos(A-B)=cosAcosB+sinAsinB=1,A-B=0,

可得A=B=C=60°,为等边三角形