由题可知:
AB^2=m^2+n^2
AC^2=m^2+t^2
BC^2=n^2+t^2
因 m,n,t∈(0,+∞)
所以,
AB^2+AC^2=2m^2+n^2+t^2=2m^2+BC^2>BC^2
AB^2+BC^2=m^2+2n^2+t^2=2n^2+AC^2>AC^2
BC^2+AC^2=m^2+n^2+2t^2=2t^2+AB^2>AB^2
即:△ABC中任意两边的平方和均大于第三边的平方
故:△ABC是锐角三角形.
由题可知:
AB^2=m^2+n^2
AC^2=m^2+t^2
BC^2=n^2+t^2
因 m,n,t∈(0,+∞)
所以,
AB^2+AC^2=2m^2+n^2+t^2=2m^2+BC^2>BC^2
AB^2+BC^2=m^2+2n^2+t^2=2n^2+AC^2>AC^2
BC^2+AC^2=m^2+n^2+2t^2=2t^2+AB^2>AB^2
即:△ABC中任意两边的平方和均大于第三边的平方
故:△ABC是锐角三角形.