解由C=90°,即A+B=90°
故a=csinA,b=csinB=ccosA
故(a+b)/c
=(csinA+ccosA)/c
=sinA+cosA
=√2sin(A+π/4)
由A属于(0,π/2)
即A+π/4属于(π/4,3π/4)
即√2/2<sin(A+π/4)≤1
即1<√2sin(A+π/4)≤√2
即1<(a+b)/c≤√2
解由C=90°,即A+B=90°
故a=csinA,b=csinB=ccosA
故(a+b)/c
=(csinA+ccosA)/c
=sinA+cosA
=√2sin(A+π/4)
由A属于(0,π/2)
即A+π/4属于(π/4,3π/4)
即√2/2<sin(A+π/4)≤1
即1<√2sin(A+π/4)≤√2
即1<(a+b)/c≤√2