sinα+cosα=√2sin(α+π/4)=1/5
sin(α+π/4)=√2/10<√2/2
∵0<α<π,
∴π/4<α<5π/4
∴3π/4<α+π/4<π
∴π/2<α<3π/4
所以三角形ABC是钝角三角形