As mathguy has helpfully pointed out, you need to resolve both velocities into the $x$ and $y$ components.
The figure below should help you to resolve the velocities.
!enter image description here
By trigonometry (using the convention that East/right is positive), the overall horizontal component is given by $$x_v=15\cos(40^\circ)-16\cos(80^\circ)=8.7123$$ So the barge is heading in an easterly direction.
Using the convention that North/up is positive, the overall vertical component is given by $$y_v=-15\sin(40^\circ)-16\sin(80^\circ)=-25.399$$ So the barge is heading in a southerly direction.
Using Pythagoras, the resultant speed of the barge is given by $$\sqrt{8.7123^2+(-25.399)^2}=26.851$$
As for the direction in which the barge is travelling, the clockwise angle with respect to the East axis will be $$\arctan\left(\frac{25.399}{8.7123}\right)=71.1^\circ$$ resulting in a heading of $161.1^\circ$