The triangle or parallelogram of forces follows the Cosine Rule. The diagram below is a representation of the problem. The blue lines represent the banks of the canal and the solid red lines represent the ropes and the tension in them. W is the vessel. WX has a tension of 200lbs and WZ a tension of 2400lbs. The green line WY is the resultant force and also the direction in which the vessel moves as it is being towed. ∠XWZ=60°, the angle between the ropes. Therefore ∠YZW=120°. WY=√(WZ2+YZ2-2WZ.YZcos120°). Length YZ=WX (parallel sides of parallelogram WZYZ).
WY=√(24002+2002+2(2400)(200)/2)=√(5800000+480000)=√6280000=2506lbs approx. The angle YWZ has the same measure as the the angle between rope WZ and the bank, and ∠XYW has the same measure as ∠YWZ. XY=WZ=2400.
(Note that the solid red lines do not necessarily represent the lengths of the ropes, which will depend on the type of material, strength, and other factors.)
Drop a perpendicular XN from X on to WZ, then XN=WXsin60° and WN=WXcos60°.
tanXYW=XN/(XY+WN)=200sin60°/(2400+200cos60°)=100√3/(2400+100)=√3/25. So ∠XYW=tan-1(√3/25)=3.96° approx.
To find the angle of the other rope to the bank we can see that this angle=180°-(∠YZW+∠XYW)=180-(120+3.96)=56.04° approx. Note that 56.04°+3.96°=60°, from ∠XWZ=∠XWY+∠YWZ, so we could have used this fact to find the required angles that the ropes make with the banks.