What Is the Net Force Using Parallelogram Method?

[Adjoint]

Here, we have to find OD from triangle ODE. But for applying cosine rule to ODE we should find the (green) angle ∠DEO, right?
So find ∠DOE

I am really sorry. I made a typing mistake the second time. It should be ∠DEO both times.

Now this is getting a bit too long discussion. And everyone can see that you have tried for long and you seem to know all relevant equations but maybe just getting lost in the details. So I hope forum Moderators won't mind if I do a little rescue mission:

In ΔOAC the $\angle$CAO = 30 degree
Using the rule of sines in this tringle,
$\frac{OC}{sin30}$ = $\frac{AC}{sinAOC}$
→ $\frac{16.1}{sin30}$ = $\frac{20}{sinAOC}$
→ $\angle$AOC = 38.4 degree
Now ∠EOC = ∠AOC + 90
→ ∠EOC = 128.4 degree
So from parallelogram EOCD, ∠DEO = 180 - ∠EOC = 180 - 128.4 = 51.6 degree (that's the green angle)

Now we can apply the law of cosines in ΔDEO to get,
OD = (ED² + OE² - 2.OE.ED.cos(51.6))^1/2
Finally, OD = 32.5

Next, to calculate direction, take ΔDEO and apply law of sines.
OD/sin∠51.5 = DE/sin∠DOE
→ 32.5/sin51.6 = 16.1/sin∠DOE
→ ∠DOE = 23

Now it can be easily seen that OD is (90-23-15) = 52 degree South of West.