Your results is partly correct. 

The problem in your solution is a confusion about angle $\theta$ from the cosine rule you are using. The triangle of forces will look like the following image:

[![enter image description here][1]][1]

where r is the angle in the cosine rule:

$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(r)$$

$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(180-\theta)$$

If this is superimposed to your initial image it will look like the following:

[![enter image description here][2]][2]

So what you calculate as 97.9 is actually equal to $180-\theta$. 

From there:

Therefore:
- $R_x = -868.75 [N]$
- $R_y =  495.25[N]$  (Negative because its pointing to the left)

If you take the magnitude of $R= \sqrt{(-868.75)^2 + 495.25^2}=1000$



---- 
## Another way to solve this


Lets write $F_1 = -800[N]$ and $F_2=500N$. Also assume the $\phi$ is angle with the positive x, therefore $\phi = 180-\theta$.

Then the resultant R will have:

$$R_x = F_1 + F_2 \cos(\phi) \qquad R_y=F_2 \sin(\phi)$$

Then $$|R|= \sqrt{(F_1 + F_2 \cos(\phi))^2 + (F_2 \sin(\phi))^2 }$$

Solving for this you get

$$\phi= 97.9[deg]\rightarrow \theta = 82.1$$


and the angle between the resultant and the horizontal +x (assuming +x is to the right) is:

*I'll put up the answer when you show what you've tried*

$$\beta = 150.31[deg]$$ 

<!--
import sympy as sp
import numpy as np
R=1000
F1=-800
F2= 500 
a = sp.symbols('a')

eq2 = sp.Eq(R**2, (F1+F2*sp.cos(sp.pi-a))**2+ (F2*sp.sin(sp.pi-a))**2)
res = sp.solve([eq2], a)
a1=res[1][0]

float(sp.sqrt((F1+float(sp.pi-a1))**2 + (F2*sp.sin(sp.pi-a1))**2))
Rx = float(F1 +F2*sp.cos(sp.pi-a1))
Rx = float(F2*sp.sin(sp.pi-a1))
print(Rx)
-->


  [1]: https://i.sstatic.net/ZOHLAm.png
  [2]: https://i.sstatic.net/ATPAH.png