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NMech
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Your results is partly correct. The angle $\theta$ is correct. I'm finding a different at the second part.

Lets write $F_1 = 800[N]$ and $F_2=500N$.

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

$$a= 97.09[deg]$$

Therefore:

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

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

$$\beta = 150.31[deg]$$

NMech
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