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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:

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:

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

$$\beta = 150.31[deg]$$

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:

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:

So what you calculate as 97.9 is actually equal to $180-\theta$. Therefore $\theta$ in your original drawing is $82.096[deg]$.

When you substitute, you can obtain the components of $R_x, R_y$ (notice the use of $\phi$). Therefore:

• $R_x = F_1 + F_2 \cos(\phi) = -868.75[N]$
• $R_y = F_2 \sin(\phi) = 495.25[N]$ (Negative because its pointing to the left)

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

Regarding the angle between R (lets denote it $\phi_R$) and the positive X axis:

$$\tan(\phi_R)=\frac{R_y}{R_x} = \frac{495}{-868.75} $$ $$\phi_R=150[deg]$$

Another way

I used the following method to find the angle $\theta$. It leads to the same values.

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

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

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

$$\beta = 150.31[deg]$$

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:

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:

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

$$\beta = 150.31[deg]$$

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

$$\beta = 150.31[deg]$$

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

$$\beta = 150.31[deg]$$