Let's break down the problem and solve it step by step.

Given:

• The magnitude of force $P = 600 \, \text{N}$
• The magnitude of the second force is $900 \, \text{N}$
• The angle between the forces $P$ and the 900 N force is $45^\circ$
• The resultant force $R$ is vertical.

We need to determine the required angle $\theta$ so that the resultant force is vertical, and the corresponding value of $R$.

Step 1: Resolve forces into components

Since the resultant $R$ is vertical, the horizontal components of the two forces must cancel each other out.

Horizontal Components:

• The horizontal component of $P$ is: $P_x = P \sin(\theta)$

• The horizontal component of the 900 N force is: $(900 \, \text{N})_x = 900 \cos(45^\circ)$

For the resultant force to be vertical, the sum of the horizontal components must be zero: $P \sin(\theta) = 900 \cos(45^\circ)$

Substitute $P = 600 \, \text{N}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}$: $600 \sin(\theta) = 900 \times \frac{1}{\sqrt{2}}$ $600 \sin(\theta) = \frac{900}{\sqrt{2}} \approx 636.396$

Solve for $\sin(\theta)$: $\sin(\theta) = \frac{636.396}{600} \approx 1.0607$

Since $\sin(\theta)$ cannot exceed 1, this suggests that the assumption needs adjustment, possibly revisiting how the components interact based on the problem's conditions. I can reprocess this by adjusting