https://file.aiquickdraw.com/mathbot/file/1729515154006gddzpnln.jpg

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

Two forces are shown in Fig. Knowing that the magnitude of P is 600 N, determine the required angle θ, if the resultant R of the two forces is to be vertical. The corresponding value of R.

Solve for the angle θ where the resultant of two forces (600 N and 900 N) becomes vertical. This trigonometry and vector resolution problem involves using horizontal and vertical components of forces and is suitable for students in Grades 11-12.

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