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To solve this problem, we need to find the magnitudes of forces \(Y\) and \(Z\), given that the three forces \(X\), \(Y\), and \(Z\) are in equilibrium.

### Step-by-Step Solution

1. **Understanding the Equilibrium Condition:**
   Since the forces are in equilibrium, the sum of all forces in both the x (horizontal) and y (vertical) directions must be zero. This gives us two equations:

   \[
   \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0
   \]

2. **Resolve Forces into Components:**

   Let's resolve the forces into their x and y components:

   - **Force \(X\):**
     - \(X\) is purely horizontal and points to the left.
     - \(X_x = -X\), \(X_y = 0\)

   - **Force \(Y\):**
     - \(Y\) makes a 30° angle with the horizontal axis.
     - \(Y_x = Y \cos(30°)\), \(Y_y = Y \sin(30°)\)

   - **Force \(Z\):**
     - \(Z\) makes a 45° angle with the horizontal axis.
     - \(Z_x = Z \cos(45°)\), \(Z_y = Z \sin(45°)\)

3. **Set Up the Equilibrium Equations:**

   - For the x-direction:
     \[
     -X + Y \cos(30°) + Z \cos(45°) = 0
     \]

   - For the y-direction:
     \[
     Y \sin(30°) - Z \sin(45°) = 0
     \]

4. **Simplify and Solve:**

   - The y-direction equation becomes:
     \[
     Y \cdot \frac{1}{2} = Z \cdot \frac{\sqrt{2}}{2}
     \]
     \[
     Y = Z \sqrt{2}
     \]

   - Substitute \(Y = Z \sqrt{2}\) into the x-direction equation:
     \[
     -X + (Z \sqrt{2}) \cos(30°) + Z \cos(45°) = 0
     \]

   - Substitute the trigonometric values:
     \[
     -X + Z \sqrt{2} \cdot \frac{\sqrt{3}}{2} + Z \cdot \frac{\sqrt{2}}{2} = 0
     \]
     \[
     -X + Z \left( \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2} \right) = 0
     \]
     \[
     -X + Z \frac{\sqrt{6} + \sqrt{2}}{2} = 0
     \]

   - Solve for \(Z\):
     \[
     Z = \frac{2X}{\sqrt{6} + \sqrt{2}}
     \]

   - To find \(Y\), substitute \(Z\) back into \(Y = Z \sqrt{2}\):
     \[
     Y = \frac{2X \sqrt{2}}{\sqrt{6} + \sqrt{2}}
     \]

### Conclusion

The magnitudes of forces \(Y\) and \(Z\) in terms of \(X\) are:

\[
Z = \frac{2X}{\sqrt{6} + \sqrt{2}}, \quad Y = \frac{2X \sqrt{2}}{\sqrt{6} + \sqrt{2}}
\]

Would you like more details or have any other questions?

### Related Questions:
1. How would the solution change if the angle between forces Y and Z was different?
2. How do trigonometric identities simplify solving for forces in equilibrium problems?
3. What is the role of vector components in determining force equilibrium?
4. Can you explain how the law of sines or cosines might be used to solve similar problems?
5. How does the concept of torque relate to force equilibrium?

**Tip:** Always check the direction and components of all forces when analyzing equilibrium problems.

The three forces, X, Y, Z shown below are in a state of equilibrium. Determine the magnitude of forces Y and Z.

Learn how to determine the magnitude of forces Y and Z in an equilibrium problem involving three forces. This solution covers vector resolution, the use of trigonometric functions, and equilibrium conditions, suitable for students in Grades 10-12.

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