Let's work through the steps to get the resultant force using both the **trigonometric method** and the **analytical method** in detail.

### 1. **Trigonometric Method:**

#### Step 1: **Determine the angle between the forces.**
- Force \( \mathbf{F} \) is at \( 45^\circ \) northwest, which corresponds to \( 135^\circ \) from the positive x-axis.
- Force \( \mathbf{Q} \) is at \( 30^\circ \) northeast, which corresponds to \( 30^\circ \) from the positive x-axis.
- The angle \( \theta \) between the forces is:
  \[
  \theta = 135^\circ - 30^\circ = 105^\circ
  \]

#### Step 2: **Use the law of cosines to find the magnitude of the resultant force \( R \).**
\[
R = \sqrt{F^2 + Q^2 - 2FQ \cos \theta}
\]
Substituting the given values:
\[
R = \sqrt{450^2 + 200^2 - 2 \times 450 \times 200 \times \cos 105^\circ}
\]
\[
R = \sqrt{202500 + 40000 - 2 \times 450 \times 200 \times (-0.2588)}
\]
\[
R = \sqrt{242500 + 46584.48} \approx 537.67 \, \text{lb}
\]

#### Step 3: **Use the law of sines to find the direction \( \alpha \) of the resultant force.**
\[
\frac{R}{\sin \theta} = \frac{Q}{\sin \alpha}
\]
\[
\sin \alpha = \frac{Q \sin \theta}{R} = \frac{200 \times \sin 105^\circ}{537.67}
\]
\[
\alpha = \arcsin \left(\frac{200 \times 0.9659}{537.67}\right) \approx \arcsin(0.3593) \approx 21.06^\circ
\]
So, the direction of \( R \) is \( 21.06^\circ \) counterclockwise from the positive x-axis.

### 2. **Analytical Method:**

#### Step 1: **Break down each force into its x and y components.**
- For \( \mathbf{F} \) at \( 45^\circ \) northwest (\(135^\circ\) from the x-axis):
  \[
  F_x = 450 \times \cos 135^\circ = 450 \times -\frac{\sqrt{2}}{2} \approx -318.2 \, \text{lb}
  \]
  \[
  F_y = 450 \times \sin 135^\circ = 450 \times \frac{\sqrt{2}}{2} \approx 318.2 \, \text{lb}
  \]

- For \( \mathbf{Q} \) at \( 30^\circ \) northeast:
  \[
  Q_x = 200 \times \cos 30^\circ = 200 \times \frac{\sqrt{3}}{2} \approx 173.2 \, \text{lb}
  \]
  \[
  Q_y = 200 \times \sin 30^\circ = 200 \times \frac{1}{2} = 100 \, \text{lb}
  \]

#### Step 2: **Sum the components to get the resultant components.**
\[
R_x = F_x + Q_x = -318.2 \, \text{lb} + 173.2 \, \text{lb} = -145 \, \text{lb}
\]
\[
R_y = F_y + Q_y = 318.2 \, \text{lb} + 100 \, \text{lb} = 418.2 \, \text{lb}
\]

#### Step 3: **Find the magnitude of the resultant vector \( R \).**
\[
R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-145)^2 + (418.2)^2}
\]
\[
R = \sqrt{21025 + 174917.64} \approx 442.62 \, \text{lb}
\]

#### Step 4: **Determine the direction \( \alpha \) of the resultant vector.**
\[
\alpha = \tan^{-1} \left(\frac{R_y}{R_x}\right) = \tan^{-1} \left(\frac{418.2}{-145}\right)
\]
\[
\alpha \approx \tan^{-1}(-2.883) \approx 109.12^\circ
\]
So, the direction of \( R \) is \( 109.12^\circ \) counterclockwise from the positive x-axis.

### Summary:
- **Magnitude of Resultant:** \( R \approx 442.62 \, \text{lb} \) (analytical) or \( 537.67 \, \text{lb} \) (trigonometric)
- **Direction of Resultant:** \( \alpha \approx 109.12^\circ \) (analytical) or \( 21.06^\circ \) (trigonometric)

If you need any clarification or have further questions, feel free to ask!

### Related Questions:
1. How does the trigonometric method differ from the analytical method in terms of application?
2. What would happen if the angle between the two forces was exactly \( 90^\circ \)?
3. Can you explain how the resultant direction differs between the two methods?
4. How would the calculations change if the forces acted in opposite directions?
5. What if the magnitudes of the forces were doubled?

**Tip:** Always verify angles and ensure you correctly account for the direction (signs) when resolving forces into components.

Learn how to calculate the resultant force of two vectors using trigonometric and analytical methods. Detailed step-by-step solutions are provided using the law of cosines, law of sines, and vector components. Suitable for advanced high school and college students.

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