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find the magnitude of R and the directional angle θ

To solve the problem, we need to find the magnitude of the resultant displacement vector \( \mathbf{R} = \mathbf{A} + \mathbf{B} \), as well as the directional angle \( \theta \) of \( \mathbf{R} \).

### Step 1: Decompose Vectors into Components
1. **Vector \( \mathbf{A} \)**:
   - Magnitude \( A = 220 \, \text{km} \)
   - Direction \( \theta_A = 30^\circ \) north of east

   The components of \( \mathbf{A} \) are:
   \[
   A_x = A \cos \theta_A = 220 \cos 30^\circ
   \]
   \[
   A_y = A \sin \theta_A = 220 \sin 30^\circ
   \]

2. **Vector \( \mathbf{B} \)**:
   - Magnitude \( B = 172 \, \text{km} \)
   - Direction \( \theta_B = 52^\circ \) west of north

   Convert this direction to standard form: 
   \( 90^\circ + 52^\circ = 142^\circ \) from the positive x-axis.

   The components of \( \mathbf{B} \) are:
   \[
   B_x = B \cos \theta_B = 172 \cos 142^\circ
   \]
   \[
   B_y = B \sin \theta_B = 172 \sin 142^\circ
   \]

### Step 2: Sum the Components
Add the corresponding components to find \( R_x \) and \( R_y \):
\[
R_x = A_x + B_x
\]
\[
R_y = A_y + B_y
\]

### Step 3: Find the Magnitude of \( \mathbf{R} \)
The magnitude of \( \mathbf{R} \) is given by:
\[
R = \sqrt{R_x^2 + R_y^2}
\]

### Step 4: Find the Directional Angle \( \theta \)
The angle \( \theta \) (relative to the positive x-axis) is found using:
\[
\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]

Let's compute the components and the final answers:

1. **Vector \( \mathbf{A} \) Components**:
   \[
   A_x = 220 \cos 30^\circ = 220 \times \frac{\sqrt{3}}{2} \approx 190.5 \, \text{km}
   \]
   \[
   A_y = 220 \sin 30^\circ = 220 \times \frac{1}{2} = 110 \, \text{km}
   \]

2. **Vector \( \mathbf{B} \) Components**:
   \[
   B_x = 172 \cos 142^\circ \approx 172 \times (-0.7880) \approx -135.5 \, \text{km}
   \]
   \[
   B_y = 172 \sin 142^\circ \approx 172 \times 0.6157 \approx 105.8 \, \text{km}
   \]

3. **Sum the Components**:
   \[
   R_x = 190.5 + (-135.5) \approx 55 \, \text{km}
   \]
   \[
   R_y = 110 + 105.8 \approx 215.8 \, \text{km}
   \]

4. **Magnitude of \( \mathbf{R} \)**:
   \[
   R = \sqrt{55^2 + 215.8^2} \approx \sqrt{3025 + 46591.64} \approx \sqrt{49616.64} \approx 222.8 \, \text{km}
   \]

5. **Directional Angle \( \theta \)**:
   \[
   \theta = \tan^{-1}\left(\frac{215.8}{55}\right) \approx \tan^{-1}(3.924) \approx 75.4^\circ \text{ north of east}
   \]

### Final Answers:
- The magnitude of \( \mathbf{R} \) is approximately **222.8 km**.
- The directional angle \( \theta \) is approximately **75.4°** north of east.

Would you like more details or have any questions?

Here are some related questions:
1. How would the calculation change if \( \mathbf{A} \) was directed south of east?
2. What if \( \mathbf{B} \) had a magnitude of 200 km instead?
3. Can the resultant angle ever be greater than \( 90^\circ \)?
4. How does the angle change if both vectors were of equal magnitude?
5. What is the significance of the angle measured counterclockwise from the positive x-axis?

**Tip**: Always convert directional angles into standard form before decomposing vectors.

A pilot flies two straight-line segments. The displacement vector A for the first segment has a magnitude of 220 km with a direction 30° north of east. The displacement vector B for the second segment has a magnitude of 172 km in a direction west of north. The resultant displacement vector R = A + B, and we need to find its magnitude and direction angle. Use the component method to find the magnitude and directional angle.

Learn how to find the magnitude and directional angle of a resultant vector when given two vector displacements. This problem demonstrates vector decomposition and the component method. Suitable for students in Grades 9-12.

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