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.