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To solve this problem, we’ll calculate the displacement vector \( D \) by determining the missing side (vector \( D \)) of this closed quadrilateral shape. We need the \( x \)- and \( y \)-components for each measured side vector and apply vector addition to find the resultant displacement.

### Given Information
1. **Vector A**: \( 3.70 \, \text{km} \) at \( 7.5^\circ \) below the negative \( x \)-axis (South of West).
2. **Vector B**: \( 1.95 \, \text{km} \) at \( 16^\circ \) above the positive \( x \)-axis (North of East).
3. **Vector C**: \( 2.38 \, \text{km} \) along the positive \( y \)-axis (due North).

### Solution Steps
1. **Calculate the \( x \)- and \( y \)-components** of vectors \( A \), \( B \), and \( C \).
2. **Add the components** of vectors \( A \), \( B \), and \( C \) to find the resultant displacement needed for vector \( D \) to close the shape.
3. **Determine the magnitude and direction** of vector \( D \).

Let’s go through these calculations:

#### Step 1: Components of Vectors

1. **Vector A**:
   \[
   A_x = -3.70 \cos(7.5^\circ), \quad A_y = -3.70 \sin(7.5^\circ)
   \]

2. **Vector B**:
   \[
   B_x = 1.95 \cos(16^\circ), \quad B_y = 1.95 \sin(16^\circ)
   \]

3. **Vector C**: Since it’s directly upward,
   \[
   C_x = 0, \quad C_y = 2.38
   \]

#### Step 2: Summing the Components

The sum of the components of vectors \( A \), \( B \), and \( C \) will provide the net displacement, which we can use to determine vector \( D \).

#### Step 3: Finding Vector \( D \)

To get the magnitude and direction of \( D \), calculate:
\[
D = \sqrt{D_x^2 + D_y^2}
\]
\[
\text{Direction of } D = \tan^{-1}\left(\frac{D_y}{D_x}\right)
\]

Let me compute these values for you.### Calculations Summary

1. **Vector Components**:
   - **Vector A**: \( A_x = -3.67 \, \text{km} \), \( A_y = -0.48 \, \text{km} \)
   - **Vector B**: \( B_x = 1.87 \, \text{km} \), \( B_y = 0.54 \, \text{km} \)
   - **Vector C**: \( C_x = 0 \, \text{km} \), \( C_y = 2.38 \, \text{km} \)

2. **Resultant Displacement (Vector D)**:
   - **Components of \( D \)**:
     \[
     D_x = 1.79 \, \text{km}, \quad D_y = -2.43 \, \text{km}
     \]
   - **Magnitude**:
     \[
     D = 3.02 \, \text{km}
     \]
   - **Direction**: \( 53.6^\circ \) west of south.

This provides the required displacement to close the quadrilateral plot of land.

Would you like more details or have any questions?

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#### Additional Related Questions

1. How would changing the angles affect the resultant displacement vector \( D \)?
2. Can the resultant direction be expressed relative to other compass directions?
3. How can vector addition be applied to solve similar real-life navigation problems?
4. What are the effects of rounding on the precision of vector component calculations?
5. How would this process change if we had more than four sides?

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#### Tip:
When calculating vector components, always double-check angle orientation relative to axes, as small changes can significantly impact results.

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A (3.70 km), B (1.95 km), and C (2.38 km) in the figure below, and then correctly calculates the length and orientation of the fourth side D. What is his result?

This problem explores vector addition and trigonometry to calculate the unknown displacement needed to close a four-sided plot of land. Given three sides with distances and directions, learn how to find the fourth side's magnitude and orientation using trigonometric formulas and vector components.

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