Find the direction of the vector sum lector A is 88.7 m long in a \( 310^{\circ} \) direction. lector B is 59.6 m long in a \( 230^{\circ} \) direction.

To determine the direction of the vector sum of **Vector A** and **Vector B**, we'll follow these steps: 1. **Decompose Each Vector into Its Components:** - **Vector A:** - **Magnitude (|A|):** 88.7 m - **Direction (θ_A):** 310° - **Components:** \[ A_x = |A| \cdot \cos(\theta_A) = 88.7 \cdot \cos(310°) \approx 88.7 \cdot 0.6428 \approx 57.0 \, \text{m} \] \[ A_y = |A| \cdot \sin(\theta_A) = 88.7 \cdot \sin(310°) \approx 88.7 \cdot (-0.7660) \approx -67.9 \, \text{m} \] - **Vector B:** - **Magnitude (|B|):** 59.6 m - **Direction (θ_B):** 230° - **Components:** \[ B_x = |B| \cdot \cos(\theta_B) = 59.6 \cdot \cos(230°) \approx 59.6 \cdot (-0.6428) \approx -38.3 \, \text{m} \] \[ B_y = |B| \cdot \sin(\theta_B) = 59.6 \cdot \sin(230°) \approx 59.6 \cdot (-0.7660) \approx -45.7 \, \text{m} \] 2. **Sum the Components:** - **Total X-Component (R_x):** \[ R_x = A_x + B_x = 57.0 \, \text{m} + (-38.3 \, \text{m}) = 18.7 \, \text{m} \] - **Total Y-Component (R_y):** \[ R_y = A_y + B_y = -67.9 \, \text{m} + (-45.7 \, \text{m}) = -113.6 \, \text{m} \] 3. **Determine the Direction of the Resultant Vector:** The direction (θ) of the resultant vector can be found using the arctangent function: \[ \theta = \arctan\left(\frac{R_y}{R_x}\right) = \arctan\left(\frac{-113.6}{18.7}\right) \approx \arctan(-6.073) \approx -80.6° \] Since the resultant vector is in the **fourth quadrant** (positive \( R_x \) and negative \( R_y \)), the standard direction is measured clockwise from the positive x-axis: \[ \theta_{\text{standard}} = 360° - 80.6° \approx 279.4° \] **Final Answer:** The direction of the vector sum is approximately **279.4°**.