Find the direction of the vector sum Vector A is 2.84 m long in a \( 23.4^{\circ} \) direction. Vector B is 1.72 m long in a \( 14.5^{\circ} \) direction.

To find the direction of the vector sum of vectors A and B, we can break each vector into its components and then sum those components. ### Step 1: Break down the vectors into components **Vector A:** - Magnitude: \( A = 2.84 \, \text{m} \) - Angle: \( \theta_A = 23.4^{\circ} \) The components of vector A are: \[ A_x = A \cdot \cos(\theta_A) = 2.84 \cdot \cos(23.4^{\circ}) \] \[ A_y = A \cdot \sin(\theta_A) = 2.84 \cdot \sin(23.4^{\circ}) \] Calculating these: \[ A_x = 2.84 \cdot \cos(23.4^{\circ}) \approx 2.84 \cdot 0.9205 \approx 2.61 \, \text{m} \] \[ A_y = 2.84 \cdot \sin(23.4^{\circ}) \approx 2.84 \cdot 0.3987 \approx 1.13 \, \text{m} \] **Vector B:** - Magnitude: \( B = 1.72 \, \text{m} \) - Angle: \( \theta_B = 14.5^{\circ} \) The components of vector B are: \[ B_x = B \cdot \cos(\theta_B) = 1.72 \cdot \cos(14.5^{\circ}) \] \[ B_y = B \cdot \sin(\theta_B) = 1.72 \cdot \sin(14.5^{\circ}) \] Calculating these: \[ B_x = 1.72 \cdot \cos(14.5^{\circ}) \approx 1.72 \cdot 0.9659 \approx 1.66 \, \text{m} \] \[ B_y = 1.72 \cdot \sin(14.5^{\circ}) \approx 1.72 \cdot 0.2504 \approx 0.43 \, \text{m} \] ### Step 2: Sum the components Now we can sum the x and y components of vectors A and B: \[ R_x = A_x + B_x = 2.61 + 1.66 \approx 4.27 \, \text{m} \] \[ R_y = A_y + B_y = 1.13 + 0.43 \approx 1.56 \, \text{m} \] ### Step 3: Find the direction of the resultant vector The direction (angle \( \theta_R \)) of the resultant vector can be found using the arctangent function: \[ \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{1.56}{4.27}\right) \] Calculating this: \[ \theta_R \approx \tan^{-1}(0.365) \approx 20.1^{\circ} \] ### Conclusion The direction of the vector sum of vectors A and B is approximately \( 20.1^{\circ} \) from the positive x-axis.