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

To find the magnitude of the vector sum, we first need to resolve vectors A and B into their x and y components. For Vector A (88.7 m at \( 310^{\circ} \)): - \( A_x = 88.7 \cos(310^{\circ}) \) - \( A_y = 88.7 \sin(310^{\circ}) \) Calculating the components: - \( A_x = 88.7 \cos(310^{\circ}) \approx 88.7 \cdot 0.953 = 84.60 \, \text{m} \) - \( A_y = 88.7 \sin(310^{\circ}) \approx 88.7 \cdot (-0.302) = -26.75 \, \text{m} \) For Vector B (59.6 m at \( 230^{\circ} \)): - \( B_x = 59.6 \cos(230^{\circ}) \) - \( B_y = 59.6 \sin(230^{\circ}) \) Calculating the components: - \( B_x = 59.6 \cos(230^{\circ}) \approx 59.6 \cdot (-0.766) = -45.69 \, \text{m} \) - \( B_y = 59.6 \sin(230^{\circ}) \approx 59.6 \cdot (-0.643) = -38.29 \, \text{m} \) Now, we add the x and y components: - Total \( X \): \( A_x + B_x = 84.60 + (-45.69) = 38.91 \, \text{m} \) - Total \( Y \): \( A_y + B_y = -26.75 + (-38.29) = -65.04 \, \text{m} \) Finally, the magnitude of the resultant vector \( R \) is given by: \[ R = \sqrt{(R_x)^2 + (R_y)^2} \] \[ R = \sqrt{(38.91)^2 + (-65.04)^2} \] Calculating: \[ R = \sqrt{1518.54 + 4228.93} = \sqrt{5747.47} \approx 75.8 \, \text{m} \] So, the magnitude of the vector sum is approximately 75.8 m.