A cruise ship is headed due north for 50 miles. Then it turns 40 degrees to the right to avoid rough waters and sails for 25 more miles. What is the ship's displacement after these two maneuvers? Round the magnitude to the nearest tenth and the angle to the nearest whole number. Write the magnitude in the first box, the units in the second box and the direction in the third box. Leave out the degree sign in your answer.

To find the ship's total displacement, we can use the law of cosines and sine. First, the ship travels due north for 50 miles. After turning 40 degrees to the right, it moves off at an angle of 40 degrees east of due north for 25 miles. 1. Calculate the displacement vector components: - Northward component (y-axis): \( d_1 = 50 \) miles - Eastward component (x-axis) from the second leg: - \( (d_2 \cos(40°) = 25 \cos(40°) \approx 19.3 \) miles (to the east) - North component from the second leg: - \( (d_2 \sin(40°) = 25 \sin(40°) \approx 16.1 \) miles (to the north) 2. Adding up the components: - Total northward displacement: \( 50 + 16.1 = 66.1 \) miles - Total eastward displacement: \( 19.3 \) miles 3. Finally, find the magnitude of the resultant displacement using Pythagorean theorem: - Magnitude: \( \sqrt{(66.1^2 + 19.3^2)} \approx \sqrt{(4363.22 + 372.49)} \approx \sqrt{4735.71} \approx 68.8 \) miles 4. Determine the angle in relation to the north: - Angle: \( \tan^{-1}\left(\frac{19.3}{66.1}\right) \approx 16.5 \) Thus, the displacement of the ship is approximately: - Magnitude: 68.8 - Units: miles - Direction: 17