Homework Score: \( 83.33 \%, 8.33 \) of 10 points Two ships leave a harbor at the same time. One ship travels on a bearing \( \mathrm{S} 12^{\circ} \mathrm{W} \) at 17 miles per hour. The other ship travels on a bearing N \( 75^{\circ} \mathrm{E} \) at 10 miles p hour. How far apart will the ships be after 2 hours? The distance is approximately \( \square \) miles. (Round to the nearest tenth as needed.)

To solve the problem, first, we find the positions of both ships after 2 hours. 1. **Calculating the Positions:** - Ship A (bearing S 12° W): This means it is heading 12° west of south. - After 2 hours at 17 mph: \[ \text{Distance} = 17 \text{ mph} \times 2 \text{ hours} = 34 \text{ miles}. \] - The angle with respect to the north is 180° + 12° = 192°. - Ship B (bearing N 75° E): This means it is heading 75° east of north. - After 2 hours at 10 mph: \[ \text{Distance} = 10 \text{ mph} \times 2 \text{ hours} = 20 \text{ miles}. \] - The angle is 75°. 2. **Finding Coordinates:** - Ship A's coordinates (using polar to rectangular conversion): \[ x_A = 34 \sin(12°) \quad \text{and} \quad y_A = -34 \cos(12°). \] - Ship B's coordinates: \[ x_B = 20 \cos(75°) \quad \text{and} \quad y_B = 20 \sin(75°). \] 3. **Calculating the Distances:** Using trigonometric values: \[ x_A \approx 34 \times 0.2079 \approx 7.07, \quad y_A \approx -34 \times 0.9659 \approx -32.83, \] \[ x_B \approx 20 \times 0.2588 \approx 5.18, \quad y_B \approx 20 \times 0.9659 \approx 19.32. \] 4. **Distance Between Ships:** The distance \(d\) between the two points is given by: \[ d = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}. \] Substitute the coordinates: \[ d = \sqrt{(7.07 - 5.18)^2 + (-32.83 - 19.32)^2}. \] Calculating: \[ d = \sqrt{(1.89)^2 + (-52.15)^2} \approx \sqrt{3.56 + 2727.82} \approx \sqrt{2731.38} \approx 52.2 \text{ miles}. \] The ships will be approximately **52.2 miles** apart after 2 hours.