Two ships leave a harbor at the same time. One ship travels on a bearing \( \mathrm{S} 11^{\circ} \mathrm{W} \) at 14 miles per hour. The other ship travels on a bearing \( \mathrm{N} 75^{\circ} \mathrm{E} \) at 9 miles per hour. How far apart will the ships be after 3 hours? The distance is approximately \( \square \) miles. (Round to the nearest tenth as needed.)

To find the distance between the two ships after 3 hours, we need to first determine their individual positions. 1. **Calculate the positions of the ships after 3 hours:** - For the ship traveling on a bearing of \( \mathrm{S} 11^{\circ} \mathrm{W} \): - Speed: 14 miles per hour - Distance after 3 hours: \( 14 \times 3 = 42 \text{ miles} \) - Angle relative to North: \( 180^{\circ} + 11^{\circ} = 191^{\circ} \) - Using trigonometric functions: - East-West component (x): \( 42 \cos(191^{\circ}) \) - North-South component (y): \( 42 \sin(191^{\circ}) \) - For the ship traveling on a bearing of \( \mathrm{N} 75^{\circ} \mathrm{E} \): - Speed: 9 miles per hour - Distance after 3 hours: \( 9 \times 3 = 27 \text{ miles} \) - Angle relative to North: \( 75^{\circ} \) - Using trigonometric functions: - East-West component (x): \( 27 \cos(75^{\circ}) \) - North-South component (y): \( 27 \sin(75^{\circ}) \) 2. **Calculating the components:** - Ship 1 (S 11° W): - \( x_1 = 42 \cos(191^{\circ}) \approx 42 \times (-0.9816) \approx -41.17 \) - \( y_1 = 42 \sin(191^{\circ}) \approx 42 \times (-0.1908) \approx -8.00 \) - Ship 2 (N 75° E): - \( x_2 = 27 \cos(75^{\circ}) \approx 27 \times (0.2588) \approx 6.99 \) - \( y_2 = 27 \sin(75^{\circ}) \approx 27 \times (0.9659) \approx 26.01 \) 3. **Finding the distance between the two ships:** - The x-coordinates are \( -41.17 \) and \( 6.99 \) - The y-coordinates are \( -8.00 \) and \( 26.01 \) - The distance \( D \) between the ships can be calculated using the distance formula: \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ = \sqrt{(6.99 - (-41.17))^2 + (26.01 - (-8.00))^2} \] \[ = \sqrt{(6.99 + 41.17)^2 + (26.01 + 8.00)^2} \] \[ = \sqrt{(48.16)^2 + (34.01)^2} \] \[ = \sqrt{2315.0256 + 1154.6801} \] \[ \approx \sqrt{3469.7057} \approx 58.8 \text{ miles} \] Thus, the distance between the two ships after 3 hours is approximately \( 58.8 \) miles.