A ship is sailing due north. At a certain point, the bearing of a lighthouse 8.8 km away is \( \mathrm{N} 35.3^{\circ} \mathrm{E} \). Later on, the capta notices that the bearing of the lighthouse has become \( \mathrm{S} 31.3^{\circ} \mathrm{E} \). How far did the ship travel between the two observations of the lighthouse? The ship travelled \( \square \mathrm{km} \) between the two observations. (Do not round until the final answer. Then round to the nearest tenth as needed.)

To solve this problem, we can use the law of cosines and some trigonometry. We will first determine the positions of the ship and the lighthouse at the two observations. 1. **Initial Position of the Ship and Lighthouse**: - Let the initial position of the ship be point \( A \). - The lighthouse is at point \( L \), which is 8.8 km away from point \( A \) at a bearing of \( \mathrm{N} 35.3^{\circ} \mathrm{E} \). - This means the angle from the north towards the east is \( 35.3^{\circ} \). We can find the coordinates of point \( L \) relative to point \( A \): \[ L_x = 8.8 \sin(35.3^{\circ}) \quad \text{(east component)} \] \[ L_y = 8.8 \cos(35.3^{\circ}) \quad \text{(north component)} \] Calculating these: \[ L_x = 8.8 \sin(35.3^{\circ}) \approx 8.8 \times 0.577 = 5.08 \text{ km} \] \[ L_y = 8.8 \cos(35.3^{\circ}) \approx 8.8 \times 0.819 = 7.21 \text{ km} \] So, the coordinates of the lighthouse \( L \) relative to \( A \) are approximately \( (5.08, 7.21) \). 2. **Second Position of the Ship**: - Let the new position of the ship be point \( B \). - The bearing of the lighthouse from point \( B \) is \( \mathrm{S} 31.3^{\circ} \mathrm{E} \), which means the angle from the south towards the east is \( 31.3^{\circ} \) or \( 180^{\circ} - 31.3^{\circ} = 148.7^{\circ} \) from the north. The coordinates of point \( L \) relative to point \( B \) can be expressed as: \[ L_x = -d + 8.8 \sin(31.3^{\circ}) \quad \text{(east component)} \] \[ L_y = -d + 8.8 \cos(31.3^{\circ}) \quad \text{(north component)} \] Where \( d \) is the distance the ship has traveled north from point \( A \) to point \( B \). We can calculate the coordinates of \( L \) from point \( B \): \[ L_x = 8.8 \sin(31.3^{\circ}) \approx 8.8 \times 0.515 = 4.54 \text{ km} \] \[ L_y = 8.8 \cos(31.3^{\circ}) \approx 8.8 \times 0.857 = 7.53 \text{ km} \] 3. **Setting Up the Equations**: Since the coordinates of \( L \) must be the same in both cases, we can set up the equations: \[ 5.08 = -d + 4.54 \] \[ 7.21 = -d + 7.53 \] From the first equation: \[ d = 4.54 - 5.08 = -0.54 \text{ km} \] From the second equation: \[ d = 7.53 - 7.21 = 0.32 \text{ km} \] Since \( d \) must be positive, we take the second equation. 4. **Distance Traveled**: The distance the ship traveled between the two observations is approximately \( 0.32 \) km. Thus, the ship traveled approximately \( \boxed{0.3} \) km between the two observations.