Question

a ship is sailing due north. at a certain point, the bearing of a lighthouse 5.9 km away is n38.8° e. later on, the captain notices that the bearing of the lighthouse has become s30.6° e. how far did the ship travel between the two observations of the lighthouse? the ship travelled □ km between the two observations. (do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Analyze the triangle

We have a triangle formed by the ship's path (due north) and the two lines of sight to the lighthouse. Let's denote the initial position of the ship as \( A \), the later position as \( B \), and the lighthouse as \( L \). We know \( AL = 5.9 \) km, \( \angle LAB = 90^\circ - 38.8^\circ = 51.2^\circ \), and \( \angle LBA = 90^\circ - 30.6^\circ = 59.4^\circ \). Then \( \angle ALB = 180^\circ - 51.2^\circ - 59.4^\circ = 69.4^\circ \).

Step2: Apply the Law of Sines

Using the Law of Sines, \( \frac{AB}{\sin \angle ALB} = \frac{AL}{\sin \angle LBA} \). So \( AB = \frac{AL \cdot \sin \angle ALB}{\sin \angle LBA} \).

Step3: Substitute the values

\( AL = 5.9 \), \( \angle ALB = 69.4^\circ \), \( \angle LBA = 59.4^\circ \). So \( AB = \frac{5.9 \cdot \sin(69.4^\circ)}{\sin(59.4^\circ)} \).

Calculate \( \sin(69.4^\circ) \approx 0.936 \), \( \sin(59.4^\circ) \approx 0.861 \). Then \( AB = \frac{5.9 \cdot 0.936}{0.861} \approx \frac{5.9 \cdot 0.936}{0.861} \approx \frac{5.5224}{0.861} \approx 6.41 \).

Answer:

\( 6.4 \) (rounded to the nearest tenth)