Question

a ship is sailing due north. at a certain point, the bearing of a lighthouse 6.3 km away is n35.7°e. later on, the captain notices that the bearing of the lighthouse has become s32.6°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: Find angles of the triangle

Let the lighthouse be point \( L \), the first - observation point be \( A \), and the second - observation point be \( B \).
The initial bearing is \( N35.7^{\circ}E \), so the angle at \( A \) with respect to the north - south line is \( 35.7^{\circ} \). The second bearing is \( S32.6^{\circ}E \), so the angle at \( B \) with respect to the north - south line is \( 32.6^{\circ} \).
The angle at \( L \), \( \angle ALB=180-(35.7 + 32.6)=111.7^{\circ} \), \( \angle LAB = 35.7^{\circ} \), and \( \angle LBA=32.6^{\circ} \), and \( AL = 6.3\) km.

Step2: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle ALB\), we want to find \( AB \). We know that \(\frac{AB}{\sin\angle ALB}=\frac{AL}{\sin\angle LBA}\).
Substitute \( AL = 6.3\) km, \(\angle ALB = 111.7^{\circ}\), and \(\angle LBA=32.6^{\circ}\) into the Law of Sines formula:
\(AB=\frac{AL\times\sin\angle ALB}{\sin\angle LBA}\)
\(\sin111.7^{\circ}=\sin(90^{\circ}+ 21.7^{\circ})=\cos21.7^{\circ}\approx0.928\), \(\sin32.6^{\circ}\approx0.539\)
\(AB=\frac{6.3\times\sin111.7^{\circ}}{\sin32.6^{\circ}}=\frac{6.3\times0.928}{0.539}\)
\(AB=\frac{5.8464}{0.539}\approx10.8\)

Answer:

\(10.8\)