## Precalculus (6th Edition) Blitzer

The distance (to the nearest tenth) after both ships are anchored is approximately $71.6\text{ miles}$.
Consider the origin at the harbor with the $y$ axis pointing north. The position of the first ship is $\text{N}42{}^\circ \text{E}$ for $23\text{ miles}$, that is ${{x}_{1}}=r\cos \theta$. Substitute, $r=23$ and $\theta =90{}^\circ -42{}^\circ$. \begin{align} & {{x}_{1}}=r\cos \theta \\ & =23\cos \left( 90{}^\circ -42{}^\circ \right) \\ & =23\cos 48{}^\circ \\ & =15.4 \end{align} And \begin{align} & {{y}_{1}}=r\sin \theta \\ & =23\sin 48{}^\circ \\ & =17.1 \end{align} The final position of the second ship is ${{x}_{2}}=r\cos \theta$. Substitute, $r=72$ and $\theta =90{}^\circ +38{}^\circ$. \begin{align} & {{x}_{2}}=r\cos \theta \\ & =72\cos \left( 90{}^\circ +38{}^\circ \right) \\ & =72\cos 128{}^\circ \\ & =-44.3 \end{align} And \begin{align} & {{y}_{2}}=r\sin \theta \\ & =72\sin 128{}^\circ \\ & =56.7 \end{align} Now, the distance between the two points is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 15.4,17.1 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -44.3,56.7 \right)$. \begin{align} & d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \\ & =\sqrt{{{\left( 15.4-\left( -44.3 \right) \right)}^{2}}+{{\left( 17.1-56.7 \right)}^{2}}} \\ & =71.6 \end{align} Hence, the two ships are approximately $71.6\text{ miles}$ apart.