Chapter 6.1, Problem 48E

To determine

Tofind:The distance $d$ from the boat to the shoreline.

Explanation of Solution

Consider the figure shown below.

  

Let the distance from point $B$ to point $N$ be $x$ .

In right triangle $LNB$ .

  $\begin{array}{c}\tan 27°=\frac{d}{x}\\ d=x\tan 27°\end{array}$

In right triangle $LNA$ .

  $\begin{array}{c}\tan 20°=\frac{d}{x+2.5}\\ d=\left(x+2.5\right)\tan 20°\end{array}$

Equate the expression of $d$ to solve for $x$ .

  $\begin{array}{c}x\tan 27°=\left(x+2.5\right)\tan 20°\\ x\tan 27°-x\tan 20°-2.5\tan 20°=\\ x=\frac{2.5\tan 20°}{\tan 27°-\tan 20°}\\ =6.25\text{miles}\end{array}$

The distance $d$ is calculated as,

  $\begin{array}{c}d=x\tan 27°\\ =6.25\tan 27°\\ =3.19\text{miles}\end{array}$

Therefore, the distance $d$ from the boat to the shoreline is $\text{3}\text{.19}\text{miles}$ .