Chapter 8, Problem 55RE

To determine

To find:The length of highway needed to go around the bay.

Explanation of Solution

Consider the figure shown below.

  

Figure (1)

In triangle $ABC$ .

  $\begin{array}{c}B=180°-60°-65°\\ =55°\end{array}$

Apply the sine law.

  $\begin{array}{c}\frac{\sin 60}{a}=\frac{\sin 55}{3}\\ a=3\times \frac{\sin 60°}{\sin 55°}\\ =3.172\end{array}$

And,

  $\begin{array}{c}\frac{\sin 55°}{3}=\frac{\sin 65°}{c}\\ c=3\times \frac{\sin 65°}{\sin 55°}\\ =3.32\end{array}$

Now, triangle EDB use the law of cosine.

  $\begin{array}{c}ED=\sqrt{{0.25}^2+{0.25}^2-2\left(0.25\right)\left(0.25\right)\cos 55°}\\ =0.25\sqrt{1+1-2\left(0.573\right)}\\ =0.378\end{array}$

Now,

  $\begin{array}{c}AE=AB-EB\\ =3.22-0.25\\ =2.97\end{array}$

And,

  $\begin{array}{c}CD=BC-DB\\ =3.172-0.25\\ =2.922\end{array}$

The total length of the road is,

  $\begin{array}{c}R=CD+ED+AE\\ =2.922+0.378+2.97\\ =6.27\text{mi}\end{array}$

Therefore, the length of the required highway $6.27\text{mi}$ .