Chapter 8.2, Problem 49AYU

To determine

What is the length of highway needed to go around the bay?

Answer to Problem 49AYU

  $\boxed{2.64miles}$

Explanation of Solution

Given information:

U.S. $41$ , a highway whose primary directions are north-south, is being constructed along the west coast of Florida. Near Naples, a bay obstructs the straight path of the road. Since the cost of the bridge is prohibitive, engineers decide to go around the bay. The illustration shows the path that they decide on and the measurement taken. What is the length of highway needed to go around the bay?

Calculation:

Consider the figure,

The length of the highway,is sum of sides, $AB,BCandCD$

  

Consider $\Delta OAD$ ,angle $\angle OAD$ is $180°-140°=40°$

Same as

  $\angle ODAis180°-135°=45°$

Now sum of all the angles of triangles,

  $\begin{array}{l}40°+\angle AOD+45°=180°\\ \angle AOD=180°-85°\\ \angle AOD=95°\end{array}$

Using law of sine, to calculate length $AO$ and $OD$

  $\begin{array}{l}\frac{\sin 95°}{2}=\frac{\sin 45°}{AO}\\ 0.4980=\frac{\sin 45°}{AO}\\ 0.4980\left(AO\right)=\sin 45°\\ AO=\frac{\sin 45°}{0.4980}\\ AO=1.42\end{array}$

Now take,

  $\begin{array}{l}\frac{\sin 95°}{2}=\frac{\sin 40°}{DO}\\ 0.4980=\frac{\sin 40°}{DO}\\ 0.4980\left(DO\right)=\sin 40°\\ DO=\frac{\sin 40°}{0.4980}\\ DO=1.29\end{array}$

Now find length $AB$ and $DC$

  $\begin{array}{l}AB=AO-\frac{1}{8}\\ 1.42-\frac{1}{8}=1.295\\ CD=DO-\frac{1}{8}\\ 1.29-\frac{1}{8}=1.165\end{array}$

Now length $BC$ can be found by using law of sine,

  $\begin{array}{l}\frac{\sin 95°}{BC}=\frac{\sin 45°}{1/8}\\ \frac{\sin 95°}{BC}=5.6568\\ BC\left(5.6568\right)=\sin 95°\\ BC=\frac{\sin 95°}{5.6568}\\ BC=0.176\end{array}$

Now the total length of high way,

  $\begin{array}{l}AB+BC+CD\\ 1.295+1.165+0.176=2.64miles\end{array}$

Hence, the length of high way, is, $\boxed{2.64miles}$