Chapter 8, Problem 33RE

Constructing a Highway 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 a bridge is prohibitive, engineers decide to go around the bay. The illustration shows the path that they decide on and the measurements taken. What is the length of highway needed to go around the bay?

To determine

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

Answer to Problem 33RE

The length of the highway is $\text{6}\text{.22}$ miles.

Explanation of Solution

Given:

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 a bridge is prohibitive, engineers decide to go around the bay. The illustration shows the path that they decide on and the measurements taken.

Formula used:

$\frac{\text{sinA}}{\text{a}}=\frac{\text{sinB}}{\text{b}}=\frac{\text{sinC}}{\text{c}}$

Calculation:

Let $\angle CAB=\alpha$, $\angle CBA=\beta$ and $\angle ACB=\gamma$.

$\alpha =180-120=60°$

$\beta =180-115=65°$

$\gamma =180-60-65=55°$

$=\frac{\text{sin}60°}{BC}=\frac{\text{sin}65°}{AC}=\frac{\sin 55°}{3}$

$BC=3*\frac{\text{sin}60°}{\text{sin}55°}$

$BC=3.17\text{ mi}$

$\frac{\sin 65°}{AC}=\frac{\sin 55}{3}$

$AC=\frac{3\text{ \&*\#x00A0;sin}65}{\text{sin}55}$

$AC=3.32\text{ mi}$

For the isosceles triangle,

$\angle \text{CDE }=\angle \text{CED }=\left(180-55\right)/2=62.5°$

$\text{BE }=3.17-.25=2.92\text{ mi}$

$\text{AD }=3.32-.25=3.07\text{ mi}$

$\frac{\text{Sin}55}{\text{DE}}=\frac{\text{sin}62.5}{0.25}$

$\text{DE }\approx 023\text{ mi}$

The length of the highway $=2.92+3.07+0.23=6.22 \text{mi}$.