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?

There is a highway constructed along the west coast of Florida and its primary directions are north-south, near Naples a bay constructs the path of the road and engineers decided to go around the bay. We have to find the length of the highway around the bay.

By taking the measurements from the illustration, we sketch a rough figure

To find the length MN,take the traingle MNO

Angle Mand$140°$are supplementary angle, by property of supplementary angles $$\begin{gathered} 140°+\angle M=180° \\ \angle M=40° \end{gathered}$$

Let's apply the sine function

role="math" localid="1647336718209" $$\begin{gathered} \sin M=\frac{NO}{MN} \\ \sin 40°=\frac{1}{a} \\ a=\frac{1}{\sin 40°} \\ a\approx 1.557mi \end{gathered}$$

To find the length RS,take the traingle SQR

Angle Sand$130°$are supplementary angle, by property of supplementary angles$$\begin{gathered} 130°+\angle S=180° \\ \angle S=50° \end{gathered}$$

Let's apply the sine function

$$\begin{gathered} \sin S=\frac{RQ}{RS} \\ \sin 50°=\frac{1}{c} \\ c=\frac{1}{\sin 50°} \\ c\approx 1.30mi \end{gathered}$$

From the figure, we can conclude that

$$\begin{gathered} MO+OQ+QS=3 \\ MO+NR+QS=3 \\ NR=3-MO-QS \end{gathered}$$

Now, we have to find the length of MOand QS

To find the length of MO, take the triangle MON

$$\begin{gathered} \tan M=\frac{NO}{MO} \\ \tan 40°=\frac{1}{x} \\ x=\frac{1}{\tan 40°} \\ x\approx 1.19mi \end{gathered}$$

To find the length of QS,take the traingle SQR

$$\begin{gathered} t\mathrm{an}S=\frac{RQ}{QS} \\ \tan 50°=\frac{1}{y} \\ y=\frac{1}{\tan 50°} \\ y\approx 0.83mi \end{gathered}$$

Thus,role="math" localid="1647337195115" $$\begin{gathered} NR=3-1.19-0.83 \\ NR\approx 0.98 \\ b\approx 0.98mi \end{gathered}$$

Now, the length of the highway is

$$\begin{gathered} a+b+c \\ =1.557+0.98+1.30 \\ =3.83mi \end{gathered}$$