Chapter 8.2, Problem 45AYU

To determine

To find: The number of more frequent flyer miles that Adam receive when he takes connecting flight rather than direct flight.

Answer to Problem 45AYU

Adam covers $100.6\text{mi}$ more frequent flyer miles.

Explanation of Solution

Given:

The total time loss is of $0.\text{01}$ .

The bearing from Oklahoma city to Kansas city is $\text{N29}\text{.6}°\text{E}$ .

The bearing from Oklahoma city to ST Louis city is $\text{N57}\text{.7}°\text{E}$ .

The bearing from ST Louis city to Oklahoma is $\text{N57}\text{.7}°\text{E}$ .

The bearing from ST Louis city to Kansas is $\text{N79}\text{.4}°\text{E}$ .

Calculation:

Consider the required diagram is shown in Figure 1

Figure 1

Consider the value of the angle $\angle ABC$ by triangle angle sum property is,

  $\begin{array}{c}\angle ABC=180°-\left(79.4°+57.7°\right)\\ =42.9°\end{array}$

Consider the angle $\angle BAC$ is,

  $\begin{array}{c}\angle BAC=\angle BAS-\angle CAS\\ =56.7°-29.6°\\ =28.1°\end{array}$

The value of the angle $C$ by sum angle property is,

  $\begin{array}{c}\angle C=180°-\left(28.1°+71°\right)\\ =109°\end{array}$

Consider the sine property,

  $\begin{array}{l}\frac{\sin 28.1°}{AC}=\frac{\sin 109°}{461.1}\\ AC=229.7\end{array}$

Consider the sine property,

  $\begin{array}{l}\frac{\sin 42.9°}{BC}=\frac{\sin 109°}{461.1}\\ BC=331.97\\ BC=332\end{array}$

Then, the total distance covered by the flight first from Oklahoma to Kansas,

  $229.7+332=561.7\text{mi}$

The total distance covered in connecting the flight rather than direct flight is,

  $561.7\text{mi}-461.1\text{mi}=100.6\text{mi}$