Chapter 8.2, Problem 46AYU

To determine

To find: The time lost due to error.

Answer to Problem 46AYU

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

Explanation of Solution

Given:

The error of the course is $10°$ .

The distance covered by the aircraft before changing the course is $50\text{miles}$ .

The distance covered by the aircraft after changing the course is $70\text{miles}$ .

The constant speed of the aircraft is $250\text{mi}/\text{hr}$ .

The given diagram is shown in Figure 1

Figure 1

Calculation:

Consider the law of sine is,

  $\frac{\sin P}{p}=\frac{\sin Q}{q}$

Then,

  $\begin{array}{l}\frac{\sin 10°}{70}=\frac{\sin Q}{50}\\ \sin Q=50\left(\frac{\sin 10°}{70}\right)\\ Q={\sin }^{-1}\left[50\left(\frac{\sin 10°}{70}\right)\right]\\ Q=7.12\end{array}$

Consider the angle sum property of triangle is,

  $\begin{array}{l}P+Q+R=180°\\ R+17.12°=180°\\ R=162.88°\end{array}$

Consider the law of sine to find the value of $r$ is,

  $\begin{array}{l}\frac{\sin 10°}{70}=\frac{\sin 162.88°}{r}\\ r=70\left[\frac{\sin 162.88°}{\sin 10°}\right]\\ r=118.67\end{array}$

Thus, the time taken by the aircraft to cover the distance of $1186.7\text{miles}$ is,

  $\begin{array}{c}t=\frac{1186.7\text{mi}}{250\text{m}/\text{h}}\\ =0.47\text{hr}\end{array}$

The time take to cover the distance of $50\text{mi}$ and $70\text{mi}$ is,

  $\begin{array}{c}t=\frac{\left(50+70\right)}{250}\\ =0.48\text{hr}\end{array}$

The total time lost is,

  $\begin{array}{c}t=0.48\text{hr}-0.47\text{hr}\\ =0.\text{01}\end{array}$