Chapter 14.1, Problem 8P

To determine

The amount of distance to be travelled and the heading of travel.

Explanation of Solution

Given information:

The distance between Mapleton to Logan beach is $600km$ and its bearing is $25°$ from Mapleton.

The plane first flies $140km$ and its heading is $325°$ , to avoid a storm.

Calculations:

To get the distance and bearing between them, first we draw the velocity diagram.

  

Here AB denotes the distance that has to be travelled by the plane. Its magnitude is $600$ and its direction is $25°$ .

Here AC denotes the distance first travelled by the plane. Its magnitude is $140km$ and its direction is $325°$ .

Therefore the remaining distance that has to be travelled is AC. From the diagram we can see that $\angle CAB=60°$

Let ,

  $\angle ACB=\alpha$

Now to determine BC we have to apply cosine law.

Therefore,

  $\begin{array}{l}B{C}^2=A{B}^2+A{C}^2-2\times AB\times AC\times \cos 60\\ B{C}^2={600}^2+{140}^2-2\times 600\times 140\times \cos 60\\ B{C}^2=295600\\ BC=\sqrt{295600}=543.69\end{array}$

Now to find the course of bearing we have to use law of sine.

  $\begin{array}{l}\frac{\sin \alpha }{600}=\frac{\sin 60}{543.69}\\ \sin \alpha =\frac{\sin 60}{543.69}\times 600\\ \sin \alpha =0.9557\\ \alpha ={\sin }^{-1}0.9557\\ \alpha =72.88°\end{array}$

Thus the distance between the two ships is $543.69km$ . The course of bearing between them is $180°-(72.88+35)°=72.12°$