Chapter 14.1, Problem 7P

To determine

The distance between two ships and bearing between them after three hours.

Explanation of Solution

Given information:

The two ships leave the port at noon.

The speed of one ship is $15km/h$ and its heading is $75°$ .

The speed of other ship is $24km/h$ and its heading is $155°$ .

Calculations:

The speed of the first ship is $15km/h$

Therefore distance travelled by it in three hours is $15km/h\times 3h=45km.$

The speed of the second ship is $24km/h$

Therefore distance travelled by it in three hours is $24km/h\times 3h=72km.$

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

  

Here AB denotes the distance travelled by first ship. Its magnitude is $45$ and its direction is $75°$ .

Here BC denotes the distance travelled by second ship. Its magnitude is $72$ and its direction is $155°$ .

Therefore the resultant speed is AC. From the diagram we can see that $\angle ABC=100°$

Let ,

  $\angle BAC=\alpha$

Now to determine AC we have to apply cosine law.

Therefore,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\times \cos 100\\ A{C}^2={45}^2+{72}^2-2\times 45\times 72\times \cos 100\\ A{C}^2=8334.24\\ AC=\sqrt{8334.24}=91.29\end{array}$

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

  $\begin{array}{l}\frac{\sin \alpha }{72}=\frac{\sin 100}{91.29}\\ \sin \alpha =\frac{\sin 100}{91.29}\times 72\\ \sin \alpha =0.77671\\ \alpha ={\sin }^{-1}0.77671\\ \alpha =50.96°\end{array}$

Thus the distance between the two ships is $91.29km$ . The course of bearing between them is $50.96°$