Chapter 14.1, Problem 6P

To determine

The ground speed and true course bearing of a plane.

Explanation of Solution

Given information:

A plane’s air speed is $270km/h$ and its heading is $70°$ .

Wind of $50km/h$ is blowing from due north.

Calculations:

To get the ground speed and true course bearing, first we draw the velocity diagram.

  

Here AB denotes the plane’s air speed. Its magnitude is $270$ and its direction is $70°$ .

Here BC denotes the wind speed. Its magnitude is $50$ and its direction is from due north.

Therefore the resultant speed is AC. From the diagram we can see that $\angle ABC=70°$

Let ,

  $\angle BAC=\alpha$

Now to determine AC we have to apply cosine law.

Therefore,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\times \cos 70\\ A{C}^2={270}^2+{50}^2-2\times 270\times 50\times \cos 70\\ A{C}^2=66165.45\\ AC=\sqrt{66165.45}=257.22\end{array}$

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

  $\begin{array}{l}\frac{\sin \alpha }{50}=\frac{\sin 70}{257.22}\\ \sin \alpha =\frac{\sin 70}{257.22}\times 50\\ \sin \alpha =0.18266\\ \alpha ={\sin }^{-1}0.18266\\ \alpha =10.524°\end{array}$

Thus the ground speed of the plane is $257.22km/h$ . The course of bearing is $10.524°$