Chapter 6, Problem 78RE

To determine

The resultant speed and direction of the airplane.

Answer to Problem 78RE

  $R\approx 740.5kph$ and $\theta ={32.1}^0$ .

Explanation of Solution

An airplane has an airspeed of 724 kilometres per hour at a bearing of 300 . The wind velocity is $32$ kilometres per hour from the west.

Calculation:

In air navigation, the bearing is measured clockwise from the north. Let the two vector of plane and wind as shown:

  

Now adding tip to tail and supplementing angles, will produce the triangle as below:

  

Where, R= resultant speed and

  $\theta =$ Bearing and is given by ${30}^0+A$

Using law of cosines, solve for R$\begin{array}{l}R=\sqrt{{724}^2+{32}^2-2(724)(32)\cos {120}^0}\\ R\approx 740.5kph\end{array}$

Now by using law of Sines, to find the value of A:

  $\begin{array}{l}\frac{\sin A}{32}=\frac{\sin {120}^0}{740.5}\\ \sin A=\frac{32\sin {120}^0}{740.5}\\ A={\sin }^{-1}\frac{32\sin {120}^0}{740.5}\\ A\approx {2.1}^o\end{array}$

The bearing therefore is:

  $\begin{array}{l}\theta ={30}^0+{2.1}^0\\ \theta ={32.1}^0\end{array}$

Therefore, the resultant speed and bearing are $R\approx 740.5kph$ and $\theta ={32.1}^0$ .