Chapter 6, Problem 77RE

To determine

The resultant speed and direction of the airplane

Answer to Problem 77RE

  $R\approx 422.3mph$ and $\theta ={130.4}^0$ .

Explanation of Solution

An airplane has an airspeed of 430 miles per hour at a bearing of the wind velocity is $35$ miles per hour in N E.

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 ${135}^0-A$

Using law of cosines, solve for R$\begin{array}{l}R=\sqrt{{430}^2+{35}^2-2(430)(35)\cos {75}^0}\\ R\approx 422.3mph\end{array}$

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

  $\begin{array}{l}\frac{\sin A}{35}=\frac{\sin {75}^0}{422.3}\\ \sin A=\frac{35\sin {75}^0}{422.3}\\ A={\sin }^{-1}\frac{35\sin {75}^0}{422.3}\\ A\approx {4.6}^o\end{array}$

The bearing therefore is:

  $\begin{array}{l}\theta ={135}^0-{4.6}^0\\ \theta ={130.4}^0\end{array}$

Therefore, the resultant speed and bearing are $R\approx 422.3mph$ and $\theta ={130.4}^0$ .