Some state trooper departments use aircraft to enforce highway speed limits. Suppose that one of the airplanes has a speed of $\mathbf{135}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}$in still air. It is flying straight north so that it is at all times directly above a north–south highway. A ground observer tells the pilot by radio that a$\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}$wind is blowing but neglects to give the wind direction. The pilot observes that in spite of the wind the plane can travel$\mathbf{135}\mathbf{}\mathbf{mi}$along the highway in$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{h}$. In other words, the ground speed is the same as if there were no wind. (a) From what direction is the wind blowing? (b) What is the heading of the plane; that is, in what direction does it point?

• The velocity of the plane with respect to the ground is,$V_{PG}=135\mathrm{mi}/\mathrm{h}$.
• The velocity of the plane with respect to air is,$V_{PA}=135\mathrm{mi}/\mathrm{h}$.
• The velocity of the air with respect to the ground is, $V_{AP}=70\mathrm{mi}/\mathrm{h}$.

The vector quantity can be resolved in the components along the x-axis and y-axis. The diagram representing the components of the vector along the x and y-axis is called a vector diagram.

We can find the direction of the wind by drawing a vector diagram with angles.

Formula:

The sine angle of an inclination,

$\sin \theta =\frac{\mathrm{frontside}\mathrm{or}\mathrm{perpendicularside}}{\mathrm{hypotenuse}}$ (i)

Since the speed of the plane in both situations remains the same, we can draw the vector diagram below. The triangle we get from this diagram is an isosceles triangle and can be divided into two right-angle triangles as shown below. Now we can apply trigonometry to the diagram to find the angle$\theta$.

From above vector diagram, we can write the expression for sin function and substitute the values of velocities to find the angle.

$$\begin{gathered} \sin \left(\frac{\theta }{2}\right)=\frac{V_{AG}}{2V_{PA}} \\ =\frac{(70\mathrm{mi}/\mathrm{h})}{(2\times 135\mathrm{mi}/\mathrm{h})} \\ \frac{\theta }{2}={\sin }^{-1}\left(0.26\right) \\ =15.{07}^0 \\ \theta =30.14° \\ \approx 30° \end{gathered}$$

Now using trigonometry, we can prove that the wind makes the same angle with the East west direction as the bisector makes in the North-South direction. So, the wind is blowing at $15°$north of west.

Due to wind, the plane cannot fly directly to the north. It has to fly into the wind, which means in the east of north direction to compensate for the effect of the wind. Using the above calculations and vector diagram, we can say that plane is heading in $30°$east of north.