91Ó°ÊÓ

• 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,

$\sin \mathrm{θ}=\frac{\mathrm{frontside}\mathrm{or}\mathrm{perpendicularside}}{\mathrm{hypotenuse}}$ (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$\mathrm{θ}$.

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

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

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.