Question

An airplane has to go from a point $\text{A}$ to another point $\text{B}$ due ${45}^{\circ }$ east of north. Wind is blowing due north at speed of $200\sqrt{2}\text{km/h}$. The steering-speed of the plane is $400\text{km/h}$. Find the direction in which the pilot should head the plane so as to reach the point $\text{B}$.

• A. ${75}^{\circ }$east of north.
• B. ${30}^{\circ }$east of north.
• C. ${60}^{\circ }$east of north.
• D. ${15}^{\circ }$east of north.

Answer 1

The correct option is A ${75}^{\circ }$east of north.

Given that,

Velocity of wind, $V_w=200\sqrt{2}\text{km/h}$,

The steering-speed of plane,$V_{aw}=400\text{km/h}$

and velocity of plane $\overrightarrow{V_a}$ should be along $\text{AB}$ or in north-east direction.

The direction of $\overrightarrow{V_{aw}}$ should be such as the resultant of $\overrightarrow{V_w}$ and $\overrightarrow{V_{aw}}$is along $\text{AB}$ or in north -east direction.

Let, $\overrightarrow{V_{aw}}$ makes an angle $\alpha$ with $\text{AB}$ as shown in figure.


Applying sine law in triangle $\text{ABC}$, we get

$\frac{AC}{\sin {45}^{\circ }}=\frac{BC}{\sin \alpha }$

$\Rightarrow \sin \alpha =\left(\frac{BC}{AC}\right)\sin {45}^{\circ }$

$\Rightarrow \sin \alpha =\left(\frac{V_w}{V_{aw}}\right)\sin {45}^{\circ }$

$\Rightarrow \sin \alpha =\left(\frac{200\sqrt{2}}{400}\right)\times \frac{1}{\sqrt{2}}$

$\Rightarrow \sin \alpha =\frac{1}{2}$

$\therefore \alpha ={30}^{\circ }$

From the figure, the direction in which the pilot should head the plane is

$=\alpha +45$

$={30}^{\circ }+{45}^{\circ }$

$={75}^{\circ }$ east of the north.

Hence, option (a) is the correct answer.