Question

An aircraft flies at $400km/h$ in still air. A wind of $200\sqrt{2}km/h$ is blowing from the south. The pilot wishes to travel from $A$ to a point $B$ north-east of $A$. The direction he must steer is

• A. ${75}^{\circ }\text{from North towards East.}$
• B. ${75}^{\circ }\text{from East towards North.}$
• C. ${45}^{\circ }\text{from North towards East.}$
• D. ${45}^{\circ }\text{from East towards North.}$

Answer 1

The correct option is A ${75}^{\circ }\text{from North towards East.}$


Given, velocity of wind $v_w=200\sqrt{2}km/h$
and velocity of aircraft w.r.t wind $v_{aw}=400km/h$
$\overrightarrow{v_a}$ should be along $AB$ or in north-east direction.
Thus, the direction of $\overrightarrow{v_{aw}}$ should be such that the resultant of $\overrightarrow{v_w}\text{and}\overrightarrow{v_{aw}}$ is along $AB$ or in north-east direction.

Let $\overrightarrow{v_{aw}}$ makes an angle $\alpha$ with $AB$ as shown in the figure.
Applying sine law in triangle ABC, we get
$\frac{AC}{\sin {45}^{\circ }}=\frac{BC}{\sin \alpha }$
or, $\sin \alpha =\left(\frac{BC}{AC}\right)\sin {45}^{\circ }=\left(\frac{200\sqrt{2}}{400}\right)\frac{1}{\sqrt{2}}=\frac{1}{2}$
$\therefore \alpha ={30}^{\circ }$
Therefore, the pilot should steer in a direction at an angle of $({45}^{\circ }+\alpha )={75}^{\circ }$ from north towards east.