Question

An aircraft flies at a speed of $400\text{km/h}$ in still air. A wind is blowing with a speed of $200\sqrt{2}\text{km/h}$ from South. The pilot wishes to travel from point $A$ to a point $B$, where point $B$ is North-East of $A$. Find the direction he must head if $AB=1000\text{km}$.

• A. ${30}^{\circ }$ from North towards East
• B. ${45}^{\circ }$ from East towards North
• C. ${75}^{\circ }$ from North towards East
• D. ${75}^{\circ }$ from East towards North

Answer 1

The correct option is C ${75}^{\circ }$ from North towards East


Given that speed of wind w.r.t ground
$v_w=200\sqrt{2}\text{km/h}$
and speed of airplane w.r.t wind
$v_{aw}=400\text{km/h}$ and

Velocity of airplane w.r.t ground $\overrightarrow{v_a}$ should be along $AB$ or in north-east direction.
Thus, the direction of $\overrightarrow{v_{aw}}$ should be such that the resultant of $\overrightarrow{v_w}$ and $\overrightarrow{v_{aw}}$ is along $AB$ or in north-east direction.

Let $\overrightarrow{v_{aw}}$ make an angle $\alpha$ with $AB$ as shown in vector triangle $ABC$:

Applying sine law in triangle $ABC$, we get:
$\frac{AC}{\sin {45}^{\circ }}=\frac{BC}{\sin \alpha }$
$\Rightarrow \sin \alpha =\left(\frac{BC}{AC}\right)\sin {45}^{\circ }=\left(\frac{200\sqrt{2}}{400}\right)\frac{1}{\sqrt{2}}=\frac{1}{2}$
$\therefore \alpha ={30}^{\circ }$

Therefore, the pilot should head in a direction at an angle of $({45}^{\circ }+\alpha )$ or ${75}^{\circ }$ from north towards east.