Question

An aeroplane pilot wishes to fly due west, the wind of $100\text{kmph}$ is flowing towards the south. If the speed of the plane in still air is $300\text{kmph}$, In which direction should the pilot head?

• A. ${\sin }^{-1}\left(\frac{1}{3}\right)$ $\text{N of W}$
• B. ${\sin }^{-1}\left(\frac{2}{3}\right)$ $\text{S of W}$
• C. ${\sin }^{-1}\left(\frac{1}{3}\right)$ $\text{W of N}$
• D. ${\cos }^{-1}\left(\frac{1}{3}\right)$ $\text{N of W}$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{1}{3}\right)$ $\text{N of W}$

Given velocity of wind is $\overrightarrow{V_W}=100(-j)$
$|\overrightarrow{V}{}_{PW}|=300$

$\overrightarrow{V}{}_P$ should be towards west as per question.
Now $\overrightarrow{V_{PW}}=\overrightarrow{V_P}-\overrightarrow{V_W}$
$\Rightarrow \overrightarrow{V_P}=\overrightarrow{V_{PW}}+\overrightarrow{V_W}$

using triangle law

Now $\sin \theta =\frac{|\overrightarrow{V_W}|}{|\overrightarrow{V_{PW}}|}$
$\sin \theta =\frac{100}{300}=\frac{1}{3}\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{3}\right)\text{N}\text{of W}$