On a particular day the velocity of the wind, in terms of $\mathbf{i}$ and $\mathbf{j}$, is $x\mathbf{i}-3\mathbf{j}$, where $x\in\mathbb{N}$.

$\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the directions East and North respectively.

To a man travelling due East the wind appears to come from a directionÂ North $\alpha^\circ$ West where $\tan\alpha=2$.

When he travels due North at the same speed as before, the wind appears to come from a direction North $\beta^\circ$ West where $\tan\beta=3/2$.

Find the actual direction of the wind.

Solution:

We start by writing

$\overrightarrow{V_{WM}}=\overrightarrow{V_W}-\overrightarrow{V_M}$.

We have two equations.

Firstly, when the man travels due East, $\overrightarrow{V_M}=a\mathbf{i}$ for some constant $a$. We have $\overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j}$ and so

$\overrightarrow{V_{WM}}=(x-a)\mathbf{i}-3\mathbf{j}$.

We know that this, with respect to the man, is coming from North $\alpha^\circ$ West. This means we have:

And furthermore:

$\displaystyle \tan\alpha=\frac{x-a}{3}\overset{!}{=}2$

$\Rightarrow x-a=6\Rightarrow a=x-6$.

Now consider when the man travels due North, $\overrightarrow{V_M}=a\mathbf{j}$. We have $\overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j}$ still and so

$\overrightarrow{V_{WM}}=x\mathbf{i}-(3+a)\mathbf{j}$.

We know that this, with respect to the man, is coming from North $\beta^\circ$ West. This means we have:

And furthermore:

$\displaystyle \tan\beta=\frac{x}{3+a}\overset{!}{=}\frac32$

$\underset{\times_{2(3+a)}}{\Rightarrow} 2x=9+3a$,

but we have $a=x-6$ and so

$2x=9+3(x-6)\Rightarrow x=9$,

so that $\overrightarrow{V_W}=9\mathbf{i}-3\mathbf{j}$.

This means that the velocity of the wind looks like:

That is the wind comes from North $\theta$ West. We have that $\tan\theta=9/3\underset{\tan^{-1}}{\Rightarrow} \theta=\tan^{-1}(3)\approx 71.57^\circ$, so the answer to the question asked is N $71.57^\circ$ W.

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