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 direction 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:

rel6.jpg

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:

rel7.jpg

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:

rel8.jpg

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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