Hello, hooke!

A fascinating problem . . . hope I got it right.

A man was walking with a velocity of 4**i** km/h

when he obvserved that the wind was blowing from the north.

When the man increased his velocity to 6**i** km/h ,

he noticed that the wind seemed to be blowing from N 45° W.

Find the speed and the direction of the wind. Evidently, the wind is blowing from the northeast.

The first phase looks like this: Code:

C
*
* |
w * |
* |
* θ |
A * - - - - * B
4

The man walks from $\displaystyle A$ to $\displaystyle B\!:\;\;|\overrightarrow{AB}| \,=\,4$

The wind has vector $\displaystyle \vec w \,=\,\overrightarrow{CA}\!:\;\;w \,=\,|\vec w|,\;\;\theta \,=\,\angle A $

The *apparent* wind vector is: .$\displaystyle \overrightarrow{CB}$

We have: .$\displaystyle \cos\theta \,=\,\frac{4}{w} \quad\Rightarrow\quad w\cos\theta \,=\,4 \quad\Rightarrow\quad w\sin\theta \,=\,\sqrt{w^2-16}$ .[1]

The second phase looks like this: Code:

C
*
* *
w * *
* *
* θ 45° *
A * - - - - - - - - - * D
6

The man walks from $\displaystyle A$ to $\displaystyle D\!:\;|\overrightarrow{AD}| \,=\,4$

The wind vector $\displaystyle \vec w$ is exactly the same.

The apparent wind vector is: .$\displaystyle \overrightarrow{CD},\;\angle D \,=\,45^o$

We have: .$\displaystyle \angle C \:=\:180^o - \theta - 45^o \:=\:135^o - \theta$

Law of Sines: .$\displaystyle \frac{w}{\sin45^o} \:=\:\frac{6}{\sin(135^o-\theta)} \quad\Rightarrow\quad w\sin(135^o-\theta) \:=\:6\sin45^o$ .[2]

The left side is: .$\displaystyle w\bigg[\sin135^o\cos\theta - \cos135^o\sin\theta\bigg] \;=\;w\bigg[\frac{1}{\sqrt{2}}\cos\theta - \left(-\frac{1}{\sqrt{2}}\right)\sin\theta\bigg]$

. . . . . . . . . . $\displaystyle =\; w\bigg[\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\bigg] \;=\; \frac{1}{\sqrt{2}}\,\bigg[w\cos\theta + w\sin\theta\bigg]$

. . Substitute [1]: .$\displaystyle \frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right) $

Then [2] becomes: .$\displaystyle \frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right) \:=\:6\left(\frac{1}{\sqrt{2}}\right) \quad\Rightarrow\quad 4 + \sqrt{w^2-16} \:=\:6$

. . . . . $\displaystyle \sqrt{w^2-16} \:=\:2\quad\Rightarrow\quad w^2-16 \:=\:4 \quad\Rightarrow\quad w^2 \:=\:20 \quad\Rightarrow\quad w \:=\:2\sqrt{5}$

The wind speed is about: .$\displaystyle \boxed{4.47\text{ km/hr}}$

Substitute into [1]: .$\displaystyle w\cos\theta \:=\:4 \quad\Rightarrow\quad 2\sqrt{5}\cos\theta \:=\:4 \quad\Rightarrow\quad \cos\theta \:=\:\frac{2}{\sqrt{5}}$

. . Hence: .$\displaystyle \theta \:\approx\:26.6^o$

The direction of the wind is: .$\displaystyle \boxed{S\,63.4^o\,W}$