Extremely difficult question on relative velocity?

Question

A man is walking in the north-east direction and wind appears to blow from north. If the man doubles his speed, wind appears at angle arccot 2(cot inverse 2) east of north. Find the actual direction of the wind.

Answer 1

actual direction is arccot 2(cot inverse 2) west of south (the opposite of the direction it appears to blow from.  duh.)

Answer 2

why would you use inverse cotangents when you could probably get rid of the tangents, and have inverse cosines.  This question is ridiculous.  

Answer 3

Attached to the ground, x-axis is in the direction WE, y-axis in the direction SN.

Let's call v and 2v the velocities of the man in 2 circumstances. The velocities of the ground with respect to the man will be -v and -2v respectively.

From what is said about the direction of v, we have:

v = vcos45 i + vsin45 j

(Unable to type an arrow on top of a vector quantity, I use v to denote both vector v and its magnitude, and hope you know which is which)

Let's call u the velocity of the wind with respect to the ground, α its direction. Then:

u = ucosα i + usinα j

If u' is the velocity of the wind relative to the man :

- when the man is moving with velocity v:

u' = u - v = (ucosα - vcos45) i + (usinα - vsin45) j

Since the direction of u' is NS, its x-component is 0:

ucosα - vcos45 = 0 (1)

-when the man is moving with velocity 2v:

u' = u - 2v = (ucosα - 2vcos45) i + (usinα - 2vsin45) j

Given that the angle made by u' and y-axis is arccot2, arctan2 is the angle made by u' and x-axis. Therefore

(usinα - 2vsin45) / (ucosα - 2vcos45) = 2

2(ucosα - 2vcos45) = (usinα - 2vsin45) (2)

We have 2 simultaneous equation to define u and α in term of v:

ucosα - vcos45 = 0 (1)

2(ucosα - 2vcos45) = (usinα - 2vsin45) (2)

The magnitude of u is not essential in this problem, for simplicity, let's put u/v = k and rewrite the equations:

kcosα - cos45 = 0 (1)

2(kcosα - 2cos45) = (ksinα - 2sin45) (2)

From (1),

k = cos45/cosα

Substituting in (2),

2(cos45 - 2cos45) = (cos45 tanα - 2sin45)

cos45.tanα = -2cos45 + 2sin45

tanα = (-2 + 2tan45) = 0

α = 0

The direction of the wind with respect to the ground is west to east

and the ratio k = u/v = cos45/cos0 = (√2)/2