Angles in right handed coordinate systems are always measured as positive counter-clockwise.

With respect to x axis the angle of the vector can be easily be found with the use of atan2 (although the result is usually $[-\pi, \pi]$ instead of 0 to $2\pi$.
So in Excel you only need to do =atan2(40,-30)
and you get the result in radians (-0.6435) or =ATAN2(40,-30) *180/PI()
if you want it degress ($\phi_x =-36.87\deg$).
In this case you could use atan
and you would get the same result, however the problem with atan
is that it gives the same result of e.g. $\langle x_0,y0\rangle$ and $\langle-x_0,-y0\rangle$ (i.e. it cannot distinguish between 1st and 3rd quadrant and 2nd and 4th).
Regarding the angle between the y-axis you only need to subtract 90 degrees from the angle with the x-axis. i.e. for this example:
$$\phi_y = -90-36.87 = -126.87 [\deg]$$
Additionally, if you wanted to use atan2
for the y-axis and the vector had coordinates $\vec{V} = (xV, yV)$ then in excel you'd need to use =atan2(yV, -xV)
. This is because you'd be expressing $\vec{V}$ in terms of coordinate system which is rotated by 90 deg ccw compared to the initial one.
Regarding the second part of the answer (unit vector) the answer kamran gave is succinct and to the point.
I could only add the following. Given the unit vector $$\vec{e}_V = \frac{4}{5}\vec{i} - \frac{3}{5}\vec{j}$$
you can estimate the angle with the x axis (let denote it $\phi_x$)by taking the dot product of the unit vector of X with the unit vector of V and then calculating the $\arccos(\vec{e}_{x}\cdot \vec{e}_{V})$. However you will only get the magnitude and you need to check which quadrant you are in. (Similary for the y-axis $\arccos(\vec{e}_{y}\cdot \vec{e}_{V})$ ).
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