confirmation about angle formula

[TrueTears]

(graph is below)
2.I want to work out the angle B, i worked it out but i dont understand why it works.
vector ZD = i (50sqrt(2)-100) +50(sqrt(2)) j  in my book it says to work out the angle a vector makes with the y axis (anti clockwise from the y axis) is cos(B) = a_2 / |a|    a_2 being the co-efficient infront of j and |a| being the magnitude of the vector (in this case , its vector ZD) .
So i plugged those values in

Cos(B) = 50(sqrt(2)) / 100sqrt(2-sqrt(2))    ( 100sqrt(2-sqrt(2)) is |a| i left out working)
B = 22.5 deg which is the right answer, however what i dont understand is that, is the line ZP here 'acting' as the y axis? because i drew that line in myself. It was not in the books diagram. Also does the vector have to be a position vector in order to use this formula, ie cos(B) = a_2 / |a|. Or can it just be any vector? Because if it does have to be a position vector, then what i did should not be right since point Z is not the origin. If it can be any vector, then does PZ "act" as the y axis?

Many thanks.

[Mao]

The principle behind this is simple, dot product.

To find the angle between the vector ZD and the y axis, your construction is quite right, ZP is parallel to the y axis. Let's assume ZP is the unit vector j

Hence,

The way you can also think of this is trigonometry, move ZD to origin, you'll find that the angle it makes with the verticle corresponds to the adjacent side and the hypotenuse.

[TrueTears]

ah yes i see now, thanks very much mao.