Direction angles: If you look at the name you will guess that the direction is the point!!!
As you know that velocity,acceleration, force etc. has magnitude as well as direction.
For example, The wind tonight will be 10 mph from the North West" then they are describing the velocity of the wind, since it has magnitude of 10 and direction as North -West.
So, with vectors you must also specify the direction, not only the magnitude.
10 mph from the North West to do this we will use the angle, the vector forms with the axes.
So now we will say 10 mph at 40° with the X-axis.
In two dimensions you have the angles α and β.
Using trigonometry and Pitagora's Theorem (to evaluate the "length" of your vector).

Direction angles

If u is a unit vector such that 'θ'(theta) is the angle measured in counterclockwise from the positive X-axis with to 'u'then the terminal point of u will lie on the unit circle and we have, u=$\left \langle x,y \right \rangle $
= $\left \langle cosθ,sinθ \right \rangle $
= (cosθ)i + (sinθ)j
According to the above diagram 'θ' is the direction angle with the vector u.
Let us consider u is a unit vector with direction angle θ. If v = ai + bj is any vector that makes an angle 'θ'with the positive X-axis. Vector v has same direction as vector u then we can write,
v = $ \left \| v \right \| \left \langle cosθ,sinθ \right \rangle $
=$ \left \| v \right \| \left \langle (cosθ)i+(sinθ)j \right \rangle $
As v = ai + bj = $ \left \| v \right \| \left \langle (cosθ)i+(sinθ)j \right \rangle $
So direction-angle 'θ' for vector v is determined by
tan θ =$\frac{sinθ}{cosθ}$
If you want to find the direction-angle in terms of a and b then
Multiply the above equation by $\left \| v \right \|$ , we will get

tan θ =$\frac{\left \| v \right \|.sinθ}{\left \| v \right \|.cosθ}$

tan θ =$\frac{b}{a}$

Examples on Direction angles

Example 1:Find the direction angle for vector u = 4i + 4j. Solution : Compare u = ai + bj and u = 4i + 4j
So we will get a= 4 and b = 4
The direction angle is determined by
tan θ =$\frac{b}{a}$
So, tan θ =$\frac{4}{4}$
tan θ = 1
∴ θ = tan$^{-1}$(1)
So, θ = 45$^{0}$
As the angle is positive and less than 45$^{0}$ so the vector uwill lie in the I quadrant.

Example 2:Find the direction angle for vector v = 5i - 2j. Solution : Compare v = ai + bj and v = 5i - 2j
So we will get a= 5 and b = -2
The direction angle is determined by
tan θ =$\frac{b}{a}$
So, tan θ =$\frac{-2}{5}$
tan θ = -0.4
As the value of tan θ is negative so θ will lie in the IV quadrant.
∴ θ = tan$^{-1}$(-0.4) = -21.8
So we will find the reference angle of -21.8.
θ = 360 -21.8 = 338.2$^{0}$