Let $\,\vec v\,$ be a vector with known analytic form $\,\langle a\,,\,b\rangle\,$:
Throughout this section, $\,a\,$ and $\,b\,$ are real numbers.
The direction of a vector can be described in different ways. For example, bearings are often used in navigation. The angle conventions used to define the trigonometric functions give a particularly convenient way to define the direction of a vector. For the reader's convenience, this scheme is repeated below: ‘Standard Direction’ Angle Conventions

a positive angle $\,\theta$ standard direction: a negative angle $\,\theta$ 
When at least one component (horizontal or vertical) of a vector equals zero,
then no formula is needed to find standard direction.
Here's a table that summarizes the results for $\,\vec v = \langle a\,,\, b\rangle\,$:
horizontal/vertical components  arrow representation  standard direction, $\,\theta$  
the zero vector: $\,a = 0\,$ and $\,b = 0\,$ 
(no arrow representation) 
the zero vector has no direction; the direction of $\,\langle 0,0\rangle\,$ is undefined 

$\,\vec v = \langle a\,,\,0\rangle\,$ with $\,a > 0\,$ 
vector points to the right 
$\,\theta = 0^\circ\,$  
$\,\vec v = \langle a\,,\,0\rangle\,$ with $\,a < 0\,$ 
vector points to the left 
$\,\theta = 180^\circ\,$  
$\,\vec v = \langle 0\,,\,b\rangle\,$ with $\,b > 0\,$ 

$\,\theta = 90^\circ\,$  
$\,\vec v = \langle 0\,,\,b\rangle\,$ with $\,b < 0\,$ 

$\,\theta = 270^\circ\,$ or $\,\theta = 90^\circ\,$ 
For an arbitrary vector, the arctangent function gives convenient formulas for
standard direction.
However, different formulas are needed in different quadrants. Why are different formulas needed? The arctangent only gives angles between $\,90^\circ\,$ and $\,90^\circ\,$, so (without adjustment) it can only cover vectors in quadrants IV and I. Keep reading! Recall that, by definition:
$\arctan x\,$ is the angle in the interval $\,(90^\circ,90^\circ)\,$ whose tangent is $\,x\,$
This is the ‘degree version’ of the definition.
Remember:
For the formulas in this section, $\,\arctan x\,$ is assumed to be in degrees.
In Quadrant I, where $\,a > 0\,$ and $\,b > 0\,$, everything is simple:
$$
\begin{gather}
\cssId{s83}{\tan\theta = \frac ba}\cr
\cssId{s84}{\theta = \arctan\bigl(\frac ba\bigr)}
\end{gather}
$$
When $\,\frac ba\,$ is positive,
$\,\arctan(\frac ba)\,$ returns the
angle between $\,0^\circ\,$ and $\,90^\circ\,$ whose tangent is $\,\frac ba\,$.
This is just the angle we want!Things are similarly easy in quadrant IV. Here, $\,\frac ba\,$ is negative, and $\,\arctan(\frac ba)\,$ returns an angle between $\,90^\circ\,$ and $\,0^\circ\,$. Again, this angle provides a correct direction for the vector. If you want a positive angle for the direction (instead of the negative one provided by the arctangent), you can use the formula $\,360^\circ + \arctan(\frac ba) \,$. 
depends upon the signs of $\,a\,$ and $\,b\,$, as indicated below:

In Quadrant II, where $\,a < 0\,$ and $\,b > 0\,$, we have to be careful:
In Quadrant III, where $\,a < 0\,$ and $\,b < 0\,$, we also have to be careful:
Instead of using the formulas for direction discussed above,
you can always find the reference angle for the vector, and then adjust.
The reference angle is the acute angle between the vector and the $x$axis.
EXAMPLE Find the standard direction of $\,\vec v = \langle 2\,,\,5\rangle\,$. Report as a positive angle in the interval $\,[0^\circ,360^\circ)\,$. Solution:


On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
