Answer:
[tex]\theta = tan^{-1}(\frac{y}{x})[/tex]
Step-by-step explanation:
If we are given components of a vector then we can find the angle between them.
Suppose we are given a vector v
[tex]v = (x, y)[/tex]
Where x is the horizontal component and y is the vertical component.
The angle can be found by using
[tex]tan(\theta)=\frac{y}{x}\\\\\theta = tan^{-1}(\frac{y}{x})[/tex]
The magnitude of the vector v can be found using
[tex]v = \sqrt{x^{2}+y^{2}}[/tex]
Example:
Lets do a quick example:
[tex]v = (2, 4)[/tex]
The angle of the vector is
[tex]tan(\theta)=\frac{4}{2}\\\\\theta = tan^{-1}(\frac{4}{2})\\\\\theta = 63.43^{\circ}[/tex]
The magnitude of the vector is
[tex]v = \sqrt{x^{2}+y^{2}}\\\\v = \sqrt{2^{2}+4^{2}}\\\\v = \sqrt{4+16}\\\\v = \sqrt{20}\\\\v = 4.47[/tex]
