# How do I find the direction angle of vector <1, -sqrt3>?

Dec 25, 2015

The direction angle of a vector is given by the formula

$\arctan \left(\frac{b}{a}\right) = \theta$

where $b = - \sqrt{3}$ and $a = 1$

because $\tan \left(\theta\right) = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t}$

The opposite is $b$ and adjacent is $a$ draw it on a paper if you don't visualize

here we have $\arctan \left(- \sqrt{3}\right) = \theta$

So $\theta = - \frac{\pi}{3}$

Here a graph of the form $y = m x$ where $m = \tan \left(\theta\right)$ you can see that when $x = 1 \implies y = - \sqrt{3}$

graph{y=-tan(22/21)x [-20, 20, -10.42, 10.42]}