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

${240}^{\circ}$

#### Explanation:

To find the direction angle $\theta$ of a vector $\underline{v}$, you can work with the definition of $\tan$ function: in a right triangle with $\alpha$ one of the non-right angles, the tangent of $\alpha$ is the length of the side opposed to $\theta$ divided by the length of the adjacent side.

If $O$ is the origin of an orthogonal coordinate system, $P$ is the terminal point of $\underline{v}$ (vector applied in $O$) and $H$ is the point obtained by projecting $P$ on the $x$-axis, the triangle $P H O$ is a right triangle because $\hat{P H O} = {90}^{\circ}$.
Then if $\alpha : = \hat{P O H}$ we get that by definition, $\tan \left(\alpha\right) = \frac{P H}{O H}$.
In our specific case $\underline{v} = < - \sqrt{3} , - 1 >$, so $O H = - \sqrt{3}$ and $P H = - 1$ and this implies that $\tan \left(\alpha\right) = \frac{- \sqrt{3}}{- 1} = \sqrt{3}$. $\tan \left(\alpha\right)$ is $\sqrt{3}$ when $\alpha = {60}^{\circ}$.

Now, since $\underline{v}$ lies in the third quadrant, the direction angle $\theta$ is given by the sum between ${180}^{\circ}$ and $\alpha$: $\theta = {180}^{\circ} + \alpha = {180}^{\circ} + {60}^{\circ} = {240}^{\circ}$.