Question

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

Answer 1

`240^@`

Explanation:

To find the direction angle `theta` of a vector `ul{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 `ul{v}` (vector applied in `O`) and `H` is the point obtained by projecting `P` on the `x`-axis, the triangle `PHO` is a right triangle because `hat{PHO}=90^@`.
Then if `alpha:=hat{POH}` we get that by definition, `tan(alpha)=(PH)/(OH)`.
In our specific case `ul{v}=<-sqrt(3),-1>`, so `OH=-sqrt(3)` and `PH=-1` and this implies that `tan(alpha)=(-sqrt(3))/(-1)=sqrt(3)`. `tan(alpha)` is `sqrt(3)` when `alpha=60^@`.

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