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

1 Answer

Answer:

#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^@#.