You can plot #-1/4, -sqrt(3/4)# on the Cartesian plane easily enough. We seek a representation of this same point in (2d) space using the polar coordinates #(r, theta)#
The r is easy enough:
#r = sqrt((-1/4)^2 + (-sqrt(3/4))^2#
# = sqrt(1/16 + 3/4)#
#= sqrt(13)/4#
The value we calculate for the angle #theta# will depend on where we choose the #theta = 0# direction to be. This is an arbitrary choice. Here I will choose #theta = 0# to be on the x-axis to the right. So, therefore, #theta = pi/2# will be on the y-axis pointing upwards, etc.
If #r = sqrt(13)/4#, then #(-1/4)/(sqrt(13)/4) = cos(theta)#
so therefore #theta = arccos(-1/sqrt(13))#
...but note that you can choose the #theta = 0# direction to be on the y axis pointing up. In this case,
#(-1/4)/(sqrt(13)/4) = sin(theta)#.
and therefore #theta# would be # arcsin(-1/sqrt(13))#
...I'm guessing your teacher may have a preferred direction for the #theta = 0# direction, so use whichever of these answers is appropriate.
GOOD LUCK.
