One way of simplifying it would be to transform the Polar coordinates into Cartesian coordinates.
In order to do that, let's take a general point #P(r,theta)# in the Polar system. If #O# is the origin of the system and #P'# is the intersection of the line going through #P# parallel to the y-axis and the x-axis, then :
In the right angled triangle #POP'#, we know the hypotenuse is #r# and one of the angles of #theta#. By this knowledge, we can deduce that, if we would transform the point P into Cartesian coordinates, the following relations take place:
#r^2 = x^2+y^2#
#x = rcostheta#
#y=rsintheta#,
where #x# and #y# are the Cartesian coordinates of #P#.
In our case:
#theta=-pi/6#
#r=5/2#
#=> color(red)x = 5/2 * cos(-pi/6) = 5/2 * sqrt(3)/2 = color(red)((5sqrt(3))/4#
#=> color(red)y = 5/2 * sin(-pi/6) = - 5/2 * 1/2 = color(red)(-5/4#.
Therefore, the point #B# has Cartesian coordinates #((5sqrt3)/4,-5/4)#.
It should be fairly easy to graph #B# now.
