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)P(r,θ) in the Polar system. If OO 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.
