Linden,
I would start by sketching a diagram.
In my diagram O is the pole, which will be the origin in the Cartesian system, and P is the point with polar coordinates (r,theta) = (6,30^{o}). I drew a perpendicular line from P to Q on the polar line, which will be the Xaxis in the Cartesian system. I now have a right triangle POQ where the length of OQ is the Xcoordinate of P and the length of PQ is the Ycoordinate of P. I can find these lengths in two ways.
First method:
I know that any triangle with angles 30, 60 and 90 degrees has side lengths in the ratio if 1 to 2 to √3. (These triangles come up so often that this is a fact I remember.) 2 is the length of the hypotenuse, 1 is the length of the shorter leg and √3 is the length of the longer leg. In the triangle above the length of the hypotenuse is 6 units so the length of the shorter leg, PQ is half of that which is 3 units. The length of the longer leg is √3 times the length of the shorter leg so OQ = √3 3. Thus P has Cartesian coordinates (3√3, 3).
Second Method:
Use trig functions.
sin(30^{o}) = ^{PQ}/_{OP} = ^{PQ}/_{6} = ^{1}/_{2} and therefore PQ = 3.
cos(30^{o}) = ^{OQ}/_{OP} = ^{OQ}/_{6} = ^{√3}/_{2} and therefore PQ = 3 √3.
Thus P has cartesian coordinates (3 √3, 3)
Penny
