Rotating rectangles

OK, this might not be 100% kosher as it's part of my college work, but I really need help, and I could always ask my teacher anyway. You can sleep with a sound mind even after answering this question.

Imagine a cartesian plane, with these points:
O=(0,0)
A=(4,0)
B=(4,3)
C=(0,3)

This forms a rectangle

Now rotate this rectangle 20 degrees anti-clockwise about the origin.

part one of the question is to find the co-ordinates of A' and C'
i did that as follows:
A' = (cos[20]*4,sin[20]*4)
A' = (3.76,1.37)
C' = (cos[90+20], sin[90+20]
C' = (-1.03,2.82)

I think that is right. Anyway, part two says:
Calculate the co-ordinates of B' from your answer to part 1.
There I'm stuck.
I could also use help on part three where is says:
By calculating OB and the angle AOB show the co-ordinates can be directly calculated.
well OB is 5 [sqrt(3^2+4^2)]
and angle AOB = 36.87 (using cosine rule)
but after that I'm stuck.
Any help on part 2 and 3 would be very helpful.

So far so good.

O' = (0,0)
A' = (3.76,1.37)
B' = (x,y)
C' = (-1.03,2.82)

From that:

(x + 0)/2 = (3.76 - 1.03)/2
x = 2.73
(y + 0)/2 = (2.82 + 1.37)/2
y = 4.19

It's just averages, the average coordinate of two opposite corners must be the same as the average coordinate of the other two corners because it's a rectangle (draw diagonals in, geometrically it's shown by the intersection of the diagonals being an equal distance from each of a pair of opposite sides).

As for the second part:

OB² = 4² + 3²
OB = 5

That's just Pythagoras theorem, as I'm sure you know.

Finally, in a separate diagram draw a diagonal line (any old diagonal line), then call the bottom end O and the top end B'.
Label the line OB' with length 5 as you just worked out. Then draw a horizontal dashed line from O accross the page, calling the acute angle between that line and the line OB' 37.whatever degrees.

From your innovative diagram, x = 5cos37.blah and y = 5sin37.blah.

I hope that helps!

thanks, it makers a lot more sense now.