Resolving forces in the $x'$ and $y'$ directions

I am trying to get my head around some introductory mechanics, however I need some help.

For this question:

Number 4

It asks me to resolve the $200kN$ force shown acting on the pin into components in the (a) $x$ and $y$ directions, (b) $x’$ and $y$ directions and (c) $x$ and $y’$ directions.

Now, I know for part (a) $R_{x}=200\cos{40}=153.21$ (2 d.p.) and $R_{y}=200\sin{40}=128.56$ (2 d.p.).

However, for parts (b) and (c), isn't it true that $R_{x'}=200\cos(40+30)$ and $R_{y'}=200\sin(40+30)$?

Also, I don't understand why $R_{x}$ and $R_{y}$ change when you are resolving them with respect to $R_{x'}$ and $R_{y'}$ and not each other.

No. The problem does not ask you to write $R$ in terms of $R_{x'}$ and $R_{y'}$. Forget everything you did in part a. What the problem asks in part b is to find two forces, one along $x'$ the other along $y$, such as the sum of them is $R$. You can write $R_{x'}$ as made up of a component along $x$: $R_{xx'}=R_{x'}\cos(30)$ and along $y$: $R_{yx'}=-R_{x'}\sin(30)$. Now write $R$ along $x$ and $y$ and you get $$R\cos(40)= R_{x'}\cos(30)$$ and $$R\sin(40)=R_y-R_{x'}\sin(30)$$ Equivalently, you can project the required components along $R$ and the direction perpendicular to $R$