Radius of Earth at specific angle,

[TheJere]

I was going to post this on Earth-forum here but I thought that you guys here can help me better with this. I'm trying to get the radius of Earth on every latitude degree from 0° to 90°, knowing 0° at Equator is ~6378,137km and 90° at North/South Pole is ~6356,7523km (source Wikipedia). In Wikipedia there is formulas to calculate radius at any angle, but I don't have a clue how to calculate with it. Now to what I'm asking here:
Can someone please explain to me (one that doesn't have even College education on mathematics) so that I could understand it? Best way to explain would be adding some examples like for 61° and 34°?

Now, I have been struggling with this for a very very long time, and I am already being really thankful for the one that explains this to me.

Wikipedia article:
http://en.wikipedia.org/wiki/Earth_radius#Radius_at_a_given_geodetic_latitude

 

[HallsofIvy]

No college math needed here, just how to use a calculator! The applicable formula is the first:
"The distance from the Earth's center to a point on the spheroid surface at geodetic latitude"
$$R= \sqrt{\frac{(a^2cos(\phi))^2+ (b^2sin(\phi))^2}{(acos(\phi))^2+ (bsin(\phi))^2}}$$
And you are given that a= 6,378.1370, b= 6,356.7523.

So if, for example, $\phi= 61^o$, $cos(\phi)= cos(61)= 0.4848096$ and $sin(\phi)= sin(61)= 0.8746197$.

So $a^2cos(\phi)= 19722361$, $b^2sin(\phi)= 35341895$, $acos(\phi)= 3092.182$, and $bsin(\phi)= 5559.741$ where I have rounded to 7 significant figures. Can you finish?

 

[TheJere]

I'm afraid and really sorry, but I think I can't finish, haha. I put numbers you had calculated on formula and I keep having number 435114,36..., double-checked. Thank you very much for the your explain this far, I get it to the point you have calculated it, but I'm afraid I have to ask you to tell me how to finish it?

 

[TheJere]

Oh, now I got it right, THANK YOU very very much, I'll be thankful for you for really long time!