Math Help!!!
 

Ok, here is a test for the engineers, I think. I would love to see if a student gets it first. Try it out and email me at lee4th@mediaone.net if your able to figure it out. Thanks


[IMG]c:\my documents\math.jpg[/IMG]

Uhmmm......
 

I dont think u can attach a file that way....u either have to use the "attach file" thing at the bottom or find some webspace, etc....until then we wont be able to see your math problemo...

:D

Sorry bout that. Here it is (I hope)

I am so confused..I totally forget geometry and, well, just about all math outside of basic arthimatic..yeah..I'm gonna die in precalc next semester..lol

ah ha!
 

I got the answer, but I won't tell how to get it until someone else do so...

So, here it is:

R = S / ( ( 8^(1/2) ) + 6*sin(120) )

hehe, so just plug 10 into S and get the answer.

I don't think I am really a student anymore.

For all you know, this could be a fake answer! ;)

???
 

Was Ken right??? did he win???

Just wondering...plus this forum seemed kinda dead
;)

Re: ???
 

Well, I can confirm that Ken was right as far as he went. I got
R = S/(2*(2^0.5) + 3*(3^0.5))
which (if I typed it correctly) is equivalent to Ken's
R = S / ( ( 8^(1/2) ) + 6*sin(120) )

But he only gets partial credit. He didn't do the second part of the problem: solving for BD. This part gets a little hairier. I get
BD = ((15+4*(6^0.5))/(35+12*(6^0.5)))^0.5*S

So if S=10, BD = 6.2056272

I agree with Ken's answer (although i got it in a slightly different form)

S = (3*sqrt(3) + 2*sqrt(2))R = Ken's answer since 3*sqrt(3) = 6*Sin120

R = S / (3*sqrt(3) + 2*sqrt(2))

and BD = S * sqrt( (237 - 40*sqrt(6)) / 361)

The solution for finding R in terms of S basically boils down to finding the sides of 3 right triangles for those who are looking for a hint. (Remember the special stuff about tangents to circles? yeah...) And to find BD, cosine law :)

Soo... do I win a prize too? :D

Anthony.