Programming Challenge: a classic trigonometry problem

[Gerson W. Barbosa]

(01-05-2014 11:50 PM)Thomas Klemm Wrote:  

(01-05-2014 09:54 PM)Gerson W. Barbosa Wrote:  '√(900-X^2)/X=15/(X-15*X/√(1600-X^2))'

Y
'√(900-X^2)=15/(1-15/√(1600-X^2))'

'1/√(900-X^2)=(1-15/√(1600-X^2))/15'

'1/√(900-X^2)=1/15-1/√(1600-X^2)'

'1/√(900-X^2)+1/√(1600-X^2)=1/15'

Nice!

Really nicer that way!

Same as Bunuel's method, I think. But it was fun to find a solution on my own.
http://m.flickr.com/photos/113638379@N08.../lightbox/

Cheers,

Gerson.