Programming Challenge: a classic trigonometry problem

[Thomas Klemm]

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

'√(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!