how is that going to help me calculate distance between 2 points?
You say you have forgotten the formular, but you seem to have foggoten even what the formular looks like. Oh Well. As a peice of advice, looksee on the net for the simplest Trig functions explinations you can find, it obviosly will not teach you Trig, but often just looking at the equations a few times, and then programming each one in helps you to remember them.
And Beaker's writing it out the long was is silly
What? Mine was more concise than yours. But having said that Link didnt understand mine, so yours was probably better. But again Beakers was probably more applicable distance# = (dx# * 0.41) + (dy# * 0.941246)
This Must by quicker than Distance# = Sqr((Dx#*Dx#) + (Dy#*Dy#))
anyone who says it isnt quicker either ment to say, its not sufficantly quicker to be worth it, or was simply Wrong.X^2 is slower than X*X. But X^2 reads better. (Well to me anyway) However I would probably change it to X*X in intensive calculations.