#math - Wed 25 Apr 2007 between 02:32 and 02:38

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timurwelcome. good luck
_kmh_sniper89 : but it is not a cotangens value
sniper89no, it is not
it's just
_kmh_ctg maps degrees to real numbers
Kirynanyone know if there is a program that can emulat a plane graph? lik what a graphing book is
_kmh_so the result cannot be in degrees
sniper89if a point lies in the 2nd quarter of the coord value, it means x will surely be negative, and y will be positive, right
and ctg a = x/y
I need to use that and maybe r = sqrt(x^2 + y^2) to solve it
_kmh_to solve what ?
sumptctg = ?
sniper89err
find the ctg 150*...
_kmh_i probably missed the earlier discussion - what is the actual problem ?
sniper89omagawd
Using the trigonometrical functions' definitions, find: ctg 150*
_kmh_well that cos (150)/sin(150)
sniper89cos a = x/r
sin a = y/r
sumptwhat is ctg?
sniper89what do you mean, sumpt
timurand you could consider r = 1
sumptwhat does it stand for?
sniper89cotangens
_kmh_and cos(150)=-cos(30)
sniper89timur: consider?
sumptwouldn't that be represented by cot?
sniper89I don't think so...
sumptthat's how my two books represent it, but ok
timurwell, r doesn't matter if you're figuring out sin, cos, tan, cot, etc, because it's the ratio that matters
sniper89oh
but still I don y x
o dam
timurdo you know cos 30, sin 30?
sniper89*but still I don't know what y, x and ctg 150 are
_kmh_sin(150)=sin(30)
timur : for that you can use a triangle
timurthat's where I was going with it ;p
_kmh_take a equilateral triangle with its height

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