timur | welcome. good luck |

_kmh_ | sniper89 : but it is not a cotangens value |

sniper89 | no, it is not it's just |

_kmh_ | ctg maps degrees to real numbers |

Kiryn | anyone 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 |

sniper89 | if 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 ? |

sumpt | ctg = ? |

sniper89 | err find the ctg 150*... |

_kmh_ | i probably missed the earlier discussion - what is the actual problem ? |

sniper89 | omagawd Using the trigonometrical functions' definitions, find: ctg 150* |

_kmh_ | well that cos (150)/sin(150) |

sniper89 | cos a = x/r sin a = y/r |

sumpt | what is ctg? |

sniper89 | what do you mean, sumpt |

timur | and you could consider r = 1 |

sumpt | what does it stand for? |

sniper89 | cotangens |

_kmh_ | and cos(150)=-cos(30) |

sniper89 | timur: consider? |

sumpt | wouldn't that be represented by cot? |

sniper89 | I don't think so... |

sumpt | that's how my two books represent it, but ok |

timur | well, r doesn't matter if you're figuring out sin, cos, tan, cot, etc, because it's the ratio that matters |

sniper89 | oh but still I don y x o dam |

timur | do 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 |

timur | that's where I was going with it ;p |

_kmh_ | take a equilateral triangle with its height |