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Math Help - Help me with Angles-Euclidean Geometry

  1. #1
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    Exclamation Help me with Angles-Euclidean Geometry

    Since I'm totally not getting any responses in the other forum... I hope its ok I posted this again... I'm not spamming!

    Anyway, Heres my problem:

    Line segment AC in bisected by ray BD at point B. Ray BT bisects angle DBC.

    a. If angle DBC= 25x+7 and angle TBC= 12x+5, find the degree measures of angle TBC, DBT and DBC

    b. If angle CBT= 10x-8 and angle TBA=15x-12, find the degree measures of both angles.

    I drew a diagram and everything, but I am still utterly confused.

    <33 Buyo
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  2. #2
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    You knwo what? forget it.. I'll just go look for it elsewhere.

    <edit> Sorry! Sorry! Ignore my frustrated ramblings~!
    Last edited by buyochan; September 3rd 2007 at 04:29 PM. Reason: I was being mean
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by buyochan View Post
    Since I'm totally not getting any responses in the other forum... I hope its ok I posted this again... I'm not spamming!

    Anyway, Heres my problem:

    Line segment AC in bisected by ray BD at point B. Ray BT bisects angle DBC.

    a. If angle DBC= 25x+7 and angle TBC= 12x+5, find the degree measures of angle TBC, DBT and DBC

    b. If angle CBT= 10x-8 and angle TBA=15x-12, find the degree measures of both angles.

    I drew a diagram and everything, but I am still utterly confused.
    ...

    You knwo what? forget it.. I'll just go look for it elsewhere.

    <33 Buyo
    Let me begin by saying that your attitude is very much not appreciated! you seem to think that people come on this site with nothing better to do than help you. if you are going to participate in a community such as this, be patient and respectful to the other members. you are only hurting yourself by acting like that! i'm sure all the other helpers will hate me for helping you, but i'll just give you your answers so you can be on your way.

    a.

    segment BT bisects \angle DBC. this means that \angle DBT = \angle TBC and \angle DBC = 2 \angle TBC

    Thus we have that:
    2(12x + 5) = 25x + 7

    \Rightarrow 24x + 10 = 25x + 7

    \Rightarrow x = 3

    Thus, \angle TBC = \angle DBT = 12(3) + 5 = 41^{\circ}

    and \angle DBC = 2 \angle TBC = 82^{\circ}




    b.
    note that \angle CBT and \angle TBA are angles on a straight line, this means they add up to 180^{\circ}

    So, (15x - 12) + (10x - 8) = 180

    \Rightarrow 25x = 200

    \Rightarrow x = 8

    \Rightarrow \angle CBT = 10(8) - 8 = 72^{\circ}

    and \angle TBA = 180 - CBT = 108^{\circ}

    and double posting is against the rules, we don't care if you are spamming or not. it wastes peoples time when the same question is floating all around the place
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  4. #4
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    Sorry!

    Ah, i'm so sorry... I was really frustrated because I had a ton other things to do for homework. @-@ I apologize for being impatient! I'm usually not so grouchy.

    And I can't thank you enough for helping me even though was being an angry little snot.

    Oh, and sorry for double posting... ^^; I'm such a nerd.
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