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Geometry Problem

  1. Feb 26, 2005 #1
    In [tex] \bigtriangleup ABC , AB = BC[/tex] and [tex]CD[/tex] bisects angle C.
    If [tex]y = \frac{1}{3}x[/tex] then z = ....

    --------------------------------------------

    Ok. I've tried to make a bunch of substitutions for each of the angles and I can't seem to solve any for z.

    I have found these two equations:

    [tex] y + z + c = 180[/tex]

    and

    [tex] x + z + \frac{c}{2} = 180[/tex]

    I know I can make substitutions/variants of these, but I still am stuck.
    Any help is appreciated.



    Jameson
     

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  2. jcsd
  3. Feb 26, 2005 #2

    cepheid

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    Notice that angle CDB = 180 - x. So:

    (180 - x) + y + C/2 = 180 (from the upper triangle).

    (180 - x) + x/3 + C/2 = 180

    C/2 = 2x/3

    Can you proceed from here?
     
  4. Feb 26, 2005 #3
    I got the same thing. [tex] \frac{c}{2} = \frac{2x}{3} [/tex] but unfortunately I still can't seem to do the right substitution.

    Substituting (2x)/3 for c/2 I get the equation:

    [tex] x + \frac{2x}{3} + z = 180 [/tex]

    This still leaves two variables. I just can't see it for some reason.
     
  5. Feb 26, 2005 #4
    Well, since you had no values in your original problem, I think that your answer will need at least two variables...
     
  6. Feb 26, 2005 #5
    This is from a published SAT book from College Board, so I think there is a solution.
     
  7. Feb 26, 2005 #6

    cepheid

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    I, too, assumed that you were meant to solve for z in terms of x, since y was given in terms of x. I don't see what else you can do with it...
     
  8. Feb 26, 2005 #7
    no no, see how AB = BC, That's a very important clue. We know triangle ABC is an isoceles triangle with angle A = angle C. You can come up with an equation to relate x directly to z and substitute it back into one of the equations to solve for z.

    Angle C = 4/3 x = Angle A = z

    you should be able to take it from here.
     
  9. Feb 27, 2005 #8

    honestrosewater

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    Eh, doesn't 2z + y = 180? So knowing y = x/3 you can solve for z in terms of x.

    Plus, if it's the CB's "10 Real SAT's", it has the answers. And any other SAT prep book should have the answers too.
     
    Last edited: Feb 27, 2005
  10. Feb 27, 2005 #9
    ok, that equation is essential to solve for z. However, an actual value can be determined here. It's not impossible. I solved it and i got z = 80 degrees. Let me know if you want me to show my work.
     
  11. Feb 27, 2005 #10
    apchemstudent... 80 degrees is the correct answer. I would be most gracious if you could show your work.

    Jameson
     
  12. Feb 27, 2005 #11
    ok, we set angle BDC = 180 - x, so that we can solve for Angle BCD

    BCD + DBC + CDB = 180 degrees
    BCD = 180 -1/3 x - 180 + x
    BCD = 2/3 x

    BCD = 1/2 BCA so BCA = 4/3 x

    Again we know that it is an isoceles triangle with angle BCA = BAC
    z = 4/3 x

    so BAC + ABC + BCA = 180
    z + 1/3 x + 4/3 x = 180
    3x = 180
    x = 60 <---- you could've substituted x in terms of z and solved for z but i find it easier this way.

    now z = 4/3 (60) = 80 degrees.
     
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