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Math Help - circle problem..

  1. #1
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    circle problem..

    In the attached figure, AB//OC, <ABC=130, Find x,y.
    Attached Thumbnails Attached Thumbnails circle problem..-2pyam9u.jpg  
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  2. #2
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    Hint:
    let angleBAO = z
    x + y = 130 - 2z
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  3. #3
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    Quote Originally Posted by Wilmer View Post
    Hint:
    let angleBAO = z
    x + y = 130 - 2z

    err...thanks wilmer...but i still cant get it ....
    would u plz be kind enough to elaborate how u got this eqn- x + y = 130 - 2z ? ?
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  4. #4
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    Quote Originally Posted by snigdha View Post
    would u plz be kind enough to elaborate how u got this eqn...
    What does "u" mean?

    Well, instead, I'll give you another hint: x - y = z - 50
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  5. #5
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    Quote Originally Posted by snigdha View Post
    In the attached figure, AB//OC, <ABC=130, Find x,y.
    x and y are "alternate" angles, if the horizontal lines are parallel.
    Now, the angle at x can be anywhere along the line segment OB, with x and y still equal. But are they parallel?
    Last edited by Archie Meade; February 7th 2010 at 11:21 AM.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    x and y are "alternate" angles, if the horizontal lines are parallel.
    Now, the angle at x can be anywhere along the line segment OB, with x and y still equal. But are they parallel?
    Ya, good question Archie.
    I don't think it's possible that they are parallel, with AD being diameter
    and CO being radius.
    The given diagram is misleading...agree?
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  7. #7
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    Exactly, Wilmer,

    we can join O to B, creating 3 internal isosceles triangles.
    We can label identical angles and discover that angles COD and BAO are 80 degrees and BCO is 50 degrees.

    This means the angle at B cannot form a parallelogram.

    The question is, where do we put the point at x ?
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    The question is, where do we put the point at x ?
    The question is probably find the sum x+y; so point at x anywhere on AO.
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  9. #9
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    That may be it, good idea!
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  10. #10
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    Quote Originally Posted by Archie Meade View Post
    That may be it, good idea!
    I'm in AA
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  11. #11
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    Think about this:
    (They are all angles, by the way)

    ABC+ADC=180
    ODC=OCD
    ODC+OCD=AOC
    ABC+BCO=180

    Do you follow?
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  12. #12
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    Quote Originally Posted by pragraphic View Post
    Think about this:
    (They are all angles, by the way)

    ABC+ADC=180
    ODC=OCD
    ODC+OCD=AOC
    ABC+BCO=180

    Do you follow?
    No...
    Well...i tried your method...check out this-

    ABC+ADC=180 -->[As ABCD is a cyclic quadrilateral so opp <s are
    supplementary]
    => 130+ADC=180
    => ADC= 50

    In isosceles triangle OCD,
    ODC=OCD=50

    AOC=OCD+ODC -->[Ext <=sum of interior opp. <s]
    =50+50
    =100

    ABC+BCO=180 -->[Co-interior angles]
    => 130+ BCO=180
    =>BCO=50

    XCO= COD+CDO -->[Ext <=sum of interior opp. <s]
    ={180-(50+50)} + 50
    = 130

    From the diagram, we see that BCO is greater than XCO
    But according to the above calculations, BCO being 50 is lesser than XCO 130.

    So what d'ya say??

    what other method can we apply??
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  13. #13
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    For this one:
    "XCO= COD+CDO -->[Ext <=sum of interior opp. <s]"
    ={180-(50+50)} + 50
    = 130

    COD+CDO not equal to XCO!

    But think this way:

    x+XCO=COD [Ext <=sum of interior opp. <s]

    (XCO is the angle next to y and that 50 degree angle, right?)
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  14. #14
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    Quote Originally Posted by pragraphic View Post
    For this one:
    "XCO= COD+CDO -->[Ext <=sum of interior opp. <s]"
    ={180-(50+50)} + 50
    = 130

    COD+CDO not equal to XCO!

    But think this way:

    x+XCO=COD [Ext <=sum of interior opp. <s]

    (XCO is the angle next to y and that 50 degree angle, right?)
    umm...yep..right!!
    thanks..!
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