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Problem about angles

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    We have ##4x+x+y=280^0##(degrees), find ##x##, ##y## and ##4x##.


    2. Relevant equations
    ##4x+x+y=280^0##


    3. The attempt at a solution
    ##4x+x+y=280^0##
    ##5x+y = 280^0##
    ##5x=280^0-y##
    ##x=\frac{280^0-y}{5}##
    ##x=56^0-y##
    ---------------------
    ##4x=4(56^0-y)=224^0-y##
    ------------------------------------
    ##(224^0-y+56^0-y)+y=280^0##
    ##y=280^0-(224^0-y+56^0-y)##
    ##y=280^0-280^0-y##
    ##y=0##
    Is that correct? thanks!
     
  2. jcsd
  3. Feb 14, 2012 #2

    tiny-tim

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    Hi -Physician! :smile:

    erm :redface:
    ##x=\frac{280^0-y}{5}##
    ##x=56^0-y/5##
    ---------------------
    ##4x=4(56^0-y/5)=224^0-4y/5##
    ------------------------------------
    ##(224^0-y/5+56^0-4y/5)+y=280^0##
    ##280^0=280^0##​
    :wink:
     
  4. Feb 14, 2012 #3
    So the ##y=0##?
     
  5. Feb 14, 2012 #4
    No ,y could be zero but that is just one out of many possible answers.The question as you presented it does not have enough information to get numerical values for x and y.
     
  6. Feb 14, 2012 #5
    This is what the book says. Not my own tasks
     
  7. Feb 14, 2012 #6
    Look at tiny-tims post above.He went with the method you started and corrected your mistakes.He expressed x and 4x in terms of a number and y.Alternatively and just as good,y can be expressed in terms of x and a number.Without extra information that's the best that can be done.
     
  8. Feb 14, 2012 #7
    I see, thank you.
     
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