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Math Help - Arc

  1. #1
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    Arc

    I need to find the arc of the circle can you please help I have a headache.
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  2. #2
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    Hello,

    I have a headache too

    Look at this : Transversal line - Wikipedia, the free encyclopedia (i don't know the name of the angles in english)

    Since AB and CE are parallel, angles BAC and ACE are equal.

    We know that CE is the bisector of ACD. And ACD is 110. So what's the value of angle BAC ?


    Now, let's calculate angle OAC.

    Use the theorem Inscribed angles with the center of the circle in their interior (Inscribed angle - Wikipedia, the free encyclopedia i really can't explain it in English, sorry...)
    So you got angle OAC.

    Then, you should know that there is a proportionality relation betwen this angle and the length of the arc :

    OAC/360 = lengthOfTheArc/(perimeter of the circle)
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  3. #3
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    what

    where are you getting oac?
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  4. #4
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    I'm sorry, replace all OAC by BOC
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  5. #5
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    bac

    can i just divide 110? or what to get the arc measurement of bc
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  6. #6
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    Yep, you can divide 110 into 2, because CE is the bisector ;-)
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  7. #7
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    your the best

    so the answer would be ACD/110=36.6 degrees
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  8. #8
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    Hey, no...

    ACD is given in the text : 110

    Read again what i told you, step by step. You should understand ^^'
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  9. #9
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    what

    i am totally confused now.
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  10. #10
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    Ok :/

    CE is the bisector of ACD (110)

    So ACE = 110/2 = 55


    ACE = BAC, because (CE)//(AB) and are defined by the same line. I don't know the name of the rule in english, i've given you a link to wikipedia that may help you.

    So BAC = 55.

    We know that BOC, the angle at the center associated with BAC is twice BAC (see the second wikipedia link - i don't know the name of the rule, but we've learnt it in high school in France).

    So BOC = 110.


    But we know that BC, the arc inscribed by BOC has a length L, given by :

    BOC/360 = L/(2 pi R), where R is the radius of the circle.

    So L is 110/360 * 2 * pi * R = 11 pi R/18
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  11. #11
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    got it

    thanks
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