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Math Help - Competitive Exam Q related to Circles

  1. #1
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    Post Competitive Exam Q related to Circles

    Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.


    Please do provide an explanation on how the answer can be obtained... Thanks in advance.
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    Re: Competitive Exam Q related to Circles

    Quote Originally Posted by SevenAces View Post
    Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.
    Find the measures of \text{arc}(AC)~\&~\text{arc}(BC).
    The measure of \text{arc}(AB) is the angle you require.
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    Re: Competitive Exam Q related to Circles

    Quote Originally Posted by Plato View Post
    Find the measures of \text{arc}(AC)~\&~\text{arc}(BC).
    The measure of \text{arc}(AB) is the angle you require.
    How do I do that???
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    Re: Competitive Exam Q related to Circles

    Quote Originally Posted by SevenAces View Post
    How do I do that???
    Look up measures of arcs in your textbook.
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    Re: Competitive Exam Q related to Circles

    Quote Originally Posted by Plato View Post
    Look up measures of arcs in your textbook.
    There aren't any measures of arcs specified..!?!? Can you please explain me step-by-step on how to obtain the solution. I have my test tomorrow!
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  6. #6
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    Re: Competitive Exam Q related to Circles

    Hi SevenAces.
    Google angles between secants then regentsprep.org
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    Re: Competitive Exam Q related to Circles

    Quote Originally Posted by SevenAces View Post
    Hey, I can't solve this Q which is in the prep book of a competitive exam that I'm taking. I have attached a pic of it.


    Please do provide an explanation on how the answer can be obtained... Thanks in advance.

    Angle CBA = Angle ACT = 48 degree

    Triangle BCT,
    Angle ACB = 180 - 48 - 36 - 48 = 48 degree

    Angle subtended by AB at the centre of circle = 2 (48) = 96 degree
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