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Math Help - Find Line Segment AB

  1. #1
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    Find Line Segment AB

    In right triangles ABC and EDC, we have line segment BCD, AC is perpendicular to EC, BD = 10, CE = 8 and the measure of angle ACB = 30 degrees. What is the length of line segment AB?

    (a) 2(sqrt{3})

    (b) 3

    (c) 3(sqrt{3})

    (d) 6

    I selected choice (d) but the book's answer is choice (a).

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  2. #2
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    Quote Originally Posted by sharkman View Post
    In right triangles ABC and EDC, we have line segment BCD, AC is perpendicular to EC, BD = 10, CE = 8 and the measure of angle ACB = 30 degrees. What is the length of line segment AB?

    (a) 2(sqrt{3})

    (b) 3

    (c) 3(sqrt{3})

    (d) 6

    I selected choice (d) but the book's answer is choice (a).
    is there a diagram with the question?
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  3. #3
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    yes..

    Quote Originally Posted by skeeter View Post
    is there a diagram with the question?
    Yes, but I don't know how to draw it here.
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    Quote Originally Posted by sharkman View Post
    Yes, but I don't know how to draw it here.
    sketch it using windows "paint" and upload as an attachment.
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  5. #5
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    Quote Originally Posted by sharkman View Post
    In right triangles ABC and EDC, we have line segment BCD, AC is perpendicular to EC, BD = 10, CE = 8 and the measure of angle ACB = 30 degrees. What is the length of line segment AB?

    (a) 2(sqrt{3})

    (b) 3

    (c) 3(sqrt{3})

    (d) 6

    I selected choice (d) but the book's answer is choice (a).
    This is my first trial of using "paint"

    If that is correct:
     \angle ECD = 60deg
    CD = 8 cos(60deg) = 4
    &
    AB = (10-4)tan(30deg) =  2\sqrt{3}
    Attached Thumbnails Attached Thumbnails Find Line Segment AB-leftangles.jpg  
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    Thanks

    Thank you all for the help.
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  7. #7
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    ok..

    Quote Originally Posted by skeeter View Post
    sketch it using windows "paint" and upload as an attachment.
    I did not know I can do that here.

    Thanks
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  8. #8
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    yes..

    Quote Originally Posted by aidan View Post
    This is my first trial of using "paint"

    If that is correct:
     \angle ECD = 60deg
    CD = 8 cos(60deg) = 4
    &
    AB = (10-4)tan(30deg) =  2\sqrt{3}
    That is correct!
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