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Math Help - Keeping the ratio of the rectangle

  1. #1
    zxe
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    Exclamation Keeping the ratio of the rectangle

    Please see picture attached:

    I need to find the size of the Green rectangle.

    I am trying to cut the Green rectangle out between the parallel lines.

    The Green rectangle has to be in the ratio of 27:10

    I attempted using trigonometry, I calculated the angle between the diagonal line of the Green rectangle and its longer adjacent line is 22 degrees.

    But now I am stuck. Please help!

    edit: The 220 cm line is perpendicular to the parallel lines. The 300 cm line is also the longer Green line extanded.
    Attached Thumbnails Attached Thumbnails Keeping the ratio of the rectangle-untitled.png  
    Last edited by zxe; April 16th 2013 at 03:33 AM.
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  2. #2
    MHF Contributor

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    Re: Keeping the ratio of the rectangle

    If you were to draw a line from the lower left corner, perpendicular to the two parallel lines, you would have a right triangle in which the 'near' side has length 220 cm and the hypotenuse has length 300 cm. Calling the angle, in the right triangle at that corner, \theta, we have cos(\theta)= \frac{220}{300}= \frac{11}{15} so that \theta= arccos\left(\frac{11}{15}\right). That means that the angle between the diagonal and the new perpendicular is arccos\left(\frac{11}{15}\right)- 28 degrees. So you now have a right triangle in which that diagonal is the hypotenuse, the angle is arccos\left(\frac{11}{15}\right)- 28 degrees, and the "near" leg has length 220 cm.
    Last edited by HallsofIvy; April 16th 2013 at 08:49 AM.
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  3. #3
    zxe
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    Re: Keeping the ratio of the rectangle

    Thanks HallsofIvy,

    I don't understand how you get 28 degrees, but I have solved the problem with similar approach.

    I will remember to label the diagram next time!
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