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Math Help - Equilateral triangle and perpendicular line

  1. #1
    MHF Contributor Unknown008's Avatar
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    Equilateral triangle and perpendicular line

    I didn't quite know what to put as title, so I'm sorry if the title doesn't give a clue of what is to be found

    Well, here's the question:

    Qu: The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR so that ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, the length of ST is

    A. \frac{1}{2}

    B. 2-\sqrt{3}

    C. 2\sqrt{3} - 3

    D. 2(2-\sqrt{3})

    E. \frac{1}{3} \sqrt{3}

    I proceeded with finding the angles, angle STR being 150 degrees and angle TSR = angle TRS = 15 degrees.

    Then, I found the length of QS, since QRS is a right angle with one of the angles being 45 degrees. QS = \frac{1}{\sqrt{2}} = SR.

    Then, fron the sine rule, I know that:
    \frac{L}{sin(15)} = \frac{1}{\sqrt{2} sin(150)}

    So,

    L = \frac{2}{\sqrt{2}sin(15)}

    Now, I don't know the exact form on sin(15) and I don't see any square root of 2 in any of the choices. Is there a quicker/simpler way to solve this?

    Thanks!
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Unknown008 View Post
    Qu: The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR so that ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, the length of ST is
    Unless I made some error, the answer is just 1/2, because consider that triangle SQT must also be equilateral, therefore length of QT = length of ST = length of TR.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    No, I don't think that triangle SQT is equilateral, because ST is perpendicular to PQ. Since one of the angles is 90 degrees, the triangle cannot be equilateral.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Unknown008 View Post
    No, I don't think that triangle SQT is equilateral, because ST is perpendicular to PQ. Since one of the angles is 90 degrees, the triangle cannot be equilateral.
    Oh yes I made the error of reading "ST perpendicular to PQ" as "ST parallel to PR". My mistake.
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  5. #5
    MHF Contributor undefined's Avatar
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    So, triangle SQT is a 30-60-90 triangle, and if we label the hypotenuse QT as having length y and TR as having length x, we get cos(pi/6) = sqrt(3)/2 = x/y which gives y = 2x/sqrt(3). So x+2x/sqrt(3) = 1. After some algebraic manipulation I get answer C.
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  6. #6
    MHF Contributor undefined's Avatar
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    Maybe it was hard to read?

    EDIT: Sorry, I switched Q and R, so my letters are all messed up... uh, let me see what I should have done.



    \cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{x}{y}

    y=\frac{2}{\sqrt{3}}x

    x+y=x+\frac{2}{\sqrt{3}}x=1

    x=\dfrac{1}{1+\frac{2}{\sqrt{3}}}

    =\dfrac{1}{\frac{\sqrt{3}+2}{\sqrt{3}}}

    =\frac{\sqrt{3}}{\sqrt{3}+2}

    =\frac{(\sqrt{3})(2-\sqrt{3})}{2^2-3}

    =2\sqrt{3}-3
    Attached Thumbnails Attached Thumbnails Equilateral triangle and perpendicular line-triaperp.png  
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Sorry, I can't follow your reasoning here.

    Why would triangle SQT be 30-60-90? It could very well be any other angle



    [sorry the 'Q' disappeared]

    EDIT: Ok, the triangle QRT is 30-60-90, so, STR is also 30-60-90.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Uh... undefined, T is on PR, not QR.
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Unknown008 View Post
    Uh... undefined, T is on PR, not QR.
    Yes, I edited my post, and many apologies for my not being able to read.

    But I think the calculations work out the same, by coincidence. Have I made yet another stupid mistake?
    Attached Thumbnails Attached Thumbnails Equilateral triangle and perpendicular line-triaperp2.png  
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Shoo! My drawing is not good. It's not TQ which is equal to ST but TR... darn...
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  11. #11
    MHF Contributor Unknown008's Avatar
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    Finally, I hope I get it right. Yes, now I understand your method.

    So,

    y + x = 1

    \frac{2x}{\sqrt{3}}  + x = 1

    x = \frac{1}{\frac{2}{\sqrt{3}} + 1}

    x = 2\sqrt{3} - 3

    Phew, thanks!
    Last edited by Unknown008; August 5th 2010 at 11:16 AM. Reason: forgot '\' in sqrt function and typo, put 'not' instead of 'now'
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  12. #12
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    Hello,
    Solvable by trig yes but I say geometry no. A quick scaled triangle shows ST = approx 1/2 of QR.


    bjh
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  13. #13
    MHF Contributor Unknown008's Avatar
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    You're right, but all of C, D and E answers are approximately 0.5. How would you choose from those three?
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  14. #14
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    Hello unknown008,
    Solve using trig to get the answer

    bjh
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  15. #15
    MHF Contributor Unknown008's Avatar
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    Well, post what you would have done using trigonometry, I'll be happy to see how you would do it.
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