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Math Help - Trigonometric system

  1. #1
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    Trigonometric system

    Hi,

    I can't solve this trigonometric system in order to y and only appearing Theta(1), h, L and w:

    sin(Theta(1))= x/L1

    sin(Theta(2))= (w-x)/L2

    cos(Theta(1))= y/L1

    cos(Theta(2))=(y-h)/L2

    L1+L2=L

    If necessary you can use sin(theta(1))=sin(theta(2))=w/L

    Thanks
    Last edited by kolg; February 21st 2010 at 07:27 AM.
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  2. #2
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    Hello, kolg!

    I can't solve this trigonometric system in order to y
    and only appearing Theta(1), h, L and w. .
    What does that mean?

    . . [1]\;\;\sin(\theta_1) \:=\: \frac{x}{L_1}

    . . [2]\;\;\sin(\theta_2) \:=\: \frac{w-x}{L_2}

    . . [3]\;\;\cos(\theta_1) \:=\: \frac{y}{L_1}

    . . [4]\;\;\cos(\theta_2) \:=\: \frac{y-h}{L_2}

    . . [5]\;\;L_1+L_2\:=\:L

    If necessary, you can use: \sin(\theta_1)\,=\,\sin(\theta_2)\,=\,\frac{w}{L}

    This doesn't make sense.
    It may be a case of TMI (too much information).

    I will assume that we are to solve for y.


    Square [1]: . \frac{x^2}{L_1^2} \:=\:\sin^2(\theta_1)

    Square [3]: . \frac{y^2}{L_1^2} \:=\:\cos^2(\theta_1)


    Add: . \frac{x^2}{L_1^2} + \frac{y^2}{L_1^2} \;=\;\underbrace{\sin^2(\theta_1) + \cos^2(\theta_1)}_{\text{This is 1}}

    We have: . \frac{x^2+y^2}{L_1^2} \:=\:1 \quad\Rightarrow\quad w^2 + y^2 \:=\:L_1^2 \quad\Rightarrow\quad y^2 \:=\:L_1^2 - x^2


    Therefore: . y \;=\;\pm\sqrt{L_1^2 - x^2}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Oh-oh!

    I think they want the answer in terms of L . . . and not L_1

    Last edited by Soroban; February 21st 2010 at 01:55 PM.
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  3. #3
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    Thanks for the reply, but yes I need that in order to L, and that is what I can't figure out

    Thanks
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