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Math Help - Solving for a

  1. #1
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    Solving for A (and D)

    Original Equation:

    \sqrt{\frac{5 \cdot G \cdot D}{M^2 \cdot sin(2 \cdot A)}} \cdot 4 = P

    Solving for D:

     \frac{M^2  \cdot P^2 \cdot sin(2 \cdot A)} { 80 \cdot G } = D

    Solving for A (Don't think is right):

     \frac {sin^{-1} \left (\frac{80 \cdot G \cdot D} { M^2 \cdot P^2} \right )} {2} = A<br />


    I changed my original post after Dan replied to it. The equation was displayed wrong. I am trying to solve for A for one equation, and D for another. This is what I got after seeing Dan's work on how to work with sine. I would love to be double checked.
    Last edited by JohnC; June 5th 2008 at 08:05 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JohnC View Post
    Original Equation
    .._____________
    98*(800(D/40)) * 4 = P
    4*(240)*SIN(2A)

    or

    4*SQRT((98 (800 (D/40))/(4(240)sin(2A))) = P


    ------------------------


    98(800(D/40)) = sin(2A)
    (P/4)*4(240)

    ------------------------

    That's about as far I got. I forgot how to get rid of sine since I haven't done it in about 2 years. If some one can tell me how I can go about getting rid of the sine, or alternatively, solving the problem for A, that would be amazing. I kept the numbers unsimplified for a reason. If you wish to solve for it and simplify it, 98 = G, and 240 = M. Otherwise, I would love to know how to get rid of the sin and solve for A.
    You have to use the inverse of the sine function: sin^{-1} or "asn" (arcsine.) Also, you didn't need to square both sides to get rid of the square root.
     \frac{\sqrt{98*(800(D/40))}}{4*(240)*sin(2A)} \cdot 4 = P

    sin(2A) =  \frac{\sqrt{98*(800(D/40))}}{4*(240)*P} \cdot 4

    Now take the inverse sine:
    2A =  sin^{-1} \left ( \frac{\sqrt{98*(800(D/40))}}{4*(240)*P} \cdot 4 \right )

    A =  \frac{1}{2} \cdot sin^{-1} \left ( \frac{\sqrt{98*(800(D/40))}}{4*(240)*P} \cdot 4 \right )

    -Dan
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  3. #3
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    I updated my main post so the equation is easier to read. Thanks for reminding me to use arcsine
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  4. #4
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    Quote Originally Posted by JohnC View Post
    Original Equation:

    \sqrt{\frac{5 \cdot G \cdot D}{M^2 \cdot sin(2 \cdot A)}} \cdot 4 = P

    Solving for D:

     \frac{M^2  \cdot P^2 \cdot sin(2 \cdot A)} { 80 \cdot G } = D

    Solving for A (Don't think is right):

     \frac {sin^{-1} \left (\frac{80 \cdot G \cdot D} { M^2 \cdot P^2} \right )} {2} = A<br />
    Are there any restrictions on G,\;D,\;M,\;\text{or}\;A? As it is, we have

    4\sqrt{\frac{5\,GD}{M^2\sin(2A)}} = P

    \Rightarrow16\left\lvert\frac{5\,GD}{M^2\sin(2A)}\  right\rvert = P^2

    \Rightarrow\frac{80|GD|}{M^2\left\lvert\sin(2A)\ri  ght\rvert} = P^2

    \Rightarrow|D| = \frac{M^2P^2\left\lvert\sin(2A)\right\rvert}{80|G|  }

    \Rightarrow D = \pm\frac{M^2P^2\left\lvert\sin(2A)\right\rvert}{80  |G|}

    and, for A,

    |\sin 2A| = \frac{80|GD|}{M^2P^2}

    \sin 2A = \pm\frac{80|GD|}{M^2P^2}

    \Rightarrow A = \frac12\left[\arcsin\left(\pm\frac{80|GD|}{M^2P^2}\right) + 2n\pi\right],\;n\in\mathbb{Z} or A = \frac12\left[(2n + 1)\pi - \arcsin\left(\pm\frac{80|GD|}{M^2P^2}\right)\right],\;n\in\mathbb{Z}

    This can all possibly be simplified quite a bit if more information is given about the variables involved.
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  5. #5
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    For the problem I am currently working on:
    A = Angle, (0 - 90)
    D = Distance, (0 - 120)
    G = Constant, 98
    M = Constant, 240
    P = Power, (0 - 4)

    On a side note, and I think this will be too complicated for me, but:
    How would one go about calculating the equation of an arc given just an image. I would guess Trial and Error to match the image?
    Last edited by JohnC; June 6th 2008 at 10:37 AM.
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