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Math Help - How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

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    How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

    How shall I transfer the equation: cos(alfa)*R-sin(alfa)*L=R-h ; to obtain angle (alfa) ?

    Values: R,L,h are given

    Thanks
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  2. #2
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    Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

    \displaystyle \begin{align*} R\cos{(\alpha)} - L\sin{(\alpha)} &= R - h \\ R\cos{(\alpha)} &= R - h + L\sin{(\alpha)} \\ \left[ R\cos{(\alpha)} \right] ^2 &= \left[ R - h + L\sin{(\alpha)} \right] ^2 \\ R^2\cos^2{(\alpha)} &= \left( R - h \right) ^2 + 2 \left( R - h \right) L\sin{(\alpha)} + L^2\sin^2{(\alpha)} \\ R^2 \left[ 1 - \sin^2{(\alpha)} \right] &= \left( R - h \right) ^2 + 2 \left( R - h \right) L \sin{(\alpha)} + L^2 \sin^2{(\alpha)} \\ R^2 - R^2\sin^2{(\alpha)} &= \left( R - h \right) ^2 + 2 \left( R - h \right) L \sin{(\alpha)} + L^2 \sin^2{(\alpha)} \\ 0 &= \left( L^2 + R^2 \right) \sin^2{(\alpha)} + 2 \left( R - h \right) L \sin{(\alpha)} + \left( R - h \right) ^2 - R^2 \\ 0 &= \left( L^2 + R^2 \right) \sin^2{(\alpha)} + 2 \left( R - h \right) L \sin{(\alpha)} + h^2 - 2\,R\,h  \end{align*}

    This is a Quadratic Equation you can now solve for \displaystyle \begin{align*} \sin{(\alpha)} \end{align*} and from there you can solve for \displaystyle \begin{align*} \alpha \end{align*}. Of course, since squaring has been required to get to this result, you will need to check if any of your solutions are extraneous.
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    Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

    Thanks!
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    Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

    I would prefer to go by the attached method.How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h-23-aug-13.png
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    Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h

    Hello, lukasz!

    \text{Solve for }\alpha\!:\;R\cos\alpha - L\sin\alpha \:=\:R-h

    \text{Divide by }\sqrt{R^2+L^2}\!:\;\frac{R}{\sqrt{R^2+L^2}} \cos\alpha - \frac{L}{\sqrt{R^2+L^2}}\sin\alpha \:=\:\frac{r-h}{\sqrt{R^2+L^2}}

    \text{Let }\,\cos\theta \,=\,\frac{R}{\sqrt{R^2+L^2}},\;\sin\theta \,=\,\frac{L}{\sqrt{R^2+L^2}}

    \text{We have: }\:\cos\theta\cos\alpha - \sin\theta\sin\alpha \:=\:\frac{R-h}{\sqrt{R^2+L^2}}

    . . . . . . . . . . . . . . . \cos(\theta + \alpha) \:=\:\frac{R-h}{\sqrt{R^2+L^2}}

    . . . . . . . . . . . . . . . . . . \theta + \alpha \:=\:\arccos\left(\frac{R-h}{\sqrt{R^2+L^2}}\right)

    . . . . . . . . . . . . . . . . . . . . \alpha \;=\;\arccos\left(\frac{R_h}{\sqrt{R^2+L^2}}\right  ) - \theta

    . . . . . . . . . . . . . . . . . . . . \alpha \;=\;\arccos\left(\frac{R-h}{\sqrt{R^2+L^2}}\right) - \arccos\left(\frac{R}{\sqrt{R^2+L^2}}\right)
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