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

• Aug 22nd 2013, 09:03 AM
lukasz
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
• Aug 22nd 2013, 09:17 AM
Prove It
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.
• Aug 23rd 2013, 12:00 AM
lukasz
Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h
Thanks!
• Aug 23rd 2013, 03:08 AM
ibdutt
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.Attachment 29055
• Aug 23rd 2013, 08:35 AM
Soroban
Re: How to obtain alfa from the equation: cos(alfa)*R-sin(alfa)*L=R-h
Hello, lukasz!

Quote:

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

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

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

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

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

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

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

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