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I am stuck on an equation I figured out:

$$x = \frac{(g - g_1)\sin(60°) + r }{g}$$

where $g_1$, $r$ and $x$ are already given. At the end I want the equation to be solved for $g$.

Thanks for the help.

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  • $\begingroup$ Welcome to MSE! We use something called MathJax here to format mathematics in questions and answers, making them easier for everyone to read (and it just looks really nice). I edited your question to include MathJax, and you can see what I did by clicking "edit" below the body of your question. If you want to learn how to do it for yourself, here's a tutorial: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up! $\endgroup$ – Robert Howard Oct 20 '18 at 22:34
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Recall that $\sin 60° = \frac{\sqrt 3}{2}$ and then

$$x = \frac{(g - g_1) \cdot \sin(60°) + r} g \iff xg=(g - g_1) \cdot \sin(60°) + r $$

$$xg- g\sin(60°) =- g_1\sin(60°) + r $$

$$g\left(x- \sin(60°)\right) =- g_1\sin(60°) + r $$

$$g =\frac{- g_1\sin(60°) + r}{x- \sin(60°)} $$

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  • $\begingroup$ I know! This is where I was multiple times, but I didnt know how to proceed. I already multiplied everything inside the brackets, tried to get rid of the sin (60°) and it failed. When I divided by x (after your last step), I wasn't able to move the (g - g1) to the left since it is chained to the sind(60°) through multiplication. I know what everything means here, just not sure about the conversion. $\endgroup$ – HeLlOwOrLd Oct 20 '18 at 22:23
  • $\begingroup$ @HeLlOwOrLd I add some step more. $\endgroup$ – gimusi Oct 20 '18 at 22:24
  • $\begingroup$ @HeLlOwOrLd Can you conlcude now? $\endgroup$ – gimusi Oct 20 '18 at 22:27
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Writing $a$ for $\sin(60)$,

$\begin{array}\\ x &= \dfrac{(g - g1) * a + r }{ g}\\ &= \dfrac{g*a - g1 * a + r }{ g}\\ &= a+\dfrac{- g1 * a + r }{ g}\\ \text{so}\\ x-a &= \dfrac{- g1 * a + r }{ g}\\ \text{or}\\ \dfrac1{x-a} &= \dfrac{ g}{- g1 * a + r }\\ \text{or}\\ g &=\dfrac{- g1 * a + r }{x-a}\\ \end{array} $

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