# How do I solve this trigonometry?

L and H are known values, how do I get (solve) $RH$:

$RH * (1 – \cos ( \arctan(L/2 / RH ))) = H$

RH (not R*H) is circle radius

H is difference in beween radius and line crossing to circle.

L is length of line.

//edit

simply put, by replacing RH with R

$R * (1 – \cos ( \arctan(L/2 / R ))) = H$

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What is L/2/RH? IS RH R*H or a variable RH? –  user45099 Jan 30 '13 at 14:15
is that $arctan\frac{\frac{L}{2}}{RH}$? –  Joseph Skelton Jan 30 '13 at 14:15
You might want to learn MathJax so that you can more easily explain your questions in the future. –  Sam DeHority Jan 30 '13 at 14:17
@user1709828, ah; no RH is NOT R*H, please se elaboration –  user247245 Jan 30 '13 at 14:20
@JosephSkelton, yes, that's correct. –  user247245 Jan 30 '13 at 14:27

You should get rid of the trigonometric functions. In fact, $\cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}$.

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Phytagoras - perfect, thanks. –  user247245 Jan 31 '13 at 8:58

Look at a right-angled triangle to see if you can simplify $\cos(\arctan(x))$, then solve for RH.

If you have a right-angled triangle where one of the non-hypotenuse sides is $L/2/RH$, and the other non-hypotenuse side is $1$, which angle corresponds to $\arctan(L/2/RH)$, and therefore what is $\cos(\arctan(L/2/RH))$?

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Yes, I belive You're right here. Think of a line crossing close ot border of a circle. H is the small (maximum) distance between the circle and line. L is the length of the line. I want the radius of the circle. –  user247245 Jan 30 '13 at 14:31
Suceeded in simplifying to: L = 2R * arctan((L/2) / (R-H)). Puzzled from here.. –  user247245 Jan 30 '13 at 21:11
If you move the $2R$ over, you have $L / 2R = \arctan((L/2) / RH)$. How do you get rid of the $\arctan$? –  ferson2020 Jan 30 '13 at 21:16
The best I can come up with is L/2 / (R-H) = tan(L / 2R) –  user247245 Jan 30 '13 at 21:27
And you're solving for $RH$, where $L$ and $R$ are known, right? –  ferson2020 Jan 30 '13 at 21:31