e.g. In this picture, you are given the length of line DE and arc DE and asked for R.It sounds easy!11but after staring at it for a few minutes, working it for a few minutes, and Googling for a few minutes it doesn't seem there's any approach that yields an exact solution ???[Edited on June 13, 2009 at 2:16 PM. Reason : ???]
6/13/2009 2:16:16 PM
6 feet
6/13/2009 2:17:55 PM
9th grade called.... it wants it's math problem back.
6/13/2009 2:19:01 PM
9th grade called.... it wants its math problem back.
ITISMATH
6/13/2009 2:19:46 PM
oh nm[Edited on June 13, 2009 at 2:26 PM. Reason : ]
6/13/2009 2:22:08 PM
nvm, you said you have the length of the arc. brb.[Edited on June 13, 2009 at 2:24 PM. Reason : ]
6/13/2009 2:22:28 PM
false
6/13/2009 2:23:33 PM
can't you find the angle at the origin between the two ends of the arc from the arc length?then you have a triangle where you know one of the angles and one of the sides and you know that the other two sides and angles are equal to each otherthen when you find that side length, that's the radiusugh, dredging up that memory had to overwrite something important, i bet i can't drive anymore now
6/13/2009 2:24:06 PM
^^ yeah i overlooked the part about arc length
6/13/2009 2:25:12 PM
basically you make a triangle with DE and two radii PD and PE
6/13/2009 2:27:16 PM
I JUST SAID THAT
6/13/2009 2:28:18 PM
yeah but i was succinct. yours was tldr[Edited on June 13, 2009 at 2:30 PM. Reason : and therefore my response was more accurate ]
6/13/2009 2:30:32 PM
i will not have my ability to drive die in vain
6/13/2009 2:31:45 PM
a=arc lengthc=cord lengthr=radiust=thetaOkay. So make the triangle, and then you get something like c^2 = 2r*(1-cos(a/r)) or sin(t) = c*t/(2a). Maybe I'm a bit blurry on my trig, but I can't think of any useful identitites for sin(x)/x or x*cos(x), except in a limit I resorted to numerical approximation at that point. I don't think exact solutions exist, I haven't found one, and I guess that's all I was saying: the "9th grade math problem" takes an unexpected turn [Edited on June 13, 2009 at 2:38 PM. Reason : .]
6/13/2009 2:35:17 PM
Chord length, C, isC = 2R * sin(theta/2)Arc length, L, isL = R * thetaC/L = 2 * sin(theta/2)/thetaSolve for theta. The solution is an infinite series. Just solve it numerically.R = L / theta]
6/13/2009 2:38:38 PM
the surest way to get a nerd to do your work for you - tell them it can't be done
6/13/2009 2:40:30 PM
http://www.1728.com/circsect.htm
6/13/2009 2:43:00 PM