How do you find a central angle theta when given the arc length cut
off by the angle and the chord length created by the angle?

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Request for Question Clarification by livioflores-ga on 14 Aug 2005 15:02 PDT

For a circumference right?

Let R be the radius of the arc, C the chord length, A the arc length,
and X the angle (in radians).

Then

A = RX
C = 2R sin (X/2).

So

CX = 2A sin (X/2),

which is a transcendental equation for X.  You can solve the equation
numerically.  One way is to plot both sides of the equation as a
function of X and find the point of intersection.

If I understand what you mean, you are talking about a minor circular
segment. If the boundaries of this segment--the minor circular arc and
the chord-- are given or known, how can we find the central angle of
the minor circular sector where the segment is a part of.

Here is one way.

Draw a circle. Draw the chord. Draw radii to the two ends of the
chord. The minor circular sector is that area bounded by the two radii
and the minor circular arc. (The chord cuts the circumference into a
major arc and a minor arc. The minor arc is the shorter arc.) The
circular segment is the area bounded by the chord and the arc.
The said chord divides the sector into an isosceles triangle and the
minor circular segment. Draw a radius that bisects the central angle,
the chord and the minor arc. This radius is perpendicular to the
chord, so the isosceles triangle is bisected into two congruent right
triangles.

Given are lengths of arc and chord. Find the central angle.
 
Let 
L = length of the minor arc.
c = length of the chord.
r = radius of the circle. 
x = theta = central angle, in radians.

>>>arc = (radius) * (central angle in radians)
So,
L = r*(x)  ----(1) 
r = L/x

>>>In any of the two equal right triangles,
hypotenuse = r
one leg = c/2
angle opposite the leg c/2 = (1/2)T = T/2

sin(x/2) = (c/2)/r  -------(2)
r = (c/2) / sin(x/2)

r from (1) = r from (2),
L/x = (c/2)/sin(x/2) 
Cross multiply,
L*sin(x/2) = x*(c/2)
sin(x/2) = x(c/2)/L
sin(x/2) = (cL/2)x  ---------(3)

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One way of doing (3) to solve for x is by graphical method.

Let
y1 = sin(x/2)
y2 = (cL/2)x

Graph y1 and y2 on the same x,y coordinate setup.
y1 is a sine curve passing (0,0).
y2 is a straight line whose slope is cL/2, and it passes the origin (0,0) too.
Their intersection point, when projected or dropped vertically down to
the x-axis, will show the value of x.
This x is the central angle you are looking for.

If you have a graphing calculator, you should be able to find this x,
this central angle.

Thank you for the comments, I believe they have answered my question
completely.  Your help is much appreciated.

Sorry, sorry,

"...sin(x/2) = x(c/2)/L
sin(x/2) = (cL/2)x  ---------(3)"

That should be
sin(x/2) = (c/2L)x  ----(3)

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y2 = (cL/2)x
should be
y2 = (c / 2L)x
Etc..

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Once I could post a comment that is all correct, man, I'd be good! :-)
But that won't come.

When they start paying comments too, I'd have less typos, less mistakes.
But that won't come either.

When I'd be embarrassed for committing these mistakes,...
Hey, that won't come, for sure.

Ain't Math good.
Mistakes or no mistakes.