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### Distance of Chord from Circumference

```
Date: 12/31/97 at 19:39:06
From: David Simpson
Subject: Trig, Distance of Chord from Circumference

Is it possible to calculate the vertical distance, at a right angle,
from a chord to the circumference of a circle?

The following details are known:

Area of Minor Segment

Dave
```

```
Date: 01/01/98 at 08:40:16
From: Doctor Jerry
Subject: Re: Trig, Distance of Chord from Circumference

Hi Dave,

See the figure:

I'm assuming that the shaded part is a minor segment.  It is known
that the area of such a segment is

A = (a^2/2)(t-sin(t))

where t is in radians. If you know A and a, then the angle can
be calculated with a numerical procedure.  So, t is known
(approximately, as accurately as you like) and you can calculate t/2.
Then, cos(t/2) = a/(a-h), where h is the distance from the chord to
the circumference. Now you can calcuate h.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 01/02/98 at 18:15:19
From: Dave
Subject: Re: Trig, Distance of Chord from Circumference

Many thanks for your prompt help it is very good of you.

However I am going to have to ask you to indulge me a little further
if you will (it is a long time since I did any of this stuff).

Given that A and a are known we have the lefthand side of the
equation, but I have no idea how to calculate t-sin(t) from this
information.

Can you, help?

Dave
```

```
Date: 01/06/98 at 08:50:12
From: Doctor Jerry
Subject: Re: Trig, Distance of Chord from Circumference

Hi Dave,

Okay, we have the equation A = (a^2/2)(t-sin(t)), where we know A and
a. As a sample problem, suppose we take a = 3 and t = pi/3.  We find
A = 0.81527...

As the sample problem, take a = 3 and A = 0.81527. When we solve the
equation A = (a^2/2)(t-sin(t)) for t we should find t = pi/3=1.047...

The equation is

2*A/a^2 = t-sin(t)

or

(1) 0.18117 = t-sin(t).

I don't know if you have a calculator and can set it in radians, but
I'll assume that this is a possibility for you. Of course, if you have
a scientific calculator with a "solver" on it, or if you can graph and
zoom, then your problems are over.

There are many procedures for locating a value of t that solves (1).
Some of these are quite sophisticated, others are more or less
informed guessing. I'm not at all sure which to choose.

Perhaps the "bisection method." First, you must locate an interval
containing the t we are looking for. How to do this?

First, let g(t) = 0.18117-t+sin(t). We want to find a value of t
(called a zero of g) for which g(t) = 0. Usually, g(t)>0 on one side
of the zero and g(t)<0 on the other side.

g(1.5) = -0.32...  So, the graph of g crosses the axis somewhere
between 1 and 1.5.

We calculate the midpoint 1.25 of the interval and say: the zero must
be in either the left half or right half. To decide, calculate
g(1.25)=-0.11...

So the zero is in the left half.  Our new interval is [1,1.25]. We now
repeat this procedure. The midpoint is 1.125; g(1.125)<0 so the new
interval is [1,1.125]. Repeat until you obtain the accuracy you need.

Here are successive midpoints:

1.25
1.125
1.0625
1.03125
1.046875
1.0546875
1.05078125
1.048828125
1.0478515625

Recall that pi/3 = 1.047...

You can see that the Bisection method isn't fast. It's sure, but not
fast. Newton's method is much faster. This require calculus.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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