Consider the following:(adsbygoogle = window.adsbygoogle || []).push({});

On a circle of radius 1, two points are marked: P1 and P2.

Two lines are drawn from the center of the circle:

one from the center to P1,

the other from the center to P2.

The angle between these two lines is [itex]\theta[/itex].

One more line is drawn: from P1 directly to P2. In other words, this third line is a chord on this circle.

For the special condition that the length of this chord equals the angle, find a simple expression.

i.e. – find a simple expression for [itex]\theta[/itex] given the special condition that chord length = [itex]\theta[/itex] = angle = [itex]\theta[/itex]

- - -

So far, all the expressions that I have worked out mix terms of [itex]\theta[/itex] and either sin([itex]\theta[/itex]) or cos([itex]\theta[/itex]); I have not been able to find an expression simply in terms of [itex]\theta[/itex], sin([itex]\theta[/itex]), or cos([itex]\theta[/itex]).

For example, following is one of my approaches:

Bisect the angle [itex]\theta[/itex], which also divides the chord in half.

The chord length is [itex]\theta[/itex].

But this value is also 2 sin([itex]\theta[/itex]/2)

Equating these two expressions: 2 sin([itex]\theta[/itex]/2) = [itex]\theta[/itex] or sin([itex]\theta[/itex]/2) = [itex]\theta[/itex]/2

I cannot find a way to simplify this expression further.

Any suggestions?

**Physics Forums - The Fusion of Science and Community**

# Simplify Condition for Chord Length Equals Angle?

Have something to add?

- Similar discussions for: Simplify Condition for Chord Length Equals Angle?

Loading...

**Physics Forums - The Fusion of Science and Community**