Howard,
Below is our diagram of your problem.
The problem is to find the measure of the angle ACB so that the area of the triangle ACB is equal to the area of the region of the sector ACB that is outside the triangle. To say it in another way, find the measure of the angle ACB if the area of the triangle ACB is half the area of the sector ACB.
Let x be the measure of the angle ACB in radians and r be the radius of the circle, then the area of the sector ACB is
AS = ^{1}/_{2} r^{2} x
Let D be the midpoint of AB, b = AB and h = CD then the area of the triangle ACB is
AT = ^{1}/_{2} b h
Using the facts that h = r cos( ^{x}/_{2} ), b = 2r sin( ^{x}/_{2} ) and sin(x) = 2 sin( ^{x}/_{2} ) cos( ^{x}/_{2} ) we get
AT = ^{1}/_{2} r^{2} sin(x).
But AS = 2 AT and hence
2 sin(x) = x
Solving this equation is not straightforward. One technique is to use Newton's method to approximate a solution. There is an outline of Newton's method in the answer to a previous problem.
Stephen and Penny
