Finding the Equation for the Angle Subtended By a Chord That Splits A Circle in a Ratio

Suppose we have a circle of radius  
  with a chord. The chord cuts the circle into minor and major segments, with areas in the ratio 1:3. We can find an angle for the angle  
  subtended by the chord AB.

Using the chord construct a sector.

The area of the minor sector is  
\[\frac{1}{2}r^2 \theta\]
  and the area of the triangle is  
\[\frac{1}{2}r^2 sin \theta\]
&nbs[; where  
  is in radians.
The difference between these is the area of the minor segment:  
\[A_{MINOR \: SEGMENT}=\frac{1}{2}r^2 \theta - \frac{1}{2}r^2 sin \theta = \frac{1}{2} r^2( \theta - \sin theta)\]
Since the area of the circle is  
\[\pi r^2\]
, the area of the major segment is  
\[\pi r^2 - \frac{1}{2} r^2( \theta - \sin theta)\]
These are in the ratio 1:3 so  
\[3 \times A_{MINOR \: SEGMENT}= A_{MAJOR \: SEGMENT}\]

\[3 \frac{1}{2} r^2( \theta - \sin theta)=\pi r^2 - \frac{1}{2} r^2( \theta - \sin theta)\]

Dividing by  
\[3 \frac{1}{2}( \theta - \sin theta)=\pi - \frac{1}{2} ( \theta - \sin theta)\]

\[\frac{1}{2} ( \theta - \sin theta)\]
  to both sides:
\[4\frac{1}{2} ( \theta - \sin theta)= \pi\]

\[2 ( \theta - \sin theta)= \pi\]