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

Suppose we have a circle of radius
$r$
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
$\theta$
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
$\theta$
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
$3 \times A_{MINOR \: SEGMENT}= A_{MAJOR \: SEGMENT}$

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

Dividing by
$r^2$
gives
$3 \frac{1}{2}( \theta - \sin theta)=\pi - \frac{1}{2} ( \theta - \sin theta)$

$\frac{1}{2} ( \theta - \sin theta)$
$4\frac{1}{2} ( \theta - \sin theta)= \pi$
$2 ( \theta - \sin theta)= \pi$