# Circular system

In circular system the unit of measurement is 'radian'.
As radian is a unit of measurement so there should be some constant quantity.
Theorem 1: Radian is a constant angle.
Proof: Let us consider a circle with center 'O' and radius 'r'. Take any point 'A' and 'P' on the circle that make an arc AP. Angle formed at the center is $\angle$ AOP = $1^{c}$. Produce AO to meet the circle at B such that $\angle$ AOB is a straight angle =$180^{0}$.

Since the angles at the center of a circle is proportional to the arc subtended in it.
∴ $\frac{\angle AOP}{\angle AOB}= \frac{arc AP}{arc APB}$

⇒ $\frac{\angle AOP}{\angle AOB}= \frac{r}{\pi r}$

⇒ $\frac{\angle AOP}{\angle AOB}= \frac{1}{\pi }$

⇒ $\angle AOP = \frac{1}{\pi }\angle AOB = \frac{straight angle}{\pi}$

⇒ $1^{c}=\frac{straight angle}{\pi}$

⇒ $1^{c}$= constant .
Hence, Radian is a constant angle.

Theorem 2 :The number of radians in an angle subtended by an arc of a circle at the center is equal to $\frac{arc}{radius}$.
Proof: Let us consider a circle with center 'O' and radius 'r'. Take any point 'A' and 'Q' on the circle that make an arc AP. Angle formed at the center is $\angle$ AOQ = $\Theta^{c}$.Let P be any point on the arc AQ and its length is 's'. Let arc AP=r.
Thus, $\angle AOP = 1^{c}$

Since the angles at the center of a circle is proportional to the arc subtended in it.
$\frac{\angle AOQ}{\angle AOP}= \frac{arc AQ}{arc AP}$

⇒ $\angle AOQ = \left ( \frac{AQ}{AP} \times 1 \right )^{c}$

⇒ $\Theta = \frac{s}{r}$ radians.

## Example on circular system

Find the length of an arc of a circle of radius 5 cm subtending central angle measuring $15^{0}$.
Solution : Let 's' be the length of the arc subtending an angle $\Theta^{c}$ at the center of a circle of radius 'r'. Then,
$\Theta = \frac{s}{r}$
r = 5 cm and $\Theta = 15^{0}$
First we have to convert $\Theta$ to radians.
$15^{0} = \left ( 15\times \frac{\pi}{180} \right )^{c} = \left ( \frac{\pi}{12} \right )^{c}$
$\frac{\pi}{12} = \frac{s}{5}$

∴ s =$\frac{5\pi}{12}$ cm.