Find the radius of a circle inscribed in a pendant

Question

The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector OAB of a circle centre O, of radius 6 cm, and angle $AOB = \frac{\pi}{3}$. The circle C, inside the sector, touches the two straight edges, OA and OB, and the arc AB as shown.

Find the radius of the circle C.

I have no clue how to approach this. Help please.

Answer 1

What about now?

still no clue?

Answer 2

Let $P$ be the center of circle $C$ and $Q$ be a point on $AO$ such that $CQ\perp AO$.

It can be shown that $OP$ is bisecting $\angle AOB$. Because of this, $\triangle OCQ$ has angles measuring $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$.

Let $r$ be the radius of the circle. From this: $CQ=r$, $CO=2r$, and $QO=r\sqrt{3}$.

From here, try finding the value of $r$ with what you know.

Answer 3

You must use the commonly abused property of a tangent line in a circle. Any tangent line touching in the circle can produce a perpendicular line. Projecting the radius of C into the side of the sector. See the picture. You can get:

$\sin \frac{\eta}{2} = \frac{r}{R-r}$