In the diagram, OAPB is a sector of circle with centre O and radius 9 cm. Angle AOB = 1.4 radians. C is the centre of the circle enclosed inside the sector. OCP is a straight line and the circle touches the sector at points P, Q and R.

1. Identify the angle used in the right-angled triangle.

Since the circle touches both straight sides of the sector, $OC$ bisects $\angle AOB$.

$\angle AOC=\frac{1.4}{2}=0.7\text{ radians}$

2. Relate the small circle radius to $OC$ using trigonometry.

Let the radius of the enclosed circle be $r$ cm.

Since the radius to a tangent is perpendicular to the tangent, $CR\perp OA$.

In right-angled triangle $OCR$,

$\sin 0.7=\frac{CR}{OC}=\frac{r}{OC}$

so

$r=OC\sin 0.7.$

3. Use internal tangency to the outer arc to express $OC$ in terms of $r$.

The sector has radius $9$ cm, and the small circle touches the arc at $P$ on the line $OCP$.

Hence

$OC+CP=OP$

$OC+r=9$

so

$OC=9-r.$

4. Solve for the radius and round to 3 decimal places.

Substitute $OC=9-r$ into $r=OC\sin 0.7$:

$r=(9-r)\sin 0.7$

$r=9\sin 0.7-r\sin 0.7$

$r(1+\sin 0.7)=9\sin 0.7$

$r=\frac{9\sin 0.7}{1+\sin 0.7}$

$r=3.526\ldots$