Answer: \(r=9.6\), \(\theta=1.25\), and the shaded area is \(81.1\text{ cm}^2\).

For a sector, use radians throughout with arc length \(s=r\theta\) and area \(A=\frac12r^2\theta\).

(a) The length of the minor arc \(AB\) is

\(r\theta=12. \tag{1}\)

The area of the minor sector is

\(\frac12r^2\theta=57.6. \tag{2}\)

Using \(r\theta=12\), equation (2) becomes

\(\frac12r(12)=57.6.\)

So

\(6r=57.6,\)

and therefore

\(r=9.6.\)

Then from (1),

\(\theta=\frac{12}{9.6}=1.25.\)

(b) Since \(AC\) is tangent to the circle at \(A\), \(OA\perp AC\). In right triangle \(OAC\),

\(\tan\theta=\frac{AC}{OA}.\)

Hence

\(AC=9.6\tan1.25\approx28.89.\)

The shaded region is the area of triangle \(OAC\) minus the area of sector \(AOB\):

\(\text{shaded area} =\frac12(9.6)(28.89)-57.6.\)

Thus

\(\text{shaded area}\approx81.1\text{ cm}^2.\)