Answer: \(\angle ACB=2.35\) radians; perimeter \(=35.8\) cm; area \(=30.6\text{ cm}^2\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Since \(OA\) and \(OB\) are tangents to the circle with centre \(C\), \(OA\perp AC\) and \(OB\perp BC\).

In right triangle \(OAC\), \(OA=12\) and \(AC=5\), so

\(\tan\angle ACO=\frac{12}{5}\).

Hence

\(\angle ACB=2\tan^{-1}\frac{12}{5}=2.352\ldots\), so \(\angle ACB=2.35\) radians to 2 decimal places.

The boundary of the shaded region consists of the two tangent lengths \(OA\) and \(OB\), and the arc \(AB\) of the smaller circle.

The arc length is \(5\times2.35=11.75\), so the perimeter is

\(12+12+11.75=35.75\ldots=35.8\text{ cm}\).

The area of kite \(OACB\) is \(2\left(\frac12\times12\times5\right)=60\).

The sector \(ACB\) has area

\(\frac12\times5^2\times2.35=29.375\).

Therefore the shaded area is

\(60-29.375=30.625\ldots=30.6\text{ cm}^2\).

This gives the required answer: \(\angle ACB=2.35\) radians; perimeter \(=35.8\) cm; area \(=30.6\text{ cm}^2\).