Answer: \(\angle AOB=1.18\text{ rad}\), perimeter \(=25.9\text{ cm}\), area \(=15.3\text{ cm}^2\).

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 \(AB\) is tangent to the circle, \(OA\perp AB\). Therefore triangle \(OAB\) is right-angled at \(A\).

Using \(OA=5\) and \(OB=13\),

\(\cos \angle AOB=\frac{5}{13}.\)

Hence

\(\angle AOB=\cos^{-1}\left(\frac{5}{13}\right)=1.176\ldots\text{ rad}.\)

So \(\angle AOB=1.18\text{ rad}\) to 3 significant figures.

Also, by Pythagoras,

\(AB=\sqrt{13^2-5^2}=12.\)

The straight part of the boundary along \(OB\) has length \(13-5=8\). The arc length is

\(5\times1.176\ldots=5.880\ldots.\)

Therefore the perimeter is

\(12+8+5.880\ldots=25.9\text{ cm}.\)

The shaded area is the area of triangle \(OAB\) minus the area of the sector:

\(\frac12(5)(12)-\frac12(5^2)(1.176\ldots).\)

This gives

\(30-14.70\ldots=15.3\text{ cm}^2.\)

This gives the required answer: \(\angle AOB=1.18\text{ rad}\), perimeter \(=25.9\text{ cm}\), area \(=15.3\text{ cm}^2\).