Answer: \(\sin\theta=\frac12\), and arc length \(AB=5\pi\).

(a) The area of triangle \(AOB\) is

\(\dfrac12(6)(6)\sin\theta.\)

This is equal to \(9\), so

\(18\sin\theta=9.\)

Therefore

\(\sin\theta=\frac12.\)

(b) The shaded segment is the sector \(AOB\) minus triangle \(AOB\).

The sector area is

\(\dfrac12(6)^2\theta=18\theta.\)

The triangle area is \(9\). Hence the segment area is

\(18\theta-9.\)

It is given that this equals \(15\pi-9\), so

\(18\theta-9=15\pi-9.\)

Thus

\(18\theta=15\pi,\)

and

\(\theta=\frac{5\pi}{6}.\)

The length of arc \(AB\) is \(r\theta\), so

\(AB=6\cdot\frac{5\pi}{6}=5\pi.\)