Answer: (a) \(36\text{ cm}\); (b)(i) \(OB=2y\cos\alpha\); (b)(ii) shown.

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

(a) Let the angle of the sector be \(\theta\) radians. The area of a sector is

\(\frac12r^2\theta.\)

Here \(r=24\), so

\(\frac12(24)^2\theta=432.\)

Thus

\(288\theta=432,\)

so

\(\theta=\frac32.\)

The arc length is \(r\theta\), hence

\(24\times\frac32=36.\)

So the arc length is \(36\text{ cm}\).

(b)(i) In triangle \(OAB\), \(OA=AB=y\). By the cosine rule,

\(y^2=y^2+OB^2-2y(OB)\cos\alpha.\)

So

\(OB^2-2y(OB)\cos\alpha=0.\)

Since \(OB\ne0\),

\(OB=2y\cos\alpha.\)

(b)(ii) The height of triangle \(OAB\) is

\(AD=y\sin\alpha.\)

Therefore the area of triangle \(OAB\) is

\(\frac12(OB)(AD) =\frac12(2y\cos\alpha)(y\sin\alpha) =y^2\sin\alpha\cos\alpha.\)

The radius of sector \(OCD\) is \(OD\). In the right triangle,

\(OD=y\cos\alpha.\)

So the area of sector \(OCD\) is

\(\frac12(y\cos\alpha)^2\alpha =\frac12y^2\alpha\cos^2\alpha.\)

The shaded area is therefore

\(y^2\sin\alpha\cos\alpha-\frac12y^2\alpha\cos^2\alpha.\)

Factorising gives

\(\frac{y^2}{2}\cos\alpha(2\sin\alpha-\alpha\cos\alpha).\)