Answer: (a) \(2.8\); (b) \(144\) cm; (c) \(439\) to \(440\text{ cm}^2\).

Answer: (a) \(2.8\); (b) \(144\) cm; (c) \(439\) to \(440\text{ cm}^2\).

(a) The length of an arc is

\(s=r\theta.\)

Here \(s=28\) and \(r=10\), so

\(28=10\theta.\)

Therefore

\(\theta=2.8.\)

(b) Since the tangents from \(C\) are equal,

\(AC=BC.\)

The line \(OC\) bisects the angle \(AOB\), so in right triangle \(AOC\),

\(\angle AOC=\frac{2.8}{2}=1.4.\)

Since \(OA=10\),

\(\tan1.4=\frac{AC}{10}.\)

Hence

\(AC=10\tan1.4.\)

The perimeter of the shaded region is

\(\text{arc }AB+AC+BC.\)

Therefore

\(\text{perimeter}=28+2(10\tan1.4).\)

This gives

\(\text{perimeter}\approx143.96\text{ cm}.\)

So the perimeter is

\(144\text{ cm}\)

to three significant figures.

(c) The area of triangle \(AOC\) is

\(\frac12\times OA\times AC =\frac12\times10\times10\tan1.4.\)

The same is true for triangle \(BOC\). Therefore the area of kite \(AOCB\) is

\(2\left(\frac12\times10\times10\tan1.4\right) =100\tan1.4.\)

The area of the minor sector \(AOB\) is

\(\frac12r^2\theta =\frac12(10)^2(2.8) =140.\)

So the shaded area is

\(100\tan1.4-140.\)

This is approximately

\(439.8\text{ cm}^2.\)

Thus the shaded area is

\(440\text{ cm}^2\)

to three significant figures.