Ans: (a) $72^\circ$ (b) $320\pi\text{ cm}^2$
- Let $\angle AOB=\theta$.
$\begin{array}{rcl}
2\pi(40) \times \dfrac{\theta}{360^\circ} & = & 16\pi \\
\theta & = & 72^\circ
\end{array}$ - The area of sector $AOB$
$\begin{array}{cl}
= & \pi (40)^2 \times \dfrac{72^\circ}{360^\circ} \\
= & 320\pi\text{ cm}^2
\end{array}$
