(a) Area of sector = $\frac{\theta }{360}\times \pi r^2$
$280=\frac{\theta }{360}\times \frac{22}{7}\times 21\times 21$
$280=\frac{1386\theta }{360}$
$\theta =\frac{280\times 360}{1386}$
$\theta =72.72°$
Perimeter of sector = $2r+\frac{\theta }{360}\times 2\pi r$
= $2(21)+\frac{72.72}{360}\times 2\times \frac{22}{7}\times 21$
= $42+(0.202\times 2\times 66)$
= $42+26.667$
= $68.667cm$
$\approxeq 68.7cm$
(b) When the sector is bent to form a cone, its radius becomes the slant height of the cone.
The radius of the base of the cone is obtained from the relation $r=\frac{R\theta }{360}$,
where r = radius of the base of the cone, R = radius of the sector, θ = angle of the sector.
Therefore, r = $\frac{21\times 800}{11\times 360}$
= $\frac{140}{33}$
If y is the vertical angle of the cone, then $sin\frac{y}{2}=\frac{r}{l}$
= $\frac{140}{33\times 21}$
= 0.2020
Hence, required angle = y = 2 x sin${}^{-1}$ (0.2020) = 23${}^o$.