Solution:
The area of a sector is expressed as
[tex]\begin{gathered} area=\frac{\theta}{360}\times\pi r^2 \\ where \\ \theta\Rightarrow angle\text{ subtended by the sector} \\ r\Rightarrow radius\text{ of the circle} \end{gathered}[/tex]
Given:
we see that
[tex]\begin{gathered} \theta=165\degree \\ r=10\text{ m} \end{gathered}[/tex]
By substitution, we have
[tex]\begin{gathered} area=\frac{165}{360}\times\pi\times10\times10 \\ =\frac{275\pi}{6}\text{ m}^2 \end{gathered}[/tex]
Hence, the area of the sector is evaluated to be
[tex]\frac{275\pi}{6}\text{ m}^2[/tex]
The correct option is A
