Answer:
The watered area is approximately 3783 square feet.
Step-by-step explanation:
The area that is watered due to the rotation of the spankler is a circular section area (
), whose formula is:
![A = \frac{\theta }{2}\times \frac{1}{360^{\circ}}\times 2\pi \times d^{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%5Ctheta%20%7D%7B2%7D%5Ctimes%20%5Cfrac%7B1%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%202%5Cpi%20%5Ctimes%20d%5E%7B2%7D)
Where:
- Water distance, measured in feet.
- Rotation angle, measured in sexagesimal degrees.
Given that
and
, the watered area is:
![A = \frac{60^{\circ}}{2} \times \frac{1}{360^{\circ}}\times 2\pi \times (85\,ft)^{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%202%5Cpi%20%5Ctimes%20%2885%5C%2Cft%29%5E%7B2%7D)
![A \approx 3783\,ft^{2}](https://tex.z-dn.net/?f=A%20%5Capprox%203783%5C%2Cft%5E%7B2%7D)
The watered area is approximately 3783 square feet.