You need to use the formula of perimeter of rectangle, such that:

`P = 2(L + w)`

`L` represents the length of rectangle

`w` represents the width of rectangle

The problem provides the information that the rectangle athletic field is twice as long as it is wide, hence, translating into equation yields:

`L = 2w`

Replacing 180 for perimeter and `2w` for `L` in equation of perimeter, yields:

`180 = 2(2w + w) => 180 = 2*3w => 180 = 6w => w = 180/6 => w = 30 yards => L = 2*30 => L = 60` yards

**Hence, evaluating the dimensions of the rectangle athletic field,
under the given conditions, yields `L = 60` yards and `w = 30`
yards.**

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