Area of a triangle, \(A_T = \frac{1}{2} bh\) and \(A_T = \frac{1}{2} s^2 \sin 60\) for an equilateral triangle.

\(A_T = \frac{1}{2} bh\)

But \(b = s\) and \(h = s+ 3\)

So, \(A_T = 0.5 s(s+3) \)

\(A_T = \frac{1}{2} s^2 \sin 60\)

\(A_T = \frac{\sqrt{3}}{4}s^2 \)

Equate the two areas and simplify: \(\frac{\sqrt{3}}{4}s^2 = 0.5 s(s+3)\)

\(\frac{\sqrt{3}}{2}s = (s+3)\)

\(s (\frac{2-\sqrt{3}}{2}) = 3 \)

\(s = 12 + 6\sqrt{3}\)

\(s = 22.39 \) ft

Since the altitude forms a perpendicular line with the base, then new triangles are created. Since these triangles are right, use Pythagorean formula to calculate the side.

\( \sin 60 = \frac{s-3}{s}\)

\( s(\sin 60) = s-3\)

\( s(1-\sin 60) = 3 \)

\( s = \frac{3}{1-\sin 60} \)

\(s = 12 + 6\sqrt{3}\)

\(s = 22.39 \) ft