To solve the problem, we need to find the length of the perpendicular dropped from C' to the ground (CC'). Consider triangle ACC' and triangle BCC'. Since each lighthouse is vertical and the land between A and B is flat, triangles ACC' and BCC' are right-angled at C. Let's denote the height of lighthouse A as hA = 60 feet and lighthouse B as hB = 40 feet. Let x be the distance from point A to point C and y be the distance from point B to point C. Since A', C', and B' form a triangle with concurrent lines at point C', the triangles are similar, which allows us to set up the equation: hA/x = hB/y. From this, it simplifies to 60/x = 40/y. Solving for y gives y = (2/3)x. The entire distance is AB = x + y. Since the lines A'B and B'A intersect, and the problem requires the height of C', substitute values back in relation to x and y, fulfilling the requirement of x + y = AB = (x + (2/3)x). However, since AB is not provided, we can consider the geometric mean relations derived between the triangles. Using properties of intersecting chords, AC' * C'B = CC'^2. With right angle triangles and similar triangle relations, C'C is found through multiplying hA and hB: 60 * 40 = 2400, yielding the geometric mean as CC'. Therefore, CC' = √(60 * 40) = √2400 ≈ 24 feet.