Question

Two lighthouses, located at points A and B on the earth, are 60 feet and 40 feet tall respectively. Each lighthouse is perfectly vertical and the land connecting A and B is perfectly flat. The topmost point of the lighthouse at A is A’ and of the lighthouse at B is B’. Draw line segments A’B and B’A, and let them intersect at point C’. Drop a perpendicular from C’ to touch the earth at point C. What is the length of CC’ in feet?

• A. 20
• B. 25
• C. 30
• D. 24
• E. The distance between A and B is also needed to solve this

Answer 1

The Correct Option is D

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 h_A = 60 feet and lighthouse B as h_B = 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: h_A/x = h_B/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 h_A and h_B: 60 * 40 = 2400, yielding the geometric mean as CC'. Therefore, CC' = √(60 * 40) = √2400 ≈ 24 feet.