From a point 30 m above the surface of a lake, the angle of elevation of an aeroplane is 30° and the angle of depression of its image in the water is 60°. What is the height of the aeroplane above the water surface?

Introduction / Context:
This problem uses the idea of reflection in water together with angles of elevation and depression. An observer stands at a point above the surface of a lake and observes both an aeroplane in the sky and its image reflected in the water. Using the angles to the aeroplane and its image, we need to determine the actual height of the aeroplane above the water surface.

Given Data / Assumptions:

• The observer is 30 m above the water surface.
• The angle of elevation from the observer to the aeroplane is 30 degrees.
• The angle of depression from the observer to the image of the aeroplane in the water is 60 degrees.
• The image of the aeroplane in the water is as far below the water level as the plane is above it.
• The lake surface is horizontal and acts as a mirror.

Concept / Approach:
Let the aeroplane be at height H above the water surface and at horizontal distance x from the vertical line through the observer. Its image will be at height −H (below the surface) but at the same horizontal distance x. The vertical distance from observer (30 m above water) to the plane is H − 30, and to the image is 30 + H. Using tan 30 = (H − 30) / x and tan 60 = (30 + H) / x, we can solve the system of equations for H and x, then report H as the height above water.

Step-by-Step Solution:
Let the height of the aeroplane above the water surface be H. Let the horizontal distance from the observer's vertical projection to the aeroplane be x. For the aeroplane: tan 30 = (H − 30) / x ⇒ 1 / √3 = (H − 30) / x. So H − 30 = x / √3. (1) For the image: angle of depression is 60 degrees, so tan 60 = (30 + H) / x ⇒ √3 = (30 + H) / x. So 30 + H = x√3. (2) From (1), x = √3(H − 30). Substitute into (2): 30 + H = √3(H − 30) × √3 = 3(H − 30). Thus 30 + H = 3H − 90 ⇒ 30 + 90 = 3H − H ⇒ 120 = 2H ⇒ H = 60 m.

Verification / Alternative check:
If H = 60 m, then from (1), H − 30 = 30 m. So x / √3 = 30 ⇒ x = 30√3. From (2), 30 + H = 90, and x√3 = 30√3 × √3 = 30 × 3 = 90, which matches perfectly. Tan 30 = 30 / (30√3) = 1 / √3 and tan 60 = 90 / (30√3) = 3 / √3 = √3, confirming that both angles are correct.

Why Other Options Are Wrong:

• 45 m and 50 m: These heights do not satisfy both tan 30 and tan 60 relations simultaneously when substituted into the system.
• 75 m and 90 m: These heights give inconsistent distances or angles relative to the given 30 m observer height and the stated angles.
• Only 60 m yields a consistent pair of right triangles with correct angles and distances.

Common Pitfalls:
Many students forget that the distance from the observer to the image is 30 + H, not H − 30, because the image is below the water surface. Another mistake is to mismatch the angles with the wrong vertical distances. Drawing a vertical axis showing the observer, water surface, aeroplane and its image often helps keep track of the correct distances and signs.

Final Answer:
The height of the aeroplane above the water surface is 60 m.