From a point 30 m above the surface of a lake, the angle of elevation of an aeroplane in the sky is 30°, and the angle of depression of its image in the water is 60°. Assuming the water surface is perfectly horizontal and reflective, what is the height of the aeroplane above the water surface (in metres)?

Introduction / Context:
This interesting height-and-distance problem involves both an aeroplane and its mirror image in a lake. We use the geometry of reflection together with angles of elevation and depression from a point above the water to determine the plane's height above the water surface.

Given Data / Assumptions:

• A point of observation is 30 m above the lake surface.
• Angle of elevation to the aeroplane from that point = 30°.
• Angle of depression to the plane's image in the lake = 60°.
• The plane is vertically above its image and the water surface acts as a perfect mirror.
• We need the height of the aeroplane from the water surface.

Concept / Approach:
Reflection in a horizontal mirror (the lake) places the image the same distance below the surface as the object is above it. Thus, if the aeroplane is H metres above the water, its image is H metres below the surface. From the observation point 30 m above the water, we can build two right triangles (to the plane and to its image) with a common horizontal distance. Using tangent relations for the two angles, we set up equations in H and solve.

Step-by-Step Solution:
Let H be the height of the plane above the water surface.Place the water surface at y = 0, the observation point O at y = 30, and the plane at y = H.The image of the plane is at y = −H.Let the horizontal distance from O to the vertical line through the plane be x.Angle of elevation to the plane = 30°:tan(30°) = (H − 30) / x = 1 / √3.Thus, x = √3(H − 30).Angle of depression to the image = 60°:tan(60°) = (30 + H) / x = √3.So, x = (30 + H) / √3.Equate the two expressions for x:√3(H − 30) = (30 + H) / √3.Multiply both sides by √3: 3(H − 30) = 30 + H.3H − 90 = 30 + H ⇒ 2H = 120 ⇒ H = 60 m.

Verification / Alternative check:
If H = 60 m, then:
H − 30 = 30 m and H + 30 = 90 m.Using x = √3(H − 30) = 30√3 m, tan(30°) = 30 / (30√3) = 1 / √3, correct.Also, tan(60°) = 90 / (30√3) = 90 / (51.96...) ≈ √3, which fits well.

Why Other Options Are Wrong:
45 m, 50 m, 75 m, 90 m: Each of these values fails to satisfy both tangent equations simultaneously. Substituting them will yield inconsistent horizontal distances x for the two angles.

Common Pitfalls:
Beware of forgetting that the image is the same distance below the water as the plane is above it, leading to wrong vertical differences. Some students also confuse angles of elevation and depression or assume H − 30 and H + 30 incorrectly. Always draw a diagram marking heights carefully.

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