If the elevation of the Sun changes from 30° to 60°, what is the difference between the lengths of the shadows of a vertical pole 15 m high?

Introduction / Context:
This problem is a standard height and distance question based on the shadow of a vertical pole when the elevation of the Sun changes. The main concept tested here is the use of trigonometric ratios, especially the tangent function, to relate the height of an object with the length of its shadow on level ground. You are asked to compare the shadow lengths at two different angles of elevation and to find the difference between those lengths for a pole of fixed height 15 m.

Given Data / Assumptions:

• The height of the vertical pole is 15 m.
• First position: elevation of the Sun is 30 degrees.
• Second position: elevation of the Sun is 60 degrees.
• The ground is assumed to be perfectly horizontal and the pole is perfectly vertical.
• We use tan θ = opposite / adjacent, where opposite is the height of the pole and adjacent is the length of the shadow.

Concept / Approach:
In a right angled triangle formed by the pole, its shadow and the line of sight to the top, the tangent of the angle of elevation equals height / shadow length. So, shadow length = height / tan θ. We calculate the shadow at 30 degrees and at 60 degrees, then subtract the smaller from the larger to obtain the required difference. The key trigonometric values used are tan 30 degrees = 1 / sqrt(3) and tan 60 degrees = sqrt(3).

Step-by-Step Solution:
Let the height of the pole be h = 15 m. When the Sun's elevation is 30 degrees, tan 30 = h / shadow_30. So shadow_30 = h / tan 30 = 15 / (1 / sqrt(3)) = 15 * sqrt(3) m. When the Sun's elevation is 60 degrees, tan 60 = h / shadow_60. So shadow_60 = h / tan 60 = 15 / sqrt(3) = 5 * sqrt(3) m. Difference in shadow lengths = shadow_30 - shadow_60 = 15√3 - 5√3 = 10√3 m.

Verification / Alternative check:
We know that as the Sun rises higher in the sky (angle increases), the shadow becomes shorter. At 30 degrees the shadow should be longer than at 60 degrees. Our calculation gives 15√3 m at 30 degrees and 5√3 m at 60 degrees, so the first shadow is indeed longer. The difference of 10√3 m is positive and dimensionally consistent (in metres), which confirms the internal logic of the solution.

Why Other Options Are Wrong:

• 7.5 m: This ignores the presence of sqrt(3) in the tangent values and is numerically too small.
• 15 m: This wrongly assumes a direct difference in height instead of using trigonometric ratios.
• 5√3 m: This corresponds roughly to the shorter shadow alone, not to the difference of the two shadows.
• 20 m: This is an arbitrary value with no correct trigonometric basis for the given angles.

Common Pitfalls:
Students often mix up tan 30 degrees and tan 60 degrees, or use sin and cos instead of tan for shadow problems. Another common mistake is to subtract in the wrong order, taking shorter minus longer and getting a negative length, which is not meaningful here. Forgetting that shadow length = height / tan θ, not height * tan θ, is also a frequent source of error.

Final Answer:
The difference between the lengths of the shadows of the 15 m high pole when the Sun's elevation changes from 30 degrees to 60 degrees is 10√3 m.