A tower is 100 m high. As the Sun’s elevation changes from 30° to 45°, the length of the tower’s shadow decreases by P metres. Find P.

Introduction / Context:Shadow length on level ground is h / tan θ. For two elevations, the difference gives the shrinkage as the Sun rises.

Given Data / Assumptions:

• Height h = 100 m.
• θ1 = 30°, θ2 = 45°.

Concept / Approach:P = shadow(30°) − shadow(45°) = h(1/tan 30° − 1/tan 45°).

Step-by-Step Solution:

Verification / Alternative check:Numerically, √3 ≈ 1.732 ⇒ P ≈ 73.2 m. As elevation increases from 30° to 45°, the shadow shortens—positive P makes sense.

Why Other Options Are Wrong:100√3 m or 100 m √3 are full shadow values at 30°, not the difference; 100 m or 50(√3 − 1) m are incorrect magnitudes.

Common Pitfalls:Using sine/cosine for shadow; reversing the subtraction order (leading to negative P).

Final Answer:100(√3 − 1) m