When the Sun’s elevation is 30°, a 50 m tall building casts a shadow of what length (in metres)?

Introduction / Context:Shadow length on level ground lies along the adjacent side when using tangent: tan θ = height / shadow.

Given Data / Assumptions:

• Height h = 50 m; θ = 30°.

Concept / Approach:shadow = h / tan θ. For θ = 30°, tan 30° = 1/√3, so shadow = 50 / (1/√3) = 50√3.

Step-by-Step Solution:

Verification / Alternative check:A lower Sun (smaller θ) makes longer shadows; 30° gives a shadow longer than the height, consistent with 50√3 > 50.

Why Other Options Are Wrong:25 m or 25√3 m underestimate; 100 m doubles the height without trigonometric basis; “50 m √3” is the same as 50√3 m but if interpreted as 50 m × √3 the correct numeric is captured by option wording b.

Common Pitfalls:Using sin or cos instead of tan; forgetting the inverse when solving for shadow.

Final Answer:50√3 m