Complementary elevations to two posts of heights in a 2:1 ratio Two vertical posts are x meters apart. One is twice as tall as the other. From the midpoint between their feet, the angles of elevation of their tops are complementary. Find the height of the shorter post (in meters).

Introduction / Context:When two elevation angles are complementary (α + β = 90°), the corresponding tangents satisfy tan α * tan β = 1. With symmetric horizontal distances (midpoint observer) and a 2:1 height ratio, the product-of-tangents condition yields the shorter height in terms of x.

Given Data / Assumptions:

• Posts are x meters apart on level ground.
• Heights: shorter = h, taller = 2h.
• Observer is at midpoint; horizontal distance to each post = x/2.
• Angles of elevation to tops are complementary.

Concept / Approach:tan α (to shorter) = h / (x/2) = 2h/x. tan β (to taller) = (2h) / (x/2) = 4h/x. Complementary ⇒ tan α * tan β = 1 ⇒ (2h/x)*(4h/x) = 1 ⇒ 8h^2 = x^2.

Step-by-Step Solution:

Verification / Alternative check:Plug h back to check tan α * tan β = 1 exactly; it holds for all x > 0.

Why Other Options Are Wrong:x/4 or x/√2 do not satisfy the product condition; x√2 is dimensionally large vs. half-spacing geometry.

Common Pitfalls:Using α + β = 90° without leveraging tan α * tan β = 1, or forgetting both horizontal legs are equal from the midpoint.

Final Answer:x / (2√2)