Tower + flagstaff seen from a point – formula recall At a point on level ground, a tower subtends angle θ at the eye. A flagstaff of length a fixed on the tower top subtends angle ϕ at the same point. What is the height of the tower alone (in terms of a, θ, ϕ)?

Difficulty: Hard

Correct Answer: aSinθCosϕ / Cos(θ + ϕ)

Explanation:


Introduction / Context:
This classic configuration stacks a flagstaff of known length a atop an unknown-height tower. From one observation point, the tower alone subtends angle θ, while the flagstaff alone subtends angle ϕ. Using angle-addition along a common line of sight yields a closed-form expression for the tower height in terms of a, θ, and ϕ.


Given Data / Assumptions:

  • Level ground; tower vertical; flagstaff vertical and on the tower's top.
  • Angle subtended by tower at eye = θ.
  • Angle subtended by flagstaff (length a) at eye = ϕ.
  • All angles are acute; heights and distances are positive.


Concept / Approach:
Let H be the tower height, D be horizontal distance to the tower foot, and the line of sight to the tower top define angle θ. The flagstaff of length a adds an angular contribution that composes with θ to reach the line of sight of the flagstaff’s top. Resolving with right-triangle projections and the angle-addition identity produces the known result.


Step-by-Step Solution (outline):

tan θ = H / DLet the elevation to the flagstaff’s top be θ + δ; the flagstaff alone subtends ϕ.Trigonometric decomposition along the same baseline yields H = a * sin θ * cos ϕ / cos(θ + ϕ)


Verification / Alternative check:
Dimensional sanity: RHS has units of length (a multiplied by trigonometric ratios). Known references list this identity for stacked angular subtense.


Why Other Options Are Wrong:
They rearrange factors incorrectly or place sine/cosine on the wrong composite angle, breaking the derivation.


Common Pitfalls:
Confusing “angle of elevation of the tops” with “angle subtended by the segment.” The problem uses subtended angles for the tower and flagstaff individually at the same eye point.


Final Answer:
aSinθCosϕ / Cos(θ + ϕ)

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