Certifying a Rooftop Telecom Tower Without Climbing It
A telecom firm has fixed a 4G tower on the flat roof of a two-storey school building in Meerut. Before the school reopens, the district office wants the tower's height recorded. The only ladder available is 3 m short, so a surveyor works from the ground. He sets his instrument on level ground, 20 m from the foot of the building, at the same level as the base; the instrument height is negligible. From that point the angle of elevation of the roof edge, which is also the foot of the tower, measures 45°, and the angle of elevation of the tower's top measures 60°. The tower stands vertically, flush with the roof edge that faces the surveyor. Two more figures are needed: the straight-line distance from the observation point to the top of the tower, for a guy-wire plan; and a check on a claim by the watchman, who says that if the surveyor simply walks 10 m further back, the angle of elevation of the tower's top will drop to exactly 30°.
1. Find the height of the building and the height of the tower itself. Explain why the 60° angle cannot be applied directly to the tower's height.
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Name the points: O = instrument, F = foot of the building, R = roof edge (foot of the tower), T = top of the tower. Both measured angles belong to right triangles standing on the same horizontal line OF.
Building (triangle OFR): tan 45° = RF / OF → 1 = RF / 20 → RF = 20 m.
Ground to tower top (triangle OFT): tan 60° = TF / OF → √3 = TF / 20 → TF = 20√3 ≈ 20 × 1.732 = 34.64 m.
Tower alone: TF − RF = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m.
Why 60° cannot be used on the tower directly: the ray that makes 60° starts at ground level, so tan 60° compares the whole vertical distance from the ground to the top of the tower with the horizontal 20 m. It never isolates the tower. If a candidate writes tan 60° = tower height / 20, the answer comes out as 34.64 m for the tower, which silently counts the 20 m building twice. Whenever an object sits on top of another, its own height is a difference of two heights measured from the same observation point.
2. The guy-wire is to run straight from the observation point to the top of the tower. Find that length and state the geometrical assumptions it depends on.
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OT is the hypotenuse of the right triangle OFT, so use Pythagoras:
OT = √(OF² + TF²) = √(20² + (20√3)²) = √(400 + 1200) = √1600 = 40 m.
Cross-check with the 60° angle: cos 60° = OF / OT → OT = 20 / 0.5 = 40 m. Both routes agree, which is a good sign that TF = 20√3 is right.
Assumptions:
- The ground from O to the building is horizontal, so OF is a true horizontal run and ∠OFT = 90°.
- The tower is vertical, i.e. parallel to the building wall; otherwise the top is displaced sideways and OT is no longer in the same plane as the angles.
- The instrument stands at ground level (or its height has been subtracted from both readings). A 1.5 m instrument would raise both target points by 1.5 m and change the computed building height from 20 m to about 21.5 m. The 40 m is a theoretical straight distance; a real guy-wire also needs an anchor offset and slack.
3. Test the watchman's claim that moving 10 m further back makes the angle of elevation of the tower's top exactly 30°. At what distance would 30° actually occur?
Read the worked answer
The height of the top above ground is fixed at TF = 20√3 ≈ 34.64 m, so the only variable is the horizontal distance d.
At d = 30 m (20 + 10): tan θ = 20√3 / 30 = (2√3)/3 ≈ 1.1547. Since tan 49° ≈ 1.150, θ ≈ 49.1°. So moving back 10 m reduces the angle from 60° only to about 49°, a change of roughly 11°.
For θ = 30°: tan 30° = 1/√3 = TF / d → d = TF × √3 = 20√3 × √3 = 60 m. The surveyor would have to stand 60 m from the foot of the building, that is 40 m further back, not 10 m; and the ground would have to remain level over all 60 m.
Conclusion: the claim is false. The error is treating the angle as inversely proportional to distance. In fact tan θ = h/d, so θ falls slowly; doubling the distance does not halve the angle. A quick sanity check is that a 30° angle of elevation always means the observer is √3 times the height away.