Tower height from two angles of elevation
During a survey project, four Praveshika students of a school in Jodhpur measure a mobile tower standing on level ground. From a point A on the ground, the angle of elevation of the top of the tower is 30°. Moving 40 m straight towards the foot of the tower along the same line, they reach point B, from where the angle of elevation is 60°. The students take √3 = 1.73 and report the height to the nearest metre. For their model they also plan a straight guy wire from the top of the tower to a peg on the ground 20 m from the foot, and they must state the angle the wire makes with the ground and the length of wire needed. Their teacher warns that if the tower is later raised by 10 m, the angle of elevation from B will not become double, and asks them to check this numerically. All observations are taken at the same level and the tower is vertical.
1. Find the height of the tower and the distance of B from the foot, using both angles of elevation.
Read the worked answer
Let the foot of the tower be F, height TF = h, and BF = d. Then AF = d + 40.
From B: tan 60° = h/d, so h = d√3.
From A: tan 30° = h/(d + 40) = 1/√3, so h√3 = d + 40.
Substituting h = d√3: (d√3)(√3) = d + 40, i.e. 3d = d + 40, giving d = 20 m.
Then h = 20√3 = 20 × 1.73 = 34.6 m ≈ 35 m to the nearest metre, and AF = 60 m.
Check: h/(d + 40) = 34.6/60 = 0.577 ≈ tan 30°, and h/d = 34.6/20 = 1.73 ≈ tan 60°. Both angles are satisfied, so the result is consistent.
2. Find the angle the guy wire makes with the ground and its length, given the peg is 20 m from the foot.
Read the worked answer
The wire runs from the top T to the peg P, with FP = 20 m and TF = 20√3 m perpendicular to the ground. The angle θ at P satisfies
tan θ = TF/FP = 20√3 / 20 = √3, so θ = 60°.
It is no accident that this equals the elevation from B: B and P are both 20 m from the foot and on the same level, so the triangles TFB and TFP are congruent right triangles.
Length of wire (hypotenuse):
TP = √(FP² + TF²) = √(20² + (20√3)²) = √(400 + 1200) = √1600 = 40 m.
Same answer via trigonometry: TP = TF/sin 60° = 20√3 ÷ (√3/2) = 40 m. The peg position 20 m away is exactly the distance that makes the wire sit at 60°, so the students should order 40 m plus a small allowance for tying.
3. The tower is later raised by 10 m. Check numerically whether the angle of elevation from B doubles.
Read the worked answer
New height = 20√3 + 10 = 34.6 + 10 = 44.6 m; B is still 20 m from the foot.
tan θ′ = 44.6/20 = 2.23, so θ′ ≈ 66° (tan 66° ≈ 2.25), not 2 × 60° = 120°.
The teacher is right. The reason is that tangent is not proportional to the angle: adding a fixed 10 m changes the ratio h/d from 1.73 to 2.23, and that ratio maps to about 66°. A doubling to 120° is also geometrically impossible, because an angle of elevation from level ground must lie between 0° and 90°. Students should state the computed angle and reject 120° on this ground as well.