Measuring a Mobile Tower from Two Perpendicular Roads
A vertical mobile tower stands at the junction O of two straight roads that meet at right angles. One road runs north–south and the other east–west. A surveyor at A, due south of the tower, measures the angle of elevation of the tower top as 45°. A second surveyor at B, due east of the tower, measures the angle of elevation as 30°. The distance AB between the two surveyors is 100 m. The ground is level and the tower is vertical. A junior surveyor claims the height cannot be found unless OA and OB are measured directly; the senior surveyor disagrees. As a check, a third surveyor proposes a reading from C, due west of the tower and 30 m from its foot, where the angle of elevation is measured as 60°. The team must decide whether the three readings can describe the same tower.
1. Let h be the tower height and O its foot. Explain why OA and OB are perpendicular, express OA and OB in terms of h, and find h.
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Because the roads meet at right angles, the north–south road is perpendicular to the east–west road. A is due south of the tower on the north–south road, so OA lies along that road. B is due east on the east–west road, so OB lies along the other road. Hence OA ⟂ OB, and triangle AOB is right-angled at O.
Let T be the top of the tower. In right triangle TOA, the angle of elevation at A is 45°, so tan 45° = h/OA. Since tan 45° = 1, OA = h.
In right triangle TOB, the angle of elevation at B is 30°, so tan 30° = h/OB. Thus OB = h/tan 30° = h√3.
In right triangle AOB, AB² = OA² + OB². Substituting gives 100² = h² + (h√3)² = h² + 3h² = 4h². Therefore h² = 10,000/4 = 2,500, so h = 50 m (taking the positive root).
2. Evaluate the junior surveyor’s claim. Find OA and OB, and check whether the computed height agrees with the 45° and 30° readings.
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The junior surveyor is wrong. The two angles determine OA and OB as multiples of h: OA = h and OB = h√3. The known distance AB then fixes h through AB² = OA² + OB². Direct measurement of OA and OB is not required.
With h = 50 m, OA = 50 m and OB = 50√3 ≈ 86.6 m. Check: AB = √(50² + (50√3)²) = √(2,500 + 7,500) = √10,000 = 100 m, as given. The angle checks also work: tan 45° = 50/50 = 1, and tan 30° = 50/(50√3) = 1/√3. So the first two readings are consistent.
3. Use the third reading at C, due west of the tower and 30 m from its foot, with angle of elevation 60°, to compute an implied height. Is this reading consistent with h = 50 m? What should the team conclude?
Read the worked answer
At C, tan 60° = h_C/OC, so h_C = OC tan 60° = 30√3 ≈ 51.96 m. This is not exactly 50 m; the difference is about 1.96 m, or roughly 4%.
A vertical tower has only one height, so strictly the three readings are not perfectly consistent. The first two readings give exactly 50 m, while the third implies about 52 m. The most likely explanation is a small measurement error in the 60° angle or in the 30 m distance at C, or that C is not exactly due west. The team should not average the three readings blindly; it should recheck the third reading. If the team’s tolerance is smaller than about 2 m, the third reading must be rejected or corrected; if a 4% error is acceptable for the survey, the readings can be treated as approximately consistent.