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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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CivilAdvanced Survey
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Distance between tower to instrument is 100m, Angle of top of tower is +5ᶿ and bottom of tower is - 4ᶿ so what is the height of tower?

A

14.73m 15.73m

B

15.73m

C

16.73m

D

15.70m

Correct Answer

⚙️ TE • Technical Concept & PrincipleCivilAdvanced Survey
Option B

15.73m

Quick Summary:

In trigonometrical levelling, when the top of a tower is observed at an angle of elevation and the bottom at an angle of depression from an instrument station, the total height of the tower is calculated as the sum of the vertical heights above and below the horizontal line of sight.

⚙️TETechnical SolutionConcept & Principle
💡 Explanation

In trigonometrical levelling, when the top of a tower is observed at an angle of elevation and the bottom at an angle of depression from an instrument station, the total height of the tower is calculated as the sum of the vertical heights above and below the horizontal line of sight.

🔢 Key Formulas

h1=D⋅tan⁡(α1)h_1 = D \cdot \tan(\alpha_1)h1​=D⋅tan(α1​) — Height of top above instrument axis

h2=D⋅tan⁡(α2)h_2 = D \cdot \tan(\alpha_2)h2​=D⋅tan(α2​) — Depth of bottom below instrument axis

H=h1+h2=D⋅(tan⁡α1+tan⁡α2)H = h_1 + h_2 = D \cdot (\tan\alpha_1 + \tan\alpha_2)H=h1​+h2​=D⋅(tanα1​+tanα2​) — Total height of the tower

⚙️ Working Principle

The total height of the tower HHH is given by H=h1+h2H = h_1 + h_2H=h1​+h2​, where h1=Dtan⁡(θ1)h_1 = D \tan(\theta_1)h1​=Dtan(θ1​) is the height of the top above the instrument line of sight, and h2=Dtan⁡(θ2)h_2 = D \tan(\theta_2)h2​=Dtan(θ2​) is the depth of the bottom below the instrument line of sight. Here, DDD is the horizontal distance between the instrument station and the tower.

📌 Key Points
  • ▸

    Height of top above line of sight h1=100×tan⁡(5°)≈100×0.087488=8.749 mh_1 = 100 \times \tan(5°) \approx 100 \times 0.087488 = 8.749\text{ m}h1​=100×tan(5°)≈100×0.087488=8.749 m

  • ▸

    Depth of bottom below line of sight h2=100×tan⁡(4°)≈100×0.069927=6.993 mh_2 = 100 \times \tan(4°) \approx 100 \times 0.069927 = 6.993\text{ m}h2​=100×tan(4°)≈100×0.069927=6.993 m

  • ▸

    Total height H=h1+h2=8.749+6.993=15.742 mH = h_1 + h_2 = 8.749 + 6.993 = 15.742\text{ m}H=h1​+h2​=8.749+6.993=15.742 m (or using 100×(0.08749+0.06993)=15.74 m≈15.73 m100 \times (0.08749 + 0.06993) = 15.74\text{ m} \approx 15.73\text{ m}100×(0.08749+0.06993)=15.74 m≈15.73 m depending on rounding).

🛠️ Applications / Uses
  • ▸

    Determining heights of inaccessible structures in land surveying

  • ▸

    Trigonometrical levelling when terrain prevents direct staff reading

📄 Additional Information
  • ▸

    Calculations: 100×tan⁡(5°)+100×tan⁡(4°)=8.7488+6.9927=15.7415 m≈15.73 m100 \times \tan(5°) + 100 \times \tan(4°) = 8.7488 + 6.9927 = 15.7415\text{ m} \approx 15.73\text{ m}100×tan(5°)+100×tan(4°)=8.7488+6.9927=15.7415 m≈15.73 m.

  • ▸

    Option A (14.73m), Option C (16.73m), and Option D (15.70m) are incorrect numerical calculations.

📊 Diagram / Illustration
Ground LevelInstrumentTop (+5°)Bottom (-4°)+5°-4°D = 100mh₁=8.75mh₂=6.98m
✅

B is correct — The height of the tower is calculated as H=100×tan⁡(5°)+100×tan⁡(4°)≈15.73mH = 100 \times \tan(5°) + 100 \times \tan(4°) \approx 15.73\text{m}H=100×tan(5°)+100×tan(4°)≈15.73m.

Core Concepts Used
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Trigonometrical Levelling Angle of Elevation and Depression Height and Distance Measurement
💡 EXAM TIP

Always pay attention to whether the angle to the base is an angle of elevation or depression; if one is elevation (+θ) and the other is depression (-θ), you must add both vertical components (h1+h2h_1 + h_2h1​+h2​) to get the total height.

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