---
title: "RL of hill top from two stations (A-B 118 m)"
subject: "Surveying II"
chapter: 3
key: "semester_4/surveying_ii:hill-top-rl-two-stations-2079j"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-top-rl-two-stations-2079j
site: BCT Engineering
---

# RL of hill top from two stations (A-B 118 m)

Surveying II · Chapter 3: Trigonometric Leveling

- Asked in: 2079 Jestha, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-top-rl-two-stations-2079j
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3

## Question

The top of the hill station 'P' was sighted from two instrument stations A and B located at very different level and in the same vertical plane to that of target and following observations were noted.

| Inst. Stn | HI (m) | Target | Zenithal angle FL | Zenithal angle FR | Distance | Target height |
|---|---|---|---|---|---|---|
| A | 1.45 | P | 59°18' | 300°36' | A to B = 118.00 m | - |
| B | 1.47 | P | 69°52' | 290°09' | | - |
| C | 1.47 | A | 85°32' | 274°26' | | 1.65 m |

If RL of B was 1590.00 m, find the RL of the top of hill station.

## Short answer

From B to A sight, RL_A = 1599.003 m, axis A = 1600.453 m. With αA=30°39', αB=20°08'30", D=118 m, solving x = 269.884 m gives RL_P = 1690.456 m.

## Answer

The observation recorded against "C" in the table is taken as the sight from B to A (instrument station C is a misprint for B; the target height 1.65 m is the vane at A).

The sight from B to A is an elevation, so A is higher than B. B sees P at the smaller angle of elevation, so B is the station farther from P. Take $D = 118$ m as the horizontal distance AB.

#### Mean vertical angles

| Station | Face left | Face right (360° − FR) | Mean zenith $z$ | Elevation $\alpha = 90^\circ - z$ |
|---|---|---|---|---|
| A to P | 59°18' | 59°24' | 59°21'00" | 30°39'00" |
| B to P | 69°52' | 69°51' | 69°51'30" | 20°08'30" |
| B to A | 85°32' | 85°34' | 85°33'00" | 4°27'00" |

#### RL of instrument axes

- Axis of B: $RL_{axis\,B} = 1590.000 + 1.47 = 1591.470$ m
- Sight B to A: the vane at A (1.65 m above the foot) is at 1591.470 + 118$\tan$(4°27'00") = 1600.653 m, so the ground at A is at 1599.003 m and
  $RL_{axis\,A} = 1599.003 + 1.45 = 1600.453$ m

#### Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 118$ m.

```text
                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|
```

The height of P above the axis of B is $x\tan\alpha_B$, and above the axis of A it is $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$
\begin{aligned}
RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\
x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A}
\end{aligned}
$$

With $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1591.470 m and axis RL of A = 1600.453 m:

$$
x = \frac{(1600.453 - 1591.470) - 118\times 0.59258}{0.36677 - 0.59258} = 269.884\ \text{m}
$$

$$
RL_{P} = 1591.470 + 269.884\times 0.36677 = 1690.456\ \text{m}
$$

Check from A: $1600.453 + (269.884 - 118)\times 0.59258 = 1690.456$ m (same).

**Answer: RL of the top of the hill P = 1690.456 m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is a face right zenith reading such as 300°36' brought to the same scale as the face left reading?
   A. Subtract it from 180°
   B. Subtract 270° from it
   C. Use it as it is
   D. Subtract it from 360°
   Answer: D. The face right zenith is changed as 360° - FR, so 300°36' becomes 59°24'. The mean with the face left 59°18' is 59°21'. (Key point: Face-right conversion)

2. The mean zenith angle from B to P is 69°51'30''. What angle of elevation is used in the calculation?
   A. 20°08'30''
   B. 69°51'30''
   C. 19°51'30''
   D. 159°51'30''
   Answer: A. Elevation = 90° - zenith = 90° - 69°51'30'' = 20°08'30''. (Key point: Elevation from zenith)

3. The sight from B to A is taken to a vane 1.65 m above the foot of the staff at A. How is the axis RL of A obtained?
   A. Axis B + D tan(elevation) + 1.65 m
   B. Axis B + 1.65 m + 1.45 m
   C. Vane level = axis B + D tan(elevation); subtract 1.65 m to get the ground at A, then add HI of A (1.45 m)
   D. RL of B + D tan(elevation) with no vane or instrument heights
   Answer: C. The sighted vane is 1.65 m above the ground, so the ground at A is 1.65 m lower. The axis is then HI (1.45 m) above that ground. (Key point: Axis RL of the upper station)

4. What condition is used to find x, the horizontal distance from B to P?
   A. The angles of elevation from A and B are equal
   B. The RL of P computed from axis B equals the RL of P computed from axis A
   C. The axis RLs of A and B are equal
   D. x is equal to the distance AB
   Answer: B. P is one point, so RL axis B + x tan(aB) must equal RL axis A + (x - D) tan(aA). Solving gives x. (Key point: Equating the RL of P)

## Close variants set in other exams

- [RL of hill top P (A to B 112 m, RL 1290)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2077c-ab-112) — Surveying II, chapter 3, asked 4×
- [RL of hill top P (A to B 112 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2073m-ab-112) — Surveying II, chapter 3, asked 4×
- [RL of hill top Q from stations P and R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-q-from-p-and) — Surveying II, chapter 3, asked 4×

## Related in Surveying II

- [RL of hill station P from stations R and Q](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-signal-r-q-p) — Surveying II, chapter 3
- [RL of hill station P, A-B 150 m, target 3.5 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-p-vane-15-ab) — Surveying II, chapter 3, asked 2×
- [RL of hill top P (A to B 120 m, vane 2.5 m)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2072m-vane-25) — Surveying II, chapter 3
- [RL of hill station P, A-B 150 m, RL B 1500](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-p-vane-175-ab) — Surveying II, chapter 3, asked 2×
- [RL of tower top Q from stations P and R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atower-top-q-from-p-and) — Surveying II, chapter 3
- [RL of temple top from stations A and B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atemple-top-rl-two-stations) — Surveying II, chapter 3