---
title: "RL of hill station P from stations R and Q"
subject: "Surveying II"
chapter: 3
key: "semester_4/surveying_ii:hill-station-signal-r-q-p"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-station-signal-r-q-p
site: BCT Engineering
---

# RL of hill station P from stations R and Q

Surveying II · Chapter 3: Trigonometric Leveling

- Asked in: 2080 Chaitra, 2+2+2 marks
- Video: [Trigonometric Levelling: Case-II Object & Instrument lies in Same vertical plane Board Exam Solution](https://www.youtube.com/watch?v=cag1wI8Qd0o) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-station-signal-r-q-p
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3

## Question

A hill station signal was sighted from two instrument stations R and Q in which stations R and Q are in the same vertical plane to that of target 'P'.

| Inst stn | Target | hi (m) | Staff Reading to BM | zenith angle | Remarks |
|---|---|---|---|---|---|
| Q | P | 1.420 | 1.250 | 62°30' | Dist. PQR = 125 m [?] |
| R | P | 1.480 | | 70°45' | Rod ht at Q = 1.75 m |
| R | Q | 1.480 | | 98°15' | |

Compute the Reduced level of top of hill station.

## Short answer

Using BM backsight and vane observation, axis RLs determined; solving height equations gives RL of top P = RL of BM + 189.929 m (1189.929 m if BM is 1000.000 m).

## Answer

Assumptions: the RL of the BM is not given, so it is taken as $RL_{BM}$; the final answer is also given for $RL_{BM} = 1000.000$ m. The stations are in line in the order R, Q, P with $QR = 125$ m (horizontal). The staff reading 1.250 m on the BM is a horizontal back-sight from Q.

#### Axis of Q and axis of R

- Axis of Q $= RL_{BM} + 1.250$
- R sights the vane at Q (rod height at Q = 1.75 m): depression $= 98^\circ15' - 90^\circ = 8^\circ15'$. Ground at Q $= (RL_{BM} + 1.250) - 1.420$. The vane is at ground Q + 1.75 $= RL_{BM} + 1.580$.
  Vane level $=$ axis R $- 125\tan 8^\circ15'$, so
  axis R $= RL_{BM} + 1.580 + 125\tan 8^\circ15' = RL_{BM} + 19.704$ m

#### Elevation angles of P

- From Q: $90^\circ - 62^\circ30' = 27^\circ30'$ ($\tan = 0.52057$)
- From R: $90^\circ - 70^\circ45' = 19^\circ15'$ ($\tan = 0.34922$)

#### Horizontal distance and RL of P

Let $x$ = horizontal distance from R to P (Q is between R and P):

$$
x = \frac{(axis_Q - axis_R) - 125\tan 27^\circ30'}{\tan 19^\circ15' - \tan 27^\circ30'} = \frac{(1.250 - 19.704) - 125\times 0.52057}{0.34922 - 0.52057} = 487.449\ \text{m}
$$

(these use $RL_{BM} = 0$ as the datum, i.e. levels are measured from the BM)

$$
RL_P = RL_{BM} + 19.704 + 487.449\times 0.34922 = RL_{BM} + 189.929\ \text{m}
$$

Check from Q: $RL_{BM} + 1.250 + (487.449 - 125)\times 0.52057 = RL_{BM} + 189.929$ m.

**Answer: RL of the top of the hill station P = RL of BM + 189.929 m. For RL of BM = 1000.000 m, RL of P = 1189.929 m.**

## Check yourself

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

1. Zenith angle from R to the vane at Q is 98°15'. What does this mean?
   A. An elevation of 98°15'
   B. A depression of 8°15', so R is higher than Q
   C. An elevation of 8°15', so Q is higher than R
   D. A depression of 81°45'
   Answer: B. A zenith angle above 90° means the line of sight points downwards: depression = 98°15' - 90° = 8°15'. (Key point: Depression angle)

2. The vane at Q is at RL(BM) + 1.580. How is the axis of R found?
   A. Vane level minus 125 tan 8°15'
   B. Vane level plus 125 tan 8°15'
   C. Vane level plus 125 sin 8°15'
   D. Axis of Q plus 1.480
   Answer: B. The vane is 125 tan(8°15') below the axis of R, so axis R = RL(BM) + 1.580 + 125 tan 8°15' = RL(BM) + 19.704. (Key point: Axis of R)

3. If x is the horizontal distance from R to P, with the stations in order R, Q, P, what is the distance from Q to P?
   A. x + 125
   B. x - 125
   C. 125 - x
   D. x / 125
   Answer: B. Q lies between R and P, 125 m from R, so QP = x - 125. (Key point: Line-up of stations)

4. If the RL of the BM is 1000.000 m, the RL of P is RL(BM) + 189.929. What is the RL of P?
   A. 189.929 m
   B. 1189.929 m
   C. 1019.704 m
   D. 1487.449 m
   Answer: B. RL of P = 1000.000 + 189.929 = 1189.929 m. (Key point: Final result)

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