---
title: "Inaccessible object, stations in different vertical plane"
subject: "Surveying II"
chapter: 3
key: "semester_4/surveying_ii:inaccessible-object-different-vertical-plane"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/inaccessible-object-different-vertical-plane
site: BCT Engineering
---

# Inaccessible object, stations in different vertical plane

Surveying II · Chapter 3: Trigonometric Leveling

- Asked in: 2070 Bhadra, 8 marks; 2069 Bhadra, 6 marks; 2065 Chaitra (old course), 7 marks (asked 3 times)
- Video: [Inaccessible case and object and instrument being in different plane | Trigonometric Surveying|](https://www.youtube.com/watch?v=Q98ixust4KA) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/inaccessible-object-different-vertical-plane
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3

## Question

Explain the method of trigonometrical levelling to determine the elevation of inaccessible object when the instrument stations and the object are in different vertical plane.

## Short answer

When instrument stations and the object are not in the same vertical plane, horizontal distances are found by triangulation from a baseline, then heights by trigonometric levelling. Using measured angles α, β and baseline b, distances XP = b sinβ / sin(α+β), then RL from two stations averaged.

## Answer

When the base of the object (tower, chimney, hill top) is inaccessible and the two instrument stations are not in the same vertical plane as the object, the horizontal distance to the object is found by the **triangulation (base line) method** and the height by trigonometric levelling.

#### Field work

1. Select two stations X and Y a convenient distance $b$ apart (base line) so that P is visible from both and the triangle XYP is well conditioned (angles at X and Y between 30° and 120°).
2. Measure the base line XY and level the instrument at X. Take a back-sight on a BM of known RL to get the RL of the instrument axis; do the same at Y.
3. At X measure the horizontal angle $\alpha$ = $\angle$YXP (face left and right) and the vertical angle $\theta_X$ to P. At Y measure $\beta$ = $\angle$XYP and the vertical angle $\theta_Y$.

```text
                 P (top)
                /  \
               /    \
          XP  /      \ YP
             /        \
            /α        β\
           X ---- b ---- Y
         (plan view)
```

#### Computation

In the plan triangle XYP, $\angle$XPY $= 180^\circ - (\alpha + \beta)$. By the sine rule:

$$
\begin{aligned}
XP &= \frac{b\sin\beta}{\sin(\alpha+\beta)} \\
YP &= \frac{b\sin\alpha}{\sin(\alpha+\beta)}
\end{aligned}
$$

These are horizontal distances. The height of P above each axis:

$$
h_X = XP\tan\theta_X, \qquad h_Y = YP\tan\theta_Y
$$

(for zenith angle $z$, use $\theta = 90^\circ - z$.)

RL of P from each station:

$$
\begin{aligned}
RL_P &= RL_{BM} + BS_X + h_X \quad (\text{from X}) \\
RL_P &= RL_{BM} + BS_Y + h_Y \quad (\text{from Y})
\end{aligned}
$$

where $BS$ is the back-sight on the BM (axis RL = RL of BM + BS). The mean of the two values is taken as the RL of P. If the two stations are on very different levels, the axis RLs are found by BM sights at each station, as above. If the angle of elevation is given from the station peg, add the height of instrument $h_i$ to the RL of the peg.

The two values should agree closely; a large difference shows an error in the angles or the base line.

## Check yourself

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

1. A base line XY is used because:
   A. The base of the object is inaccessible so distance is found by triangulation
   B. The object is on the line XY
   C. The angles are not needed
   D. Levelling is impossible
   Answer: A. The horizontal distance cannot be taped, so a triangle XYP is solved. (Key point: Why base line)

2. XP is:
   A. b sinβ / sin(α+β)
   B. b sinα / sin(α+β)
   C. b / sin(α+β)
   D. b cosβ
   Answer: A. Opposite of angle at Y is XP, angle at P is 180 - (α+β). (Key point: Sine rule)

3. Height of P above axis from X:
   A. XP tanθ_X
   B. XP sinθ_X
   C. b tanθ_X
   D. YP tanθ_X
   Answer: A. h_X = XP tanθ_X, using the horizontal distance. (Key point: Height)

4. Two RLs of P from X and Y should:
   A. Agree closely, mean is taken
   B. Differ by b
   C. Be added
   D. Be ignored
   Answer: A. A large difference shows an error. (Key point: Check)

## Related in Surveying II

- [Inaccessible object, stations at different levels](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ainaccessible-object-different-levels-same-plane) — Surveying II, chapter 3
- [Importance of trig levelling, inaccessible RL](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atrig-leveling-importance-inaccessible-object) — Surveying II, chapter 3
- [Reciprocal trigonometrical levelling formula](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Areciprocal-trig-leveling-formula) — Surveying II, chapter 3
- [RL of tower top, base inaccessible](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atower-inaccessible-base-rl-two-stations) — Surveying II, chapter 3
- [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

## Related in other subjects

- [Reciprocal levelling; RL of closed traverse stations](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-closed-traverse-bcdefgb) — Surveying I, chapter 5