---
title: "Gradient A to B, RL of P = 1234.560"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:gradient-ab-tacheometer-100-0-p"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/gradient-ab-tacheometer-100-0-p
site: BCT Engineering
---

# Gradient A to B, RL of P = 1234.560

Surveying II · Chapter 2: Tacheometry

- Asked in: 2076 Baisakh, 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/gradient-ab-tacheometer-100-0-p
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

Compute the gradient between two instrument stations A and B from the following observation having tacheometric constant 100 & 0.

| Inst st'n | Hi | Sighted to | Bearing | Zenith Angle | Staff Readings | Remarks |
|---|---|---|---|---|---|---|
| A | 1.42 | P | 60°00' | 75°35'25" | 1.025, 1.525, 2.025 | RL of P = 1234.560 m |
| B | 1.48 | P | 345°00' | 96°52'45" | 0.925, 1.525, 2.125 | |

## Short answer

Using stadia formula with K=100, C=0 and zenith angles: RL of A=1210.562m, RL of B=1248.875m, horizontal distance AB=130.56m, gradient rising from A to B 1 in 3.41 (29.3%).

## Answer

Staff is vertical, $k = 100$, $c = 0$, angles are zenith angles $z$, so the vertical angle is $\theta = 90^\circ - z$.

$$
H = k\,s\sin^2 z, \qquad V = \frac{k\,s}{2}\sin 2\theta
$$

#### From station A

- $s = 2.025 - 1.025 = 1.000$ m, $z = 75^\circ 35' 25''$, $\theta = +14°24'35"$ (elevation)
- $H_A = 100\times 1.000\times\sin^2 z = 93.807$ m
- $V_A = 50\times 1.000\times\sin 2\theta = +24.103$ m
- RL of P = RL of A + HI + V − r, so
  $RL_A = 1234.560 - 24.103 + 1.525 - 1.42 = 1210.562$ m

#### From station B

- $s = 2.125 - 0.925 = 1.200$ m, $z = 96^\circ 52' 45''$, $\theta = -6°52'45"$ (depression, so $V$ is negative)
- $H_B = 100\times 1.200\times\sin^2 z = 118.278$ m
- $V_B = 50\times 1.200\times\sin 2\theta = -14.270$ m
- $RL_B = 1234.560 - (-14.270) + 1.525 - 1.48 = 1248.875$ m

#### Horizontal distance AB

Bearing of AP = 60°, bearing of BP = 345°, so the bearing of PA = 240° and of PB = 165°. The angle at P is $240^\circ - 165^\circ = 75^\circ$.

```text
            P
           /|\
    H_A   / | \  H_B
         /  75°\
        A ------ B
```

$$
AB = \sqrt{H_A^2 + H_B^2 - 2H_AH_B\cos 75^\circ} = \sqrt{93.807^2 + 118.278^2 - 2(93.807)(118.278)\cos 75^\circ} = 130.561\ \text{m}
$$

#### Gradient

Difference in RL = $RL_B - RL_A = 1248.875 - 1210.562 = 38.312$ m

$$
\text{Gradient} = \frac{38.312}{130.561} = 0.2934 = 1\text{ in }3.41
$$

**Answer: the ground rises from A to B at a gradient of 1 in 3.41 (about 29.3 %); AB = 130.56 m and B is 38.31 m higher than A.**

## Check yourself

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

1. With zenith angle z, horizontal distance is:
   A. H = ks sin^2 z
   B. H = ks cos^2 z
   C. H = ks sin z
   D. H = ks cos z
   Answer: A. cos^2θ with θ = 90° - z equals sin^2 z. (Key point: Horizontal distance with zenith angle)

2. RL of station A is found from the known RL of P as:
   A. RL_P + V + HI - r
   B. RL_P - V - HI + r
   C. RL_P - V + HI + r
   D. RL_P + V - r
   Answer: B. RL_P = RL_A + HI + V - r gives RL_A = RL_P - V - HI + r. (Key point: Station RL)

3. The angle at P between PA and PB is found using:
   A. Bearings of PA (240°) and PB (165°), difference 75°
   B. Bearings AP and BP directly, 45°
   C. Zenith angles
   D. 90°
   Answer: A. Reverse the bearings AP and BP by 180° and subtract to get 75°. (Key point: Angle at P)

4. For station B the zenith angle is 96°52'45 sec, so V is:
   A. Positive
   B. Negative, since the line of sight is depressed
   C. Zero
   D. Equal to H
   Answer: B. θ = -6°52'45 sec, so V is negative. (Key point: Sign for depression)

## Close variants set in other exams

- [Gradient P to Q, constants 100 and 0](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-pq-tacheometric-constants-100-0) — Surveying II, chapter 2, asked 2×

## Related in Surveying II

- [Tacheometric RL of B, C, gradient BC](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-b-c-gradient-bc) — Surveying II, chapter 2, asked 2×
- [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×
- [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
- [Gradient P to Q from common point R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-pq-common-point-r) — Surveying II, chapter 2
- [Tacheometric RL of B, C, gradient BC](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-b-c-gradient-bc-2) — Surveying II, chapter 2, asked 2×
- [Subtense bar, RL difference, gradient, bearing AB](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-c-difference-rl-gradient) — Surveying II, chapter 2

## Related in other subjects

- [Level book, staff readings at B and C (1251.250)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-250) — Surveying I, chapter 5, asked 3×
- [Level book, staff readings at B and C (1251.000)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-000) — Surveying I, chapter 5, asked 3×
- [Level book: profile levelling readings](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aprofile-leveling-readings-3rd-7th-9th) — Surveying I, chapter 5