---
title: "Chainage of R across a river"
subject: "Surveying I"
chapter: 3
key: "semester_3/surveying_i:obstacle-river-chainage-of-r"
source: https://www.bctengineer.com/q/semester_3/surveying_i/obstacle-river-chainage-of-r
site: BCT Engineering
---

# Chainage of R across a river

Surveying I · Chapter 3: Chain Survey

- Asked in: 2057 Chaitra, 10 marks
- Video: [Chain surveying: finding width of river-2(numerical)](https://www.youtube.com/watch?v=0M57jCMpW4o) — civil engineering basics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/obstacle-river-chainage-of-r
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-3

## Question

A chain line PQR crosses a river, Q and R being on the near and distant banks respectively. A perpendicular QS, 90 m long, is set out at Q on the left of the chain line. The respective bearings of R and P taken at S are 87°30′20″ and 177°30′20″. Find the chainage of R given the PQ is 45 m and the chainage of Q is 650 m.

## Short answer

From bearings, ∠PSR = 90°. In right triangle SQP, tan(∠QSP) = 45/90 → ∠QSP = 26°33′54″. Then ∠QSR = 63°26′06″. QR = 90 × tan(∠QSR) = 180 m. Chainage of R = 650 + 180 = 830 m.

## Answer

A chain line crosses a river, so QR cannot be chained; it is found by trigonometry from the perpendicular QS set out at Q (the bearings at S are measured with a theodolite or compass).

**Given:** QS = 90 m (perpendicular to PQR), PQ = 45 m, chainage of Q = 650 m, bearing of SR = 87°30′20″, bearing of SP = 177°30′20″.

```text
              S
            / | \
          /   |    \
        /     | 90   \
      /       |        \
 ---P---------Q-----------R---
       45 m        ?    (river between Q and R)
```

#### Step 1: Angle PSR

$$
\angle PSR = 177^\circ 30'20'' - 87^\circ 30'20'' = 90^\circ 00'00''
$$

#### Step 2: Angle QSP from the right triangle SQP

$$
\tan \angle QSP = \frac{PQ}{QS} = \frac{45}{90} = 0.5 \;\Rightarrow\; \angle QSP = 26^\circ 33'54''
$$

#### Step 3: Angle QSR

Q lies between P and R, so

$$
\angle QSR = \angle PSR - \angle QSP = 90^\circ - 26^\circ 33'54'' = 63^\circ 26'06''
$$

#### Step 4: Length QR from the right triangle SQR

$$
QR = QS \tan \angle QSR = 90 \times \tan 63^\circ 26'06'' = 90 \times 2.000 = 180.00\ \text{m}
$$

(Check: in the right-angled triangle PSR, $QS^{2} = PQ\times QR$, so $QR = 8100/45 = 180$ m.)

#### Step 5: Chainage of R

$$
\text{Chainage of R} = 650 + 180 = 830\ \text{m}
$$

(Chainage of P = 650 − 45 = 605 m.)

**Answer: width of river QR = 180 m; chainage of R = 830 m.**

## Check yourself

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

1. Angle PSR is found as:
   A. Sum of the bearings
   B. Their average
   C. Difference of the bearings = 90 degrees
   D. Always 45 degrees
   Answer: C. 177 30 20 minus 87 30 20 = 90. (Key point: Angle PSR)

2. Angle QSP is:
   A. arctan(QS / PQ) = 63 26 06
   B. 45 degrees
   C. 90 degrees
   D. arctan(PQ / QS) = 26 33 54
   Answer: D. tan = 45/90 = 0.5. (Key point: Angle QSP)

3. QR is found by:
   A. QS tan(angle QSR) = 180 m
   B. QS cos(QSR)
   C. QS / tan(QSR)
   D. PQ + QS
   Answer: A. SQR is right-angled at Q, QSR = 63 26 06 and tan = 2. (Key point: QR)

4. Chainage of R is:
   A. 740 m
   B. 830 m
   C. 650 m
   D. 605 m
   Answer: B. 650 + 180 = 830 m. (Key point: Chainage)

## Related in Surveying I

- [Length of PQ across a pond](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aobstacle-pond-chain-line) — Surveying I, chapter 3
- [Adjusted angles and bearings, angles to the left](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aadjusted-angles-bearings-left-angles-aw-dx) — Surveying I, chapter 4
- [Width of river from bearings](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Ariver-width-from-bearings-pq-400m) — Surveying I, chapter 4
- [Local attraction and true bearings, 1°30′E](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Acompass-closed-traverse-declination-1-30-east) — Surveying I, chapter 4
- [Area from perpendicular offsets to a hedge](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-hedge-5-5-12-7) — Surveying I, chapter 9
- [Area enclosed by chain line and hedge](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-hedge-5m-interval-simpson-trapezoidal) — Surveying I, chapter 9

## Related in other subjects

- [Transition curve 6°, Ls=90 m, chainage 13+700](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-t1v-t2v-chainage-13700) — Surveying II, chapter 6
- [Transition curve 6°, Ls=90 m, chainage (13+7) chains](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-t1v-t2v-chainage-13-7) — Surveying II, chapter 6
- [Is treatment necessary? Industry near river](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atreatment-industry-river-1p72) — Sanitary Engineering, chapter 6
- [Building near river bank; stop construction?](https://www.bctengineer.com/question.md?key=semester_8%2Fengineering_professional_practice%3Abuilding-on-river-bank-flood) — Engineering Professional Practice, chapter 7