---
title: "Width of river from bearings"
subject: "Surveying I"
chapter: 4
key: "semester_3/surveying_i:river-width-from-bearings-pq-400m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/river-width-from-bearings-pq-400m
site: BCT Engineering
---

# Width of river from bearings

Surveying I · Chapter 4: The Compass

- Asked in: 2066 Jestha, 8 marks
- Video: [Width of river_(SUR-17) by tarun sir (IIT Delhi)_GATE AIR -15_FOR SSC_JEN/GATE/ESE](https://www.youtube.com/watch?v=fRup3ykSDuc) — Rinchtar classes for civil engineering (RCC)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/river-width-from-bearings-pq-400m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4

## Question

P and Q are two points 400 m apart on the same bank of the river. The bearings of a tree on the other bank observed from P and Q are N40°30′E and N37°45′W respectively. Find the width of the river if the bearing of PQ is S85°30′E.

## Short answer

Using triangle PQT with angles TPQ=54°, TQP=47°45′, PTQ=78°15′. PT = 400 sin47°45′ / sin78°15′ = 302.42 m. Width = 302.42 sin54° = 244.67 m, approximately 244.7 m.

## Answer

Let T be the tree on the opposite bank and PQ the base line of 400 m on the near bank. The width of the river is the perpendicular distance of T from the line PQ.

#### Step 1: Bearings in whole circle form

| Line | Reduced bearing | WCB |
|---|---|---|
| PQ | S 85°30′ E | 180° − 85°30′ = 94°30′ |
| PT | N 40°30′ E | 40°30′ |
| QT | N 37°45′ W | 360° − 37°45′ = 322°15′ |
| QP (back bearing of PQ) | | 94°30′ + 180° = 274°30′ |

```text
         T (tree, far bank)
        / \
       /   \
  ----P-----Q---- near bank
        400 m
```

#### Step 2: Angles of the triangle PQT

$$
\angle TPQ = 94^\circ 30' - 40^\circ 30' = 54^\circ 00'
$$

$$
\angle TQP = 322^\circ 15' - 274^\circ 30' = 47^\circ 45'
$$

$$
\angle PTQ = 180^\circ - 54^\circ 00' - 47^\circ 45' = 78^\circ 15'
$$

#### Step 3: Sine rule

$$
PT = \frac{PQ \sin \angle TQP}{\sin \angle PTQ} = \frac{400 \times \sin 47^\circ 45'}{\sin 78^\circ 15'} = 302.42\ \text{m}
$$

#### Step 4: Width of river

$$
\text{Width} = PT \sin \angle TPQ = 302.42 \times \sin 54^\circ = 244.67\ \text{m}
$$

(Check with the single formula: $\dfrac{400 \sin 54^\circ \sin 47^\circ 45'}{\sin 78^\circ 15'} = 244.67$ m.)

**Answer: width of the river = 244.67 m ≈ 244.7 m.**

## Check yourself

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

1. What is the whole circle bearing of PQ, S 85°30′ E?
   A. 94°30′
   B. 85°30′
   C. 265°30′
   D. 274°30′
   Answer: A. In the SE quadrant WCB = 180° - 85°30′ = 94°30′. (Key point: Bearing conversion)

2. What is the angle TPQ?
   A. 134°00′
   B. 40°30′
   C. 85°30′
   D. 54°00′
   Answer: D. Angle TPQ = bearing of PQ - bearing of PT = 94°30′ - 40°30′ = 54°00′. (Key point: Angle at P)

3. Which expression gives PT in triangle PQT?
   A. 400 sin 54° / sin 78°15′
   B. 400 sin 78°15′ / sin 47°45′
   C. 400 sin 47°45′ / sin 78°15′
   D. 400 sin 47°45′ / sin 54°
   Answer: C. PT is opposite angle Q (47°45′) and PQ is opposite angle T (78°15′), so PT = PQ sin Q / sin T. (Key point: Sine rule)

4. Why is the width 302.42 x sin 54° = 244.67 m and not 302.42 m?
   A. Width equals the distance PQ times sin 54°
   B. Width is the perpendicular distance of T from PQ, not the slant distance PT
   C. PT must be reduced by 400 m
   D. The river is measured along the bearing of PQ
   Answer: B. PT is slanting; the perpendicular from T to PQ is PT sin(angle TPQ). (Key point: Width of river)

## Related in Surveying I

- [Local attraction, bearings AB 305°00′ (EA 45°)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-305-00) — Surveying I, chapter 4, asked 2×
- [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
- [Local attraction, bearings AB 305°30′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-bb-44-00) — Surveying I, chapter 4, asked 2×
- [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
- [Chainage of R across a river](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aobstacle-river-chainage-of-r) — Surveying I, chapter 3
- [Corrected bearings and interior angles, PQRST](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Acorrected-bearings-quadrantal-pq-s37-30e) — Surveying I, chapter 4

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