---
title: "Prolonging line past a high building"
subject: "Surveying I"
chapter: 3
key: "semester_3/surveying_i:obstacle-high-building-prolong-line"
source: https://www.bctengineer.com/q/semester_3/surveying_i/obstacle-high-building-prolong-line
site: BCT Engineering
---

# Prolonging line past a high building

Surveying I · Chapter 3: Chain Survey

- Asked in: 2062 Baisakh, 10 marks
- Video: [Problem No - 10 in  Obstacles in Chain Surveying from previous university questions](https://www.youtube.com/watch?v=28JpPXvhxiM) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/obstacle-high-building-prolong-line
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-3

## Question

A survey line AB is obstructed by a high building. To prolong the line beyond the building, a perpendicular BC 121.92 m long is set out at B. From C two lines CD and CE are set out at angles of 30° and 40° with CB respectively. Determine the lengths CD and CE so that D and E may be on the prolongation of AB. If the chainage of B is 95.10 m, find the chainage of D. Draw a sketch showing all the points.

## Short answer

From perpendicular BC=121.92 m, using right triangles: CD = BC/cos30° = 140.78 m, CE = BC/cos40° = 159.16 m. BD = BC tan30° = 70.39 m. Chainage of D = 95.10 + 70.39 = 165.49 m.

## Answer

#### Data
BC = 121.92 m, perpendicular to AB at B. At C, $\angle BCD = 30^\circ$ and $\angle BCE = 40^\circ$. D and E are to lie on the prolongation of AB beyond the building. Chainage of B $= 95.10$ m.

```text
 A ------ B ---[ building ]--- D ----- E ---->
          |                  /       /
 121.92   |              /      /
          |          /     /
          C ----------------
          (30 deg at C to D, 40 deg at C to E)
```
Triangles CBD and CBE are right-angled at B.

#### Length CD
$$
CD = \frac{BC}{\cos 30^\circ} = \frac{121.92}{0.8660} = 140.78\ \text{m}
$$

#### Length CE
$$
CE = \frac{BC}{\cos 40^\circ} = \frac{121.92}{0.7660} = 159.16\ \text{m}
$$

#### Distances BD and BE
$$
BD = BC\tan 30^\circ = 121.92\times 0.5774 = 70.39\ \text{m}
$$

$$
BE = BC\tan 40^\circ = 121.92\times 0.8391 = 102.30\ \text{m}
$$

#### Chainage of D
$$
\text{Chainage of D} = 95.10 + 70.39 = 165.49\ \text{m}
$$

(Chainage of E $= 95.10 + 102.30 = 197.40$ m.)

**Answer: CD = 140.78 m, CE = 159.16 m; chainage of D = 165.49 m.**

## Check yourself

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

1. Which formula gives CD?
   A. CD = BC x cos 30
   B. CD = BC tan 30
   C. CD = BC / cos 30
   D. CD = BC / sin 30
   Answer: C. Triangle CBD is right-angled at B so BC is adjacent to the 30 degree angle. (Key point: Method for CD)

2. CD is:
   A. 105.58 m
   B. 70.39 m
   C. 243.8 m
   D. 140.78 m
   Answer: D. 121.92 / 0.8660 = 140.78 m. (Key point: CD value)

3. BD is obtained from:
   A. BC tan 30 = 70.39 m
   B. BC cos 30 = 105.6 m
   C. BC / tan 30 = 211.2 m
   D. BC sin 30 = 60.96 m
   Answer: A. BD is opposite the 30 degree angle. (Key point: Distance BD)

4. Chainage of D is:
   A. 95.10 m
   B. 165.49 m
   C. 235.49 m
   D. 140.78 m
   Answer: B. 95.10 + 70.39 = 165.49 m. (Key point: Chainage of D)

## Related in Surveying I

- [Two peg test, AB = 40 m, near B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-40m-inner-side-b) — Surveying I, chapter 5
- [Local attraction, S 40°30′ W traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-quadrantal-s40-30w) — Surveying I, chapter 4
- [Local attraction, BC 79°30′ and DE 82°15′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-191-30-bc-79-30) — 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×
- [Local attraction: AB 69°30′ compass traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-69-30-reordered) — Surveying I, chapter 4
- [Local attraction ABCDEA by included angles](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-abcdea-included-angle) — Surveying I, chapter 4, asked 2×

## Related in other subjects

- [Design pipelines AB, BC and CD, C = 120](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c74) — Water Supply Engineering, chapter 7
- [Hardy-Cross network discharge AB, BC, AD, CD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardy-cross-as69) — Water Supply Engineering, chapter 7
- [Head loss and velocity in loop pipes AB, BC, AC, CD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c73) — Water Supply Engineering, chapter 7
- [Design pipes AB and BC, 75 lpcd, tank 6 m high](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-b81) — Water Supply Engineering, chapter 7