---
title: "Circular curve through D, 5 m from B"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:circular-curve-through-point-d-5m"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/circular-curve-through-point-d-5m
site: BCT Engineering
---

# Circular curve through D, 5 m from B

Surveying II · Chapter 6: Curves

- Asked in: 2075 Bhadra, 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/circular-curve-through-point-d-5m
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Two straights intersecting at a point B have the following bearings. AB 90°, CB 290°. They are to be joined by a circular curve which must pass through a point D which is 5 m from B and the bearing of BD is 190°. Find the required radius, tangent lengths, length of curve and setting-out angle for a 20-m chord.

## Short answer

The deflection angle is 20°, BD = 5 m lies on the bisector, so external distance E=5 m. Radius R = E/(sec(Δ/2)–1) = 324.12 m, tangent length T = R tan(Δ/2) = 57.15 m, curve length L = πRΔ/180 = 113.14 m. Setting-out angle for a 20 m chord = 1718.87×20/R = 106.1′ = 1°46′04″.

## Answer

#### Step 1: Deflection angle
Direction of travel A→B: $90^\circ$. Bearing of CB is $290^\circ$, so the bearing of B→C is $290^\circ - 180^\circ = 110^\circ$.

$$
\Delta = 110^\circ - 90^\circ = 20^\circ\ \text{(to the right)}
$$

#### Step 2: Position of D
Back bearing BA $= 270^\circ$ and bearing BC $= 110^\circ$. The bisector of the interior angle ABC has bearing $\dfrac{270^\circ + 110^\circ}{2} = 190^\circ$, which is the bearing of BD. Hence D lies on the line from B to the centre, and BD is the external distance:

$$
E = 5\ \text{m}
$$

```text
   A ------- B ------- (to C at 110°)
              \
               D  (BD = 5 m, 190°)
               |
               O
```

#### Step 3: Radius
$$
E = R\left(\sec\frac{\Delta}{2} - 1\right)\ \Rightarrow\ R = \frac{5}{\sec 10^\circ - 1} = \frac{5}{1.01543-1} = 324.12\ \text{m}
$$

#### Step 4: Tangent length and curve length
$$
\begin{aligned}
T &= R\tan\frac{\Delta}{2} = 324.12\tan 10^\circ = 57.15\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 113.14\ \text{m}
\end{aligned}
$$

#### Step 5: Setting-out angle for a 20 m chord
$$
\delta = \frac{1718.87\,c}{R} = \frac{1718.87\times 20}{324.12} = 106.07' = 1°46′04″
$$

**Answer:** $R = 324.12$ m, $T = 57.15$ m, $L = 113.14$ m, setting-out angle for a 20 m chord $= 1°46′04″$ (about 106.1′).

## Check yourself

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

1. AB has bearing 90 deg and CB has bearing 290 deg. What is the deflection angle of the curve?
   A. 160 deg
   B. 20 deg
   C. 180 deg
   D. 110 deg
   Answer: B. Bearing of BC = 290 - 180 = 110 deg, so deflection = 110 - 90 = 20 deg to the right. (Key point: Deflection angle)

2. Why can BD = 5 m be taken as the external distance E?
   A. Because BD is perpendicular to AB
   B. Because D is on the tangent
   C. Because BD equals the tangent length
   D. Because the bearing 190 deg of BD equals the bisector of the interior angle, so D lies on the line B to centre
   Answer: D. The bisector of angle ABC has bearing (270+110)/2 = 190 deg, so D lies on the line from B to the centre, making BD the external distance. (Key point: Meaning of BD)

3. Which relation gives the radius from E?
   A. R = E(sec(Δ/2) - 1)
   B. R = E/(sec(Δ/2) - 1)
   C. R = E/(1 - cos Δ)
   D. R = E tan(Δ/2)
   Answer: B. E = R(sec(Δ/2) - 1), so R = 5/(sec 10° - 1) = 324.12 m. (Key point: Radius from external distance)

4. With R = 324.12 m and Δ = 20 deg, which pair of results is correct?
   A. T = 114.3 m, L = 57.15 m
   B. T = 57.15 m, L = 56.57 m
   C. T = 324.12 m, L = 113.14 m
   D. T = 57.15 m, L = 113.14 m
   Answer: D. T = R tan 10° = 57.15 m and L = πRΔ/180 = 113.14 m. (Key point: Tangent and curve length)

## Related in Surveying II

- [Simple curve data, roads intersecting at 150°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-two-roads-150-4deg) — Surveying II, chapter 6
- [Composite curve R=400 m, Ls=90 m, IP 1+400](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-r400-ls90-ip-1400) — 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
- [Highway curve through M 16 m from IP](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahighway-curve-pqr-140-point-m) — Surveying II, chapter 6
- [Circular curve with cubic spiral transitions R=450](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acubic-spiral-transition-450m-18-36) — Surveying II, chapter 6
- [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

## Related in other subjects

- [Transition curve 625 m radius, 90 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75b-tc-625) — Transportation Engineering I, chapter 3
- [Vertical curve +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — Transportation Engineering I, chapter 3
- [Transition curve for 310 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-tc-310) — Transportation Engineering I, chapter 3
- [Transition curve 400 m radius, 85 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-tc-400-85) — Transportation Engineering I, chapter 3, asked 2×