---
title: "Curve by tangential angle, deflection 48°30'"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:tangential-angle-curve-radius-r-48"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/tangential-angle-curve-radius-r-48
site: BCT Engineering
---

# Curve by tangential angle, deflection 48°30'

Surveying II · Chapter 6: Curves

- Asked in: 2077 Chaitra, 6 marks
- Video: [Problems on Rankine method||Surveying ||Deflection angle||Simple circular curve||curve||Civil engg.](https://www.youtube.com/watch?v=vK7JSjaJ-pc) — TECHNICAL LEARNING
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tangential-angle-curve-radius-r-48
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

It is necessary to design a circular curve by tangential angle method by selecting radius 'R' in such a way that the tangent length should be provided within the length of 35 m, having deflection angle 48°30' and chainage of corresponding IP is 1+945.55 km. Take normal chord = 20 m. Prepare a setting out table.

## Short answer

Radius selected = 75 m to keep tangent length within 35 m. Tangent length = 33.79 m, curve length = 63.49 m. Chainages: T1 = 1911.76 m, T2 = 1975.25 m. Setting out uses normal chord 20 m and cumulative deflection angles from T1.

## Answer

Data: $\Delta = 48^\circ30'$, chainage of IP = 1+945.55 (1945.55 m), tangent length not more than 35 m, normal chord $C = 20$ m.

#### 1. Selecting the radius

$$
T = R\tan\frac{\Delta}{2} \le 35 \quad\Rightarrow\quad R \le \frac{35}{\tan 24^\circ15'} = 77.70\ \text{m}
$$

Adopt **R = 75 m** (the next lower round value). Then

$$
T = 75\tan 24^\circ15' = 33.785\ \text{m} \ (< 35\ \text{m})
$$

#### 2. Curve elements

$$
L_c = \frac{\pi R\Delta}{180^\circ} = \frac{\pi\times 75\times 48.5}{180} = 63.486\ \text{m}
$$

- Chainage of T1 $= 1945.55 - 33.785 = 1911.765$ m
- Chainage of T2 $= 1911.765 + 63.486 = 1975.251$ m

#### 3. Setting out table (tangential angles)

$$
\delta = \frac{1718.9\,c}{R}\ \text{minutes} \quad (\delta_{20} = 7°38'22")
$$

Pegs are at the multiples of 20 m; the first and the last chords are sub-chords.

| Peg chainage (m) | Chord (m) | Deflection for chord $\delta=1718.9\,c/R$ | Cumulative deflection |
|---|---|---|---|
| 1920.00 | 8.24 | 3°08'44" | 3°08'44" |
| 1940.00 | 20.00 | 7°38'22" | 10°47'07" |
| 1960.00 | 20.00 | 7°38'22" | 18°25'29" |
| 1975.25 | 15.25 | 5°49'32" | 24°15'01" |

Check: the last cumulative angle 24°15'01" $= \Delta/2 = 24^\circ15'00''$ ✓.

**Procedure:** set up at T1, set the horizontal circle to zero on the IP, and set out each peg by turning the cumulative angle in the table and measuring the chord from the previous peg with a tape; the last peg must fall on T2, which checks the work.

**Answer: R = 75 m, T = 33.79 m, $L_c$ = 63.49 m, T1 = 1911.76 m, T2 = 1975.25 m.**

## Check yourself

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

1. How is the radius chosen for a tangent length not exceeding 35 m?
   A. From T = R tan(Delta/2) <= 35, so R <= 77.7 m, adopt 75 m
   B. R >= 77.7 m so adopt 80 m
   C. R = 35/tan(48.5 deg)
   D. R = 35 m
   Answer: A. T grows with R, so R must not exceed 35/tan24 deg15' = 77.7 m; the next lower round value 75 m is adopted. (Key point: Choosing R)

2. What formula gives the curve length?
   A. Lc = R tan(Delta/2) = 33.79 m
   B. Lc = pi R Delta/180 = 63.49 m
   C. Lc = 2R sin(Delta/2)
   D. Lc = R Delta in degrees
   Answer: B. The arc length is pi R Delta/180 with Delta in degrees. (Key point: Curve length)

3. What is the tangential angle for a chord c?
   A. delta = 1718.9 R/c minutes
   B. delta = 573 c/R minutes
   C. delta = 1718.9 c/R minutes
   D. delta = 3437.7 c/R minutes
   Answer: C. The tangent-chord angle is half the central angle: delta = 1718.9 c/R minutes. (Key point: Tangential angle)

4. What check confirms the table is right?
   A. Last cumulative deflection equals Delta
   B. Sum of chords equals T
   C. Last cumulative deflection equals 90 deg
   D. Last cumulative deflection equals Delta/2 = 24 deg 15' and the last peg falls on T2
   Answer: D. The final chord is at the end of the curve, where the deflection from the tangent is Delta/2. (Key point: Check of work)

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