---
title: "Highway curve through M 16 m from IP"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:highway-curve-pqr-140-point-m"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/highway-curve-pqr-140-point-m
site: BCT Engineering
---

# Highway curve through M 16 m from IP

Surveying II · Chapter 6: Curves

- Asked in: 2076 Baisakh, 6 marks
- Video: [Surveying Horizontal Curve Simple Curve Problem 1 (PH)](https://www.youtube.com/watch?v=62nj5A4VKjg) — CE BOX
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/highway-curve-pqr-140-point-m
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Two tangents PQ & QR of a highway curve meet at an angle to right of 140°. The circular curve which will pass through a point M 16 m away from IP(Q) with an angle PQM being 70°. Compute the suitable radius of curve then compute the first three deflection angles on the curve if chainage of IP is 2250.00 m. Take normal chord = 30 m.

## Short answer

M lies on the bisector 16 m from IP, so external E=16 m. Deflection Δ=40°, R=16/(sec20°−1)=249.31 m. First three deflection angles (30 m chords) from T₁: 0°05′06″, 3°31′57″, 6°58′47″.

## Answer

#### Step 1: Deflection angle and position of M
The angle to the right between the tangents is the interior angle $\angle PQR = 140^\circ$, so the deflection angle is $\Delta = 180^\circ - 140^\circ = 40^\circ$.

Since $\angle PQM = 70^\circ = 140^\circ/2$, the point M lies on the line joining the IP (Q) to the centre of the curve (the bisector of the interior angle). M is on the curve, so $QM$ is the external (apex) distance:

```text
        Q (IP)
        /|\
      P/ M \R      QM = E = 16 m
        \|/
         O
```

#### Step 2: Radius
$$
E = R\left(\sec\frac{\Delta}{2} - 1\right)\ \Rightarrow\ R = \frac{E}{\sec 20^\circ - 1} = \frac{16}{1.06418 - 1} = 249.31\ \text{m}
$$

#### Step 3: Tangent length, curve length, chainages
$$
\begin{aligned}
T &= R\tan 20^\circ = 90.74\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 174.05\ \text{m}\\
\text{Ch. of T}_1 &= 2250.00 - 90.74 = 2159.26\ \text{m}\\
\text{Ch. of T}_2 &= 2159.26 + 174.05 = 2333.31\ \text{m}
\end{aligned}
$$

#### Step 4: First three deflection angles
The first sub-chord runs from T₁ to the next 30 m peg, $c_1 = 0.74$ m, followed by full chords of 30 m. $\delta = 1718.87\,c/R$ minutes.

| Point (chainage, m) | Chord c (m) | δ | Cumulative deflection |
|---|---|---|---|
| 2160.00 | 0.74 | 0°05′06″ | 0°05′06″ |
| 2190.00 | 30.00 | 3°26′50″ | 3°31′57″ |
| 2220.00 | 30.00 | 3°26′50″ | 6°58′47″ |

**Answer:** $R = 249.31$ m; the first three deflection angles are 0°05′06″, 3°31′57″ and 6°58′47″.

## Check yourself

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

1. Interior angle PQR is 140°. The deflection angle is
   A. 140°
   B. 70°
   C. 20°
   D. 40°
   Answer: D. Δ = 180° - 140° = 40°. (Key point: Deflection angle)

2. Since angle PQM = 70° = half of 140° and M is on the curve, QM is
   A. the external distance E
   B. the tangent length
   C. the versine at the centre
   D. the long chord
   Answer: A. QM lies on the bisector of the interior angle, from IP to the curve, which is the external distance. (Key point: Meaning of M)

3. Which formula gives R from E = 16 m?
   A. R = E/tan(Δ/2)
   B. R = E/(sec(Δ/2) - 1) = 249.31 m
   C. R = E/(1 - cos Δ)
   D. R = E x sec Δ
   Answer: B. E = R(sec(Δ/2) - 1), so R = 16/0.06418 = 249.31 m. (Key point: Radius)

4. T1 = 2159.26 m and pegs are every 30 m. The first chord is
   A. 29.26 m
   B. 90.74 m
   C. 0.74 m
   D. 30 m
   Answer: C. Next peg is 2160, which is 0.74 m ahead; its δ = 1718.87 x 0.74/249.31 = 5.1 min = 0°05′06″. (Key point: First chord)

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