---
title: "Anti-clockwise traverse PQRSP missing data"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:anticlockwise-traverse-pqrsp-missing"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/anticlockwise-traverse-pqrsp-missing
site: BCT Engineering
---

# Anti-clockwise traverse PQRSP missing data

Surveying II · Chapter 1: Traversing

- Asked in: 2078 Chaitra, 10 marks
- Video: [The length and bearings of a closed traverse PQRSP are given in the table.The missing length and](https://www.youtube.com/watch?v=CkQNnP2pZh4) — Unique Series - Civil Mantraa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anticlockwise-traverse-pqrsp-missing
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

In a four sided anti-clockwise traverse PQRSP the following information are given.

| Side | Length (m) | Deflection angle | Bearing | Co-ordinates |
|---|---|---|---|---|
| PQ | 436.80 | ? | S45°W | ? |
| QR | ? | 85°00' Left | ? | ? |
| RS | 499.20 | 100°00' Left | ? | R = 2345 mE, 6789 mN |
| SP | 516.084 | ? | ? | ? |

a) Compute all the missing figures.
b) Compute the co-ordinates of other points with respect to R.

## Short answer

QR = 561.600 m; SP bearing = 313°43' (N46°17'W); deflection at P = 88°43' left, at S = 86°17' left. Coordinates relative to R: S (7171.409 mN, 2665.880 mE), P (7528.075 mN, 2292.875 mE), Q (7219.210 mN, 1984.011 mE).

## Answer

The traverse P-Q-R-S-P is anticlockwise, so each deflection is to the left and **bearing of a line = bearing of previous line − left deflection**. Unknowns: length QR, bearing SP (length SP is given), deflections at P and S, and the coordinates.

#### Step 1: Bearings that can be fixed
- PQ = S45°W = 180° + 45° = 225°00'
- QR = 225°00' − 85°00' = 140°00'
- RS = 140°00' − 100°00' = 40°00'

Latitude and departure of PQ and RS:

$$
\begin{aligned}
\text{PQ}:&\ 436.80\cos225^\circ=-308.86,\quad 436.80\sin225^\circ=-308.86\\
\text{RS}:&\ 499.20\cos40^\circ=+382.41,\quad 499.20\sin40^\circ=+320.88
\end{aligned}
$$

Their sums are $a=73.545$ m and $b=12.015$ m.

#### Step 2: Length of QR and bearing of SP
Closure requires $\text{PQ}+\text{QR}+\text{RS}+\text{SP}=0$. With QR of unknown length $x$ at $140^\circ$, the vector SP is $-(a+x\cos140^\circ,\ b+x\sin140^\circ)$ and its length is 516.084 m:

$$
(a+x\cos140^\circ)^2+(b+x\sin140^\circ)^2=516.084^2
$$

$$
x^2+2(a\cos140^\circ+b\sin140^\circ)x+(a^2+b^2-516.084^2)=0
$$

$$
x^2+(-97.231)x+(-260789.4)=0\ \Rightarrow\ x=561.600\ \text{m}\ \ (\text{the negative root is rejected})
$$

So **QR = 561.600 m**.

Then SP: latitude $=-(-356.665)=+356.665$, departure $=-(+373.005)=-373.005$.

$$
L_{SP}=\sqrt{+356.665^2+(-373.005)^2}=516.084\ \text{m (agrees with the given 516.084 m)},\qquad
\theta_{SP}=\tan^{-1}\frac{-373.005}{+356.665}=313^\circ 43'
$$

(North-west quadrant, N 46°17' W.)

| Line | Length (m) | Bearing | Latitude | Departure |
|---|---|---|---|---|
| PQ | 436.800 | 225°00' | -308.864 | -308.864 |
| QR | 561.600 | 140°00' | -430.210 | +360.989 |
| RS | 499.200 | 40°00' | +382.409 | +320.880 |
| SP | 516.084 | 313°43' | +356.665 | -373.005 |
| **Sum** |  |  | +0.000 | +0.000 |

#### Step 3: Missing deflection angles
- At P: bearing SP − bearing PQ = 313°43' − 225°00' = **88°43' left**
- At S: bearing RS − bearing SP = 40°00' − 313°43' + 360° = **86°17' left**

Check: 88°43' + 85°00' + 100°00' + 86°17' = 360°00', as required for a closed anticlockwise traverse.

| Station | Left deflection |
|---|---|
| P (SP to PQ) | 88°43' (computed) |
| Q (PQ to QR) | 85°00' (given) |
| R (QR to RS) | 100°00' (given) |
| S (RS to SP) | 86°17' (computed) |

#### Step 4: Coordinates with respect to R (R = 2345 mE, 6789 mN)
Going round R to S to P to Q and back to R:

| Station | Northing (m) | Easting (m) |
|---|---|---|
| R | 6789.000 | 2345.000 |
| S | 7171.409 | 2665.880 |
| P | 7528.075 | 2292.875 |
| Q | 7219.210 | 1984.011 |
| R (check) | 6789.000 | 2345.000 |

The traverse closes on R, so the work is checked.

**Answer:** QR = 561.600 m; SP bearing = 313°43'; deflection at P = 88°43' L, at S = 86°17' L; bearings PQ 225°00', QR 140°00', RS 40°00'. Coordinates: S (7171.409 N, 2665.880 E), P (7528.075 N, 2292.875 E), Q (7219.210 N, 1984.011 E).

## Check yourself

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

1. For an anticlockwise traverse with left deflection angles, how is the bearing of a line obtained?
   A. Bearing of previous line minus the left deflection angle
   B. Bearing of previous line plus the left deflection angle
   C. Bearing of previous line plus 180° minus the deflection
   D. Back bearing of the previous line plus the deflection
   Answer: A. For anticlockwise travel with left deflection, bearings decrease by the deflection: QR = 225° - 85° = 140°. (Key point: Bearing rule)

2. What is the whole circle bearing of PQ = S45°W?
   A. 135°00'
   B. 225°00'
   C. 315°00'
   D. 45°00'
   Answer: B. South-west quadrant: WCB = 180° + 45° = 225°. (Key point: Whole circle bearing of PQ)

3. How is the length QR found when SP's length is given but its bearing is not?
   A. Using the sine rule on the angles at P and Q
   B. Taking QR equal to RS
   C. Write closure with QR = x at 140°, impose |SP| = 516.084 m, and solve the quadratic, rejecting the negative root
   D. Using Bowditch's rule
   Answer: C. Latitude and departure of SP follow from the closure sums; its length condition gives a quadratic in x with one meaningful positive root (561.600 m). (Key point: Finding QR)

4. What check confirms the deflection angles of a closed anticlockwise traverse?
   A. They sum to 180°
   B. They sum to (2n-4) x 90° = 360° only for clockwise traverses
   C. They sum to 720°
   D. They sum to 360°
   Answer: D. 88°43' + 85° + 100° + 86°17' = 360°, as required for external turning of a closed figure. (Key point: Deflection check)

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