---
title: "Explain one method of analytical resection"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:analytical-resection-one-method"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/analytical-resection-one-method
site: BCT Engineering
---

# Explain one method of analytical resection

Surveying II · Chapter 5: Orientation

- Asked in: 2070 Bhadra, 8 marks; 2068 Bhadra, 7 marks; 2065 Chaitra (old course), 8 marks (asked 3 times)
- Video: [maurieo INVESTIGATES THE COLLINS POINT RESECTION METHOD](https://www.youtube.com/watch?v=607d5HbkQrc) — maurieo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/analytical-resection-one-method
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

Explain any one method of analytical resection to calculate the position of instrument station.

## Short answer

Pothenot's tangent method solves three-point resection. From known stations A, B, C, measure angles α=∠APB and β=∠BPC at unknown P. Compute sides a=BC, c=AB and angle B from coordinates. Let x=∠PAB, y=∠PCB; then x+y = S = 360°-(α+β+B). Compute k = (a sinα)/(c sinβ), then tan((x-y)/2) = ((k-1)/(k+1)) tan(S/2). Solve for x,y, then find bearing and distance from A or C to get P coordinates.

## Answer

**Resection** is the method of finding the coordinates of an unknown station P by measuring the horizontal angles at P to three (or two) stations of known coordinates. The instrument is set up only at the unknown point. When the unknown station is observed to three known stations, it is called the **three-point problem**.

#### Three-point resection (Pothenot's tangent method)

**Problem:** A, B and C are known stations (B is the middle one). The instrument is at the unknown station P, and the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are observed. Find the coordinates of P.

```text
         B
        /|\
     c / | \ a
      /  |  \
     A   |   C
      \  |  /
       \ | /
        \|/ 
         P      angles at P: alpha (APB), beta (BPC)
```

**Derivation**

From the coordinates, find the sides $c = AB$, $a = BC$ and the angle $B = \angle ABC$ (from the bearings of BA and BC). Let $x = \angle PAB$ and $y = \angle PCB$ be the unknown angles. In the quadrilateral PABC the sum of the angles is $360^\circ$:

$$
x + y + \alpha + \beta + B = 360^\circ \quad\Rightarrow\quad x + y = S = 360^\circ - (\alpha + \beta + B)
$$

In triangle APB, by the sine rule, $PB = \dfrac{c\sin x}{\sin\alpha}$. In triangle BPC, $PB = \dfrac{a\sin y}{\sin\beta}$. Equating,

$$
\frac{\sin x}{\sin y} = \frac{a\sin\alpha}{c\sin\beta} = k
$$

Using $\dfrac{\sin x - \sin y}{\sin x + \sin y} = \dfrac{\tan\frac{x-y}{2}}{\tan\frac{x+y}{2}}$:

$$
\tan\frac{x-y}{2} = \frac{k-1}{k+1}\tan\frac{S}{2}
$$

So $\dfrac{x-y}{2}$ is found, and with $x + y = S$ we get $x$ and $y$ separately.

**Coordinates of P**

1. In triangle APB: $\angle ABP = 180^\circ - \alpha - x$, so $AP = \dfrac{c\sin\angle ABP}{\sin\alpha}$ (or $BP = \dfrac{c\sin x}{\sin\alpha}$).
2. Bearing of $AP$ = bearing of $AB$ $\pm\, x$ (the sign depends on the side on which P lies).
3. $E_P = E_A + AP\sin(\text{bearing }AP)$, $\;N_P = N_A + AP\cos(\text{bearing }AP)$.
4. Check by computing P from C: $CP = \dfrac{a\sin\angle CBP}{\sin\beta}$, with the bearing of $CP$ = bearing of $CB \mp y$. Both must give the same coordinates.

**Note:** If P lies on the circle through A, B and C (the "danger circle"), then $\alpha + \beta + B = 180^\circ$, $\tan\frac{S}{2}$ is infinite, and the problem has no unique solution. The point P should be chosen away from this circle.

#### Practical check

If $\alpha + \beta + B$ is close to $180^\circ$, the point is near the danger circle and the result is weak; the observations should then be repeated with a different choice of known stations.

## Check yourself

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

1. In resection the instrument is set up at:
   A. Only the known stations
   B. Only the unknown station, with angles measured to known stations
   C. Both unknown and known stations
   D. The midpoint of the base line
   Answer: B. Resection finds the coordinates of an unknown station by measuring angles at that station to stations of known coordinates. (Key point: Resection definition)

2. In quadrilateral PABC, with x = PAB, y = PCB, the relation for x + y is:
   A. x + y = 180 - (alpha + beta)
   B. x + y = 360 - (alpha + beta + B)
   C. x + y = alpha + beta + B
   D. x + y = 90 - B
   Answer: B. The angles of the quadrilateral sum to 360 degrees, so x + y = S = 360 - (alpha + beta + B). (Key point: Angle sum relation)

3. The key result giving (x - y)/2 is:
   A. tan((x-y)/2) = (k-1)/(k+1) tan(S/2)
   B. tan((x-y)/2) = (k+1)/(k-1) tan(S/2)
   C. sin((x-y)/2) = k tan(S/2)
   D. tan((x-y)/2) = k tan(S)
   Answer: A. From sin x / sin y = k and (sin x - sin y)/(sin x + sin y) = tan((x-y)/2)/tan((x+y)/2). (Key point: Tangent formula)

4. In Pothenot's method, k = sin x / sin y equals:
   A. c sin alpha / (a sin beta)
   B. a sin alpha / (c sin beta)
   C. a sin beta / (c sin alpha)
   D. a / c
   Answer: B. Equating PB = c sin x / sin alpha and PB = a sin y / sin beta gives sin x / sin y = a sin alpha / (c sin beta). (Key point: Value of k)

5. When does the three-point problem have no unique solution?
   A. When P lies on the circle through A, B and C, so alpha + beta + B = 180 degrees
   B. When angles alpha and beta are equal
   C. When B is the middle station
   D. When P lies inside triangle ABC
   Answer: A. On the danger circle tan(S/2) becomes infinite, so the position is indeterminate; P must be chosen away from it. (Key point: Danger circle)

## Related in Surveying II

- [Three point analytical resection, average coordinates of O](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-resection-p-w-d) — Surveying II, chapter 5
- [Analytical resection, derive coordinate expression](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanalytical-resection-derive-three-known) — Surveying II, chapter 5
- [Coordinates of instrument station O and station A](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-pwd-instrument-station-a) — Surveying II, chapter 5
- [Derive three point analytical resection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-analytical-resection-derivation) — Surveying II, chapter 5, asked 2×
- [Distances OA and OC by resection on A, B, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-abc-sides-oa-oc) — Surveying II, chapter 5
- [Resection point O inside triangle N, P, G](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-nag-triangle-2078c) — Surveying II, chapter 5

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- [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3×
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