---
title: "Derive three point analytical resection"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:three-point-analytical-resection-derivation"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/three-point-analytical-resection-derivation
site: BCT Engineering
---

# Derive three point analytical resection

Surveying II · Chapter 5: Orientation

- Asked in: 2066 Magh (old course), 8 marks; 2065 Kartik (old course), 8 marks (asked 2 times)
- Video: [Analytical method of Solving three point problem | Collin's method | Surveying | Prashant YT | BE |](https://www.youtube.com/watch?v=i-Af3CXWY4w) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/three-point-analytical-resection-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

Derive the expression for the three point resection problem.

## Short answer

From known coordinates of A,B,C compute AB=c, BC=a and ∠ABC=B. With measured α=∠APB, β=∠BPC, let x=∠PAB, y=∠PCB. In quadrilateral sum, x+y = 360°-(α+β+B)=S. Using sine rules in triangles APB and BPC, equating PB gives sin x/sin y = (a sinα)/(c sinβ)=k. Apply identity to get tan((x-y)/2) = ((k-1)/(k+1)) tan(S/2), whence x and y. Then compute AP or BP length and bearing to obtain P coordinates.

## Answer

In the three-point resection problem the coordinates of three stations A, B, C are known and the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are measured at the unknown station 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 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 $AB$ $\pm\, x$ (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.

## Check yourself

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

1. In the three-point problem, what is measured and what is required?
   A. Angles APB and BPC at P are measured; coordinates of P are required
   B. Distances PA, PB, PC are measured; angles are required
   C. Coordinates of P are known; those of A, B, C are required
   D. Only the angle ABC is measured
   Answer: A. A, B, C have known coordinates, alpha and beta are observed at P, and P's coordinates are computed. (Key point: Given and required)

2. x + y = S where S is:
   A. 180 - (alpha + beta)
   B. 360 - (alpha + beta + B)
   C. alpha + beta - B
   D. B - alpha - beta
   Answer: B. The four interior angles of PABC total 360 degrees. (Key point: Sum of x and y)

3. How is sin x / sin y = k obtained?
   A. Sine rule in triangles APB and BPC, equating the two expressions for PB
   B. Cosine rule in triangle ABC
   C. Angle sum of triangle APB
   D. Bearing of AB and BC
   Answer: A. PB = c sin x / sin alpha = a sin y / sin beta, so sin x / sin y = a sin alpha / (c sin beta). (Key point: Equating PB)

4. After finding x, the coordinates of P are obtained from:
   A. E_P = E_A + AP cos(bearing), N_P = N_A + AP sin(bearing)
   B. E_P = E_A + AP sin(bearing AP), N_P = N_A + AP cos(bearing AP)
   C. E_P = E_B, N_P = N_C
   D. E_P = AP sin x, N_P = AP cos x
   Answer: B. Easting uses sin and northing uses cos of the bearing of AP, where bearing AP = bearing AB +/- x. (Key point: Coordinates from AP)

5. How is the computed position of P checked?
   A. By computing P again from C using CP and bearing CB -/+ y; both must agree
   B. By repeating the same computation from A
   C. By measuring PB with a tape only
   D. No check is possible
   Answer: A. An independent computation from C should give the same coordinates. (Key point: Check)

## Related in Surveying II

- [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
- [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
- [Resection, two point problem, intersection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-two-point-problem-ab-p) — Surveying II, chapter 5
- [Intersection vs resection, three point method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-vs-resection-three-point) — Surveying II, chapter 5
- [Explain one method of analytical resection](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanalytical-resection-one-method) — Surveying II, chapter 5, asked 3×
- [Two point resection, point C from A, B, D](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-acb-dca-adb) — Surveying II, chapter 5

## Related in other subjects

- [Resection by three point problem](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Athree-point-resection-sketches) — Surveying I, chapter 6
- [Three-point problem: A 800 m N, C 600 m E (600, 200, 500)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-800-m) — Engineering Geology I, chapter 5, asked 4×
- [Three-point problem: A 600 m N, C 400 m W (400, 100, 300)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-600-m-n) — Engineering Geology I, chapter 5, asked 4×
- [Three-point problem: A 800 m N, C 400 m W (500, 100, 300)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-800-m-n) — Engineering Geology I, chapter 5, asked 4×