---
title: "Two point resection, point C from A, B, D"
subject: "Surveying II"
chapter: 5
key: "semester_4/surveying_ii:two-point-resection-acb-dca-adb"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/two-point-resection-acb-dca-adb
site: BCT Engineering
---

# Two point resection, point C from A, B, D

Surveying II · Chapter 5: Orientation

- Asked in: 2072 Magh, 6 marks
- Video: [Two Point Problem In Plane Table Surveying](https://www.youtube.com/watch?v=LJXWZrobg-c) — Ganesh Shegar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/two-point-resection-acb-dca-adb
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5

## Question

In two point resection problem, if two known points A and B having coordinates (6928.474 mN, 7464.418 mE) and (5363.275 mN, 9602.054 mE) are given. From two points C and D located south and west of AB, angles observations are: ∠ACB = 70°35'48", ∠DCA = 52°25'35", ∠ADB = 65°27'35" and ∠BDC = 32°16'42". Determine the coordinates of resection point "C".

## Short answer

Using two-point resection: AB=2649.403 m, bearing AB=126°12'43". Solving the quadrilateral gives x=∠DAB=67°24'16", y=∠CBA=71°50'04", AD=2134.766 m, BC=1712.650 m. Coordinates of C = (8209.889 mE, 4365.754 mN).

## Answer

This is the two-point resection (Hansen's problem), solved by the sine rule and the tangent relation.

Coordinates are (Easting, Northing): A = (7464.418 E, 6928.474 N), B = (9602.054 E, 5363.275 N). C and D are on the south-west side of AB (C south, D west). Observed: $\angle ACB = 70°35'48"$, $\angle DCA = 52°25'35"$ at C and $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'42"$ at D.

```text
        A ___________________ B
          \ \          __---'
           \  \   __--'
            D --- C
```

#### Step 1: Base AB

$$
c = AB = \sqrt{(9602.054-7464.418)^2 + (5363.275-6928.474)^2} = 2649.403\ \text{m},\qquad \text{bearing of AB} = 126°12'43"
$$

#### Step 2: Angles of the quadrilateral ADCB

$$
\angle ADC = \angle ADB + \angle BDC = 97°44'17",\qquad \angle DCB = \angle ACB + \angle DCA = 123°01'23"
$$

$$
S = x + y = \angle DAB + \angle CBA = 360^\circ - 97°44'17" - 123°01'23" = 139°14'20"
$$

#### Step 3: Solve for x = ∠DAB

$P_0 = 180^\circ - \angle ADB = 114°32'25"$ and

$$
K = \frac{\sin\angle ADB\,\sin\angle DCA}{\sin\angle ACB\,\sin\angle ADC} = 0.771417
$$

$$
\tan x = \frac{\sin P_0 - K\sin S}{\cos P_0 - K\cos S} \quad\Rightarrow\quad x = \angle DAB = 67°24'16"
$$

$$
y = \angle CBA = S - x = 71°50'04"
$$

#### Step 4: Distances AD and BC

$\angle ABD = 180^\circ - \angle ADB - x = 47°08'09"$ and $\angle BAC = 180^\circ - \angle ACB - y = 37°34'08"$

$$
AD = \frac{c\sin\angle ABD}{\sin\angle ADB} = 2134.766\ \text{m},\qquad BC = \frac{c\sin\angle BAC}{\sin\angle ACB} = 1712.650\ \text{m}
$$

#### Step 5: Coordinates

Bearing of AD = bearing of AB + x $= 193°36'59"$; bearing of BC = bearing of BA − y $= 234°22'39"$ (C and D are on the right-hand side of the direction A to B).

$$
E_D = 7464.418 + 2134.766\sin 193°36'59" = 6961.853,\qquad N_D = 6928.474 + 2134.766\cos 193°36'59" = 4853.708
$$

$$
E_C = 9602.054 + 1712.650\sin 234°22'39" = 8209.889,\qquad N_C = 5363.275 + 1712.650\cos 234°22'39" = 4365.754
$$

**Check:** the angles recomputed from these coordinates are $\angle ACB = 70°35'48"$, $\angle DCA = 52°25'35"$, $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'42"$, equal to the observed values. The length CD = 1340.035 m from the coordinates agrees with triangle ADC.

**Answer: C = (8209.889 E, 4365.754 N); D = (6961.853 E, 4853.708 N).**

## Check yourself

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

1. How is S = x + y = angle DAB + angle CBA found?
   A. S = 360° - (angle ADC + angle DCB) from the quadrilateral ADCB
   B. S = 180° - angle ACB
   C. S = angle ADB + angle ACB
   D. S = 90°
   Answer: A. The angles of quadrilateral ADCB sum to 360°, so with ADC = 97°44'17" and DCB = 123°01'23" we get S = 139°14'20". (Key point: Angle sum S)

2. How is the angle ADC obtained?
   A. Angle ADB + angle BDC = 97°44'17"
   B. Angle ADB - angle BDC
   C. 180° - angle ADB
   D. Angle DCA + angle ACB
   Answer: A. Ray DB lies between DA and DC, so ADC = ADB + BDC = 65°27'35" + 32°16'42". (Key point: Composite angles)

3. In triangle ABD, AD equals:
   A. c sin(ADB)/sin(ABD)
   B. c sin(ABD)/sin(ADB), with c = AB
   C. c sin(x)
   D. c cos(ABD)
   Answer: B. AD is opposite angle ABD and AB (=c) is opposite ADB, so AD = c sin(ABD)/sin(ADB) = 2134.766 m. (Key point: Distance AD by sine rule)

4. Bearing of AD = bearing of AB + x because:
   A. D lies on the right-hand side of the direction A to B
   B. x is always added
   C. D lies north of AB
   D. The bearing of AB is zero
   Answer: A. C and D lie south-west of AB, to the right of the direction A to B, so x is added clockwise: 193°36'59". (Key point: Bearing of AD and side)

## Related in Surveying II

- [Two point resection, point C (A, B at 7928.474 N)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-c-a-b) — 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
- [Coordinates of X from S, D, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-sdc-theodolite-2077c) — 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

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