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

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).

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".

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: ∠ACB=70°35′48"\angle ACB = 70°35'48", ∠DCA=52°25′35"\angle DCA = 52°25'35" at C and ∠ADB=65°27′35"\angle ADB = 65°27'35", ∠BDC=32°16′42"\angle BDC = 32°16'42" at D.

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

Step 1: Base AB

c=AB=(9602.054−7464.418)2+(5363.275−6928.474)2=2649.403 m,bearing of AB=126°12′43"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

∠ADC=∠ADB+∠BDC=97°44′17",∠DCB=∠ACB+∠DCA=123°01′23"\angle ADC = \angle ADB + \angle BDC = 97°44'17",\qquad \angle DCB = \angle ACB + \angle DCA = 123°01'23" S=x+y=∠DAB+∠CBA=360∘−97°44′17"−123°01′23"=139°14′20"S = x + y = \angle DAB + \angle CBA = 360^\circ - 97°44'17" - 123°01'23" = 139°14'20"

Step 3: Solve for x = ∠DAB

P0=180∘−∠ADB=114°32′25"P_0 = 180^\circ - \angle ADB = 114°32'25" and

K=sin⁡∠ADB sin⁡∠DCAsin⁡∠ACB sin⁡∠ADC=0.771417K = \frac{\sin\angle ADB\,\sin\angle DCA}{\sin\angle ACB\,\sin\angle ADC} = 0.771417 tan⁡x=sin⁡P0−Ksin⁡Scos⁡P0−Kcos⁡S⇒x=∠DAB=67°24′16"\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=∠CBA=S−x=71°50′04"y = \angle CBA = S - x = 71°50'04"

Step 4: Distances AD and BC

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

AD=csin⁡∠ABDsin⁡∠ADB=2134.766 m,BC=csin⁡∠BACsin⁡∠ACB=1712.650 mAD = \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"= 193°36'59"; bearing of BC = bearing of BA − y =234°22′39"= 234°22'39" (C and D are on the right-hand side of the direction A to B).

ED=7464.418+2134.766sin⁡193°36′59"=6961.853,ND=6928.474+2134.766cos⁡193°36′59"=4853.708E_D = 7464.418 + 2134.766\sin 193°36'59" = 6961.853,\qquad N_D = 6928.474 + 2134.766\cos 193°36'59" = 4853.708 EC=9602.054+1712.650sin⁡234°22′39"=8209.889,NC=5363.275+1712.650cos⁡234°22′39"=4365.754E_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 ∠ACB=70°35′48"\angle ACB = 70°35'48", ∠DCA=52°25′35"\angle DCA = 52°25'35", ∠ADB=65°27′35"\angle ADB = 65°27'35", ∠BDC=32°16′42"\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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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