Two point resection, point C (A, B at 7928.474 N)

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

Coordinates are (Easting, Northing): A = (8464.418 E, 7928.474 N), B = (10602.054 E, 6363.275 N). C and D are on the south-west side of AB (C south, D west). Observed: $\angle ACB = 70°35'50"$, $\angle DCA = 52°25'40"$ at C and $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'40"$ at D.

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

Step 1: Base AB

$$c = AB = \sqrt{(10602.054-8464.418)^2 + (6363.275-7928.474)^2} = 2649.403\ \text{m},\qquad \text{bearing of AB} = 126°12'43"$$

Step 2: Angles of the quadrilateral ADCB

$$\angle ADC = \angle ADB + \angle BDC = 97°44'15",\qquad \angle DCB = \angle ACB + \angle DCA = 123°01'30"$$ $$S = x + y = \angle DAB + \angle CBA = 360^\circ - 97°44'15" - 123°01'30" = 139°14'15"$$

Step 3: Solve for x = ∠DAB

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

$$K = \frac{\sin\angle ADB\,\sin\angle DCA}{\sin\angle ACB\,\sin\angle ADC} = 0.771427$$ $$\tan x = \frac{\sin P_0 - K\sin S}{\cos P_0 - K\cos S} \quad\Rightarrow\quad x = \angle DAB = 67°24'14"$$ $$y = \angle CBA = S - x = 71°50'01"$$

Step 4: Distances AD and BC

$\angle ABD = 180^\circ - \angle ADB - x = 47°08'11"$ and $\angle BAC = 180^\circ - \angle ACB - y = 37°34'09"$

$$AD = \frac{c\sin\angle ABD}{\sin\angle ADB} = 2134.785\ \text{m},\qquad BC = \frac{c\sin\angle BAC}{\sin\angle ACB} = 1712.655\ \text{m}$$

Step 5: Coordinates

Bearing of AD = bearing of AB + x $= 193°36'57"$; bearing of BC = bearing of BA − y $= 234°22'42"$ (C and D are on the right-hand side of the direction A to B).

$$E_D = 8464.418 + 2134.785\sin 193°36'57" = 7961.869,\qquad N_D = 7928.474 + 2134.785\cos 193°36'57" = 5853.684$$ $$E_C = 10602.054 + 1712.655\sin 234°22'42" = 9209.871,\qquad N_C = 6363.275 + 1712.655\cos 234°22'42" = 5365.771$$

Check: the angles recomputed from these coordinates are $\angle ACB = 70°35'50"$, $\angle DCA = 52°25'40"$, $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'40"$, equal to the observed values. The length CD = 1339.988 m from the coordinates agrees with triangle ADC.

Answer: C = (9209.871 E, 5365.771 N); D = (7961.869 E, 5853.684 N).