Two point resection method

Two-point resection is the method of finding the coordinates of two unknown stations C and D, from which two stations A and B of known coordinates are visible, by observing the horizontal angles at C and D. It is used when only two known points are available (Hansen's problem).

Problem: Two known stations A and B are visible from two unknown stations C and D, which can also see each other. The angles $\angle ACB$ and $\angle DCA$ are observed at C, and $\angle ADB$ and $\angle BDC$ at D. Find the coordinates of C and D.

        A _________________ B
          \      \      /
           \       \  /
            D ------ C

Steps

1. From the coordinates find $c = AB$ and the bearing of AB.
2. In the quadrilateral $A\,D\,C\,B$ the interior angles at the two observed stations are $\angle ADC = \angle ADB + \angle BDC$ and $\angle DCB = \angle ACB + \angle DCA$.
3. Let $x = \angle DAB$ and $y = \angle CBA$. Their sum is $S = x + y = 360^\circ - \angle ADC - \angle DCB$.
4. From triangle ABD: $AD = \dfrac{c\sin p}{\sin\angle ADB}$ with $p = \angle ABD = 180^\circ - \angle ADB - x$. From triangle ABC: $AC = \dfrac{c\sin y}{\sin\angle ACB}$.
5. In triangle ADC: $\dfrac{AD}{\sin\angle DCA} = \dfrac{AC}{\sin\angle ADC}$. Combining, $$\sin(P_0 - x) = K\sin(S - x), \qquad K = \frac{\sin\angle ADB\,\sin\angle DCA}{\sin\angle ACB\,\sin\angle ADC}, \quad P_0 = 180^\circ - \angle ADB$$ which gives $$\tan x = \frac{\sin P_0 - K\sin S}{\cos P_0 - K\cos S}$$
6. With $x$ known, find $y = S - x$, then $AD$, $BC$ (from triangle ABC: $BC = \dfrac{c\sin q}{\sin\angle ACB}$, $q = 180^\circ - \angle ACB - y$).
7. Bearing of $AD$ = bearing of $AB \pm x$; bearing of $BC$ = bearing of $BA \mp y$ (according to the side of AB on which C and D lie). Then $E = E_A + AD\sin\theta,\; N = N_A + AD\cos\theta$, and similarly for C from B.
8. Check: the distance CD computed from the coordinates must agree with the distance found from triangle ADC.