Derive three point analytical resection

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.

         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.