Define intersection and resection, derive

Definitions

Intersection is the method of finding the coordinates of an unknown point P by measuring the horizontal angles to P from two (or more) stations of known coordinates. The instrument is set up at the known stations, not at P.

Resection is the method of finding the coordinates of an unknown station P by measuring the horizontal angles at P to three (or two) stations of known coordinates. The instrument is set up only at the unknown point.

Three-point resection (Pothenot's tangent method)

Problem: A, B and C are known stations (B is the middle one). The instrument is at the unknown station P, and the angles $\alpha = \angle APB$ and $\beta = \angle BPC$ are observed. Find the coordinates of 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.