Three point analytical resection, average coordinates of O

Angles at O

Readings: P = 90°00'00", W = 230°58'51", D = 313°17'05", P (closing) = 90°00'30".

• $\angle POW = 140^\circ58'51''$, $\angle WOD = 82^\circ18'14''$, $\angle DOP$ from the first and the closing reading of P = $136^\circ42'55''$ and $136^\circ43'25''$.

The closing reading differs by 30", so the third angle can be taken in two ways. Each pair of observed angles (the third being 360° minus the sum) gives a solution by Tienstra's formula, and the average of the three solutions is taken.

Tienstra's formula. Let the known stations be $S_1, S_2, S_3$ with interior angles $A, B, C$ of the triangle $S_1S_2S_3$ at them. Let $\alpha, \beta, \gamma$ be the observed angles at P subtended by the opposite sides: $\alpha = \angle S_2PS_3$, $\beta = \angle S_3PS_1$, $\gamma = \angle S_1PS_2$ ($\alpha+\beta+\gamma = 360^\circ$). Then

$$K_1 = \frac{1}{\cot A - \cot\alpha},\quad K_2 = \frac{1}{\cot B - \cot\beta},\quad K_3 = \frac{1}{\cot C - \cot\gamma}$$ $$E_P = \frac{K_1E_1 + K_2E_2 + K_3E_3}{K_1+K_2+K_3}, \qquad N_P = \frac{K_1N_1 + K_2N_2 + K_3N_3}{K_1+K_2+K_3}$$

Detailed solution for set 1 ($\angle POW = 140^\circ58'51''$, $\angle WOD = 82^\circ18'14''$, $\angle DOP = 136^\circ42'55''$)

Known stations: P (2876.240 E, 8754.110 N), W (3810.800 E, 7997.250 N), D (2959.390 E, 7487.090 N).

Sides and angles of triangle PWD from the coordinates: $P = 47°14'34"$, $W = 69°55'56"$, $D = 62°49'30"$.

$K_P = 1.26664$, $K_W = 0.70075$, $K_D = 0.57227$ giving O = (3152.842 E, 8259.775 N).

Three solutions

Average

$$E_O = \frac{3152.842 + 3152.793 + 3152.777}{3} = 3152.804\ \text{m},\qquad N_O = \frac{8259.775 + 8259.754 + 8259.837}{3} = 8259.788\ \text{m}$$

Answer: average coordinates of O = (3152.804 E, 8259.788 N).