Resection point O from stations A, B, C

The three-point problem is solved by Tienstra's method.

Observed angles at O

Readings: A = 90°00'00", B = 230°58'51", C = 313°17'05", A (closing) = 90°00'30".

• $\angle AOB = 140^\circ58'51''$, $\angle BOC = 82^\circ18'14''$, $\angle COA = 136^\circ43'25''$ (sum 360°00'30")
• The closing error of 30" is shared equally (−10" each): $\angle AOB = 140^\circ58'41''$, $\angle BOC = 82^\circ18'04''$, $\angle COA = 136^\circ43'15''$.

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

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

Data

Angles at O: $\angle AOB = \gamma = 140°58'41"$, $\angle BOC = \alpha = 82°18'04"$, $\angle COA = \beta = 136°43'15"$ (sum = 360°00'00").

Step 1: Angles of the triangle ABC from the coordinates

Interior angles: $A$ (at A) $= 47°14'34"$, $B$ (at B) $= 69°55'56"$, $C$ (at C) $= 62°49'30"$ (sum 180°00'00").

Step 2: Tienstra constants

$$\begin{aligned} K_1 &= \frac{1}{\cot 47°14' - \cot 82°18'04"} = 1.26672 \\ K_2 &= \frac{1}{\cot 69°55' - \cot 136°43'15"} = 0.70065 \\ K_3 &= \frac{1}{\cot 62°49' - \cot 140°58'41"} = 0.57231 \end{aligned}$$

$\Sigma K = 2.53967$

Step 3: Coordinates of O

$$E_O = \frac{(1.26672)(2876.240) + (0.70065)(3810.800) + (0.57231)(2959.390)}{2.53967} = 3152.804\ \text{m}$$ $$N_O = \frac{(1.26672)(8754.110) + (0.70065)(7997.250) + (0.57231)(7487.090)}{2.53967} = 8259.788\ \text{m}$$

Check: the angles computed back from these coordinates are 140°58'41", 82°18'04" and 136°43'15", equal to the observed angles.

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