Link traverse A to X, Y by transit method

Coordinates are listed as E, N in the question; in the working below they are written as (N, E). The reference bearings come from the coordinates of the control stations:

$$\begin{aligned} \theta_{AB}&=\tan^{-1}\frac{E_B-E_A}{N_B-N_A}=\tan^{-1}\frac{-711.87}{+397.67}\ \Rightarrow\ 299^\circ 11'20''\\ \theta_{XY}&=\tan^{-1}\frac{E_Y-E_X}{N_Y-N_X}=\tan^{-1}\frac{+661.00}{-131.14}\ \Rightarrow\ 101^\circ 13'18'' \end{aligned}$$

A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station.

Known data: A (2094.72 N, 1769.15 E), B (2492.39 N, 1057.28 E), X (1747.32 N, 2334.85 E), Y (1616.18 N, 2995.85 E).

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line AB (299°11'20" from the coordinates) and using the observed angles, the bearing of the last line comes out as 101°11'10", while its known value is 101°13'18" (X-Y from the coordinates).

Angular misclosure = computed − known = 101°11'10" − 101°13'18" = -0°02'08".

The error is spread equally over the 6 angles: correction per angle = +0°00'21". The k-th bearing therefore receives k times this correction.

Step 2: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Starting from A (2094.720 N, 1769.150 E), the unadjusted coordinates of X are (1747.249 N, 2334.614 E), whereas the known coordinates are (1747.320 N, 2334.850 E).

Step 3: Linear misclosure

$$\begin{aligned} e_N&=1747.249-1747.320=-0.071\ \text{m}\\ e_E&=2334.614-2334.850=-0.236\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.247\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.247}{1262.380}=\frac{1}{5121} \end{aligned}$$

Step 4: Adjustment by the transit rule

The transit rule: the correction to latitude (or departure) of each line is proportional to the size of that latitude (or departure):

$$C_{lat}=-e_N\frac{|\text{lat}|}{\Sigma|\text{lat}|},\qquad C_{dep}=-e_E\frac{|\text{dep}|}{\Sigma|\text{dep}|}$$

Step 5: Final coordinates

The computed coordinates of X now equal its known coordinates, so the traverse is balanced.

Answer: 1 (2214.743 N, 1939.400 E); 2 (2355.934 N, 2071.721 E); 3 (2090.461 N, 2262.173 E); 4 (1815.654 N, 2120.640 E) (linear misclosure 0.247 m, 1 in 5121).