Length, bearing of EF between midpoints

The positions of E (midpoint of CD) and F (midpoint of AB) are found from the coordinates of A, B, C and D. The traverse is first computed and balanced, and EF is then found from the coordinates of its two end points. A is taken as the origin (1000 N, 1000 E) to keep the coordinates positive.

Step 1: Latitudes and departures

The sums are not zero, so the traverse has a closing error of $e=\sqrt{(-25.512)^2+(-277.378)^2}=278.55$ m on a perimeter of 2700 m. The error is very large (about 1 in 10), so the field data should be rechecked; for the computation it is distributed by Bowditch's rule so that the traverse closes.

Step 2: Adjusted lines and coordinates

Step 3: Midpoints

$$\begin{aligned} E&=\left(\frac{N_C+N_D}{2},\frac{E_C+E_D}{2}\right)=(1520.535,\ 1332.168)\\ F&=\left(\frac{N_A+N_B}{2},\frac{E_A+E_B}{2}\right)=(1054.971,\ 1331.853) \end{aligned}$$

Step 4: Length and bearing of EF

$$\begin{aligned} \Delta N&=N_F-N_E=-465.565,\qquad \Delta E=E_F-E_E=-0.316\\ EF&=\sqrt{\Delta N^2+\Delta E^2}=465.56\ \text{m}\\ \theta_{EF}&=\tan^{-1}\frac{-0.316}{-465.565}=180^\circ 02' \end{aligned}$$

Both differences are negative, so EF points almost due south (S 0°02' W), whole circle bearing 180°02'.

Answer: EF = 465.56 m, bearing 180°02' (S 0°02' W). (If the unbalanced values of AB, BC and CD are used directly, i.e. EF taken as the line closing F-B-C-E, then EF = 474.12 m at 162°38'.)