Mean angles, traverse ABCDA (2074 Asoj)

Assumption: closed traverse ABCDA; the angle sum must be 360°.

Method. For each target the mean direction is found from the two faces:

$$\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

Field book and mean angles

Check and adjustment

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 369°23′20″. Angular error = 369°23′20″ - 360°00′00″ = +9°23′20″. Correction per angle = -2°20′50″ (equal for all 4 angles).

Answer: corrected horizontal angles: A = 112°05′00″, B = 98°15′00″, C = 76°48′15″, D = 72°51′45″.

Note on the data: the angular error computed from the readings as printed is about 9°23′, far larger than any acceptable error for a four-sided traverse (a few tens of seconds). So there must be a recording or printing mistake in the readings (for example in 204°25′40″ or 190°36′10″). In the field, such an error would mean the angles must be re-observed rather than adjusted. The equal-correction adjustment above is shown only to demonstrate the method on the values as printed; with correct data the same steps give a small error that is adjusted in the same way.