Mean angles, traverse ABCDA (2068 Baisakh)

Assumption: closed traverse ABCDA; the sum of the interior angles 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

Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 369°03′00″. Angular error = 369°03′00″ - 360°00′00″ = +9°03′00″.

An error of about 9° cannot come from observation (a theodolite traverse normally closes within a minute or so), so it must not simply be distributed. It shows a misprint in the data: the readings on B from A are probably 200°25′40″ (face left) and 20°25′30″ (face right), which make every other reading consistent.

Adjustment with the corrected readings (200°25′40″, 20°25′30″)

Mean direction AB = 200°25′35″, so angle at A = 200°25′35″ - 90°00′00″ = 110°25′35″. Observed sum = 110°25′35″ + 90°16′15″ + 89°08′35″ + 70°12′35″ = 360°03′00″. Angular error = +0°03′00″; correction per angle = -0°00′45″.

Answer: corrected horizontal angles: A = 110°24′50″, B = 90°15′30″, C = 89°07′50″, D = 70°11′50″.

If the readings are used exactly as printed: the error of +9°03′00″ shared equally (-2°15′45″ each) would give A = 117°09′50″, B = 88°00′30″, C = 86°52′50″, D = 67°56′50″. This is shown only for completeness; an error this large would mean the observations must be repeated, not adjusted.