Mean angles; VCR and VA from A

Assumption: closed pentagon ABCDEA, so the sum of the interior angles is $(5-2) \times 180^\circ = 540^\circ$.

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 = $(5-2) \times 180^\circ$ = 540°00′00″. Observed sum = 539°57′40″. Angular error = 539°57′40″ - 540°00′00″ = -0°02′20″. Correction per angle = +0°00′28″ (equal for all 5 angles).

Answer: corrected horizontal angles: A = 128°47′38″, B = 102°06′48″, C = 108°52′58″, D = 91°00′28″, E = 109°12′08″.

VCR and vertical angles from A

For a zenith-angle vertical circle, FL + FR = 360°.

Answer: VCR (FR) to E = 294°49′30″; VCR (FL) to B = 62°34′20″; VA to E = 24°49′30″ (elevation); VA to B = 27°25′40″ (elevation).