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Horizontal angles from field notes, triangle ABC

Short answer

Mean angles from two-face readings: A = 30°00′05″, B = 59°59′55″, C = 90°00′02.5″. Sum 180°00′02.5″, excess +2.5″; equal correction -0.83″ gives A = 30°00′04.17″, B = 59°59′54.17″, C = 90°00′01.67″.

The following are field notes of theodolite survey. Compute horizontal angles and apply check.

StationSighted toDirect (D M S)Reverse (D M S)
AC0 00 10180 00 00
B30 00 20210 00 00
BA0 00 00180 00 20
C60 00 00240 00 10
CB0 00 10180 00 00
A90 00 05270 00 10

Check. The three stations A, B, C form a triangle, so the sum of the three angles must be 180°.

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

Mean direction=Face L+(Face R±180∘)2\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 nn sides, the sum of the interior angles must be (n−2)×180∘(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

Inst. stnSighted toFace LFace RMean directionMean angle
AC0 00 10180 00 000°00′05″
B30 00 20210 00 0030°00′10″30°00′05″
BA0 00 00180 00 200°00′10″
C60 00 00240 00 1060°00′05″59°59′55″
CB0 00 10180 00 000°00′05″
A90 00 05270 00 1090°00′07.5″90°00′02.5″

Check and adjustment

Required sum = (3−2)×180∘(3-2) \times 180^\circ = 180°00′00″. Observed sum = 180°00′02.5″. Angular error = 180°00′02.5″ - 180°00′00″ = +0°00′02.5″. Correction per angle = -0°00′00.83″ (equal for all 3 angles).

Angle atMean angleCorrectionCorrected angle
A30°00′05″-0°00′00.83″30°00′04.17″
B59°59′55″-0°00′00.83″59°59′54.17″
C90°00′02.5″-0°00′00.83″90°00′01.67″
Sum180°00′02.5″-0°00′02.5″180°00′00″

Answer: corrected horizontal angles: A = 30°00′04.17″, B = 59°59′54.17″, C = 90°00′01.67″.

✓ Answer checked by Claude Opus 5.5 (high effort)

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