Why is the angle at Q left out of the correction?

It comes from the correct bearings of PQ and QR, so it is already exact

It is the smallest angle of the traverse

It was measured with a better compass

Corrections are only applied to acute angles

With least count 30′, what is the permissible angular error for the four observed angles, and how does the 1°00′ error compare?

30′ x sqrt(4) = 60′; the error is just permissible

30′ x 4 = 120′; the error is well within it

30′ x sqrt(5) = 67′; the error is exceeded

30′; the error is exceeded and the work must be repeated

2080 Bhadra · 8 marks

Corrected bearings of closed traverse PQRST

Short answer

Interior angle sum error +1°00′ distributed equally (−15′) to R, S, T, P. Corrected bearings: RS 300°15′, ST 80°30′, TP 300°45′ (check on PQ closes at 180°00′).

Calculate the corrected bearings of legs RS, ST and TP of closed traverse PQRST where the correct bearings of legs PQ and QR are 180°00′ and 90°30′ respectively. The included angles observed at stations R, S, T and P are 30°, 320°30′, 40°30′ and 59°30′ respectively. Take least count of prismatic compass 0°30′.

The bearings of PQ (180°00′) and QR (90°30′) are correct, so the angle at Q is known exactly and the other included angles are corrected to make the total $(2n-4)\times 90^\circ$ .

Step 1: Angle at Q and angle sum

At Q, the back bearing of PQ is QP $= 180^\circ + 180^\circ = 0^\circ$ (i.e. 360°00′) and the bearing of QR is 90°30′, so the included angle at Q $= 90^\circ 30'$ (the traverse runs anticlockwise, with the interior on the left).

Station	P	Q	R	S	T
Observed angle	59°30′	90°30′ (from correct bearings)	30°00′	320°30′	40°30′

Observed sum $= 59°30′ + 90°30′ + 30°00′ + 320°30′ + 40°30′ = 541°00′$ .

Required sum $= (2 \times 5 - 4) \times 90^\circ = 540°00′$ .

Error $= +1°00′$ (excess). The angle at Q is already fixed by correct bearings, so the error is distributed equally over the other four angles: $-60'/4 = -15'$ each. (Least count 30′: the permissible error is $30'\sqrt{4} = 60'$ , so the error is just permissible.)

Station	Observed	Correction	Corrected
R	30°00′	−0°15′	29°45′
S	320°30′	−0°15′	320°15′
T	40°30′	−0°15′	40°15′
P	59°30′	−0°15′	59°15′

New sum $= 59°15′ + 90°30′ + 29°45′ + 320°15′ + 40°15′ = 540°00′$ (check).

Step 2: Corrected bearings

For this anticlockwise traverse, the FB of the next line = BB of the previous line + included angle at the station between them.

Line	Working	Corrected FB
PQ	given	180°00′
QR	given	90°30′
RS	(90°30′ + 180°) + 29°45′ = 300°15′	300°15′
ST	(300°15′ − 180°) + 320°15′ = 440°30′ − 360°	80°30′
TP	(80°30′ + 180°) + 40°15′	300°45′
PQ (check)	(300°45′ − 180°) + 59°15′	180°00′ ✓

Answer: RS = 300°15′, ST = 80°30′, TP = 300°45′ (their back bearings are 120°15′, 260°30′ and 120°45′). The check on PQ closes exactly at 180°00′.

Assumption: the included angles are interior angles of an anticlockwise traverse, the angle at Q (90°30′) being taken from the correct bearings.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Corrected bearings of closed traverse PQRST

Step 1: Angle at Q and angle sum

Step 2: Corrected bearings

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