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Question 1
ప్రశ్న 1
For establishing control along a proposed water-main route, survey stations A–B–C–D are connected by measured lengths and bearings, but D is not tied to any independently known position. Which description and limitation are correct?
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Explanation:
A traverse consists of connected lines whose lengths and directions establish successive station positions. This route does not return to A and does not finish on an independently known control point. It is therefore an open traverse. Its calculated terminal coordinates cannot be compared with a separately established terminal position. Check observations, additional ties or independent control would improve reliability. Merely measuring each course does not make the traverse geometrically closed.
వివరణ:
Question 2
ప్రశ్న 2
A waterworks survey starts at control station P with known coordinates and ends at a DIFFERENT station Q whose coordinates have already been independently established. The intermediate course lengths and bearings are observed. How should the survey be classified for checking purposes?
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Explanation:
A loop traverse returns to its starting station. A connecting or link traverse instead joins two independently known control stations. The known coordinates of P and Q provide a terminal comparison. The sum of course departures can be checked against the known easting difference. The sum of course latitudes can likewise be checked against the northing difference. A different terminal station does not prevent a traverse from providing coordinate-closure checks.
వివరణ:
Question 3
ప్రశ్న 3
In traverse A–B–C, the fore whole-circle bearings of AB and BC are 070° and 145° respectively, both referred to a common meridian. What is the SMALLER included angle ABC at B?
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Explanation:
The included angle ABC is formed by the rays BA and BC meeting at B. The fore bearing of AB is 070°, so the back direction BA is 250°. The direction BC is 145° from the same north reference. The absolute difference between 250° and 145° is 105°. This difference is already less than 180°, so it is the smaller included angle. Subtracting the two fore bearings directly would give the traverse's change of direction instead.
వివరణ:
Question 4
ప్రశ్న 4
The fore whole-circle bearing of AB is 300° and that of BC is 040°, referred to the same meridian. What is the SMALLER included angle ABC?
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Explanation:
The angle at B must be measured between the reverse direction BA and the forward direction BC. The back bearing of AB is 300° − 180° = 120°. The forward bearing of BC is 040°. The difference between the two rays is 120° − 40° = 80°. Thus the smaller included angle is 080°. Directly comparing the fore bearings of AB and BC without reversing AB would give the wrong vertex angle.
వివరణ:
Question 5
ప్రశ్న 5
A compass traverse proceeds along AB at a whole-circle bearing of 065°. At B it makes a RIGHT deflection of 55° relative to the forward prolongation of AB. What is the fore bearing of BC, assuming a common meridian and no local attraction?
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Explanation:
A right deflection is a clockwise change from the forward prolongation of the incoming course. The incoming direction AB has bearing 065°. The new course BC is therefore obtained by adding 55° to that bearing. The result is 120°. It lies in the southeast quadrant as expected after the stated turn. Reversing AB first would instead compute an angle referenced to the ray BA rather than the forward prolongation.
వివరణ:
Question 6
ప్రశ్న 6
At station B, the fore bearing of incoming line AB is 035°. The horizontal angle is measured CLOCKWISE from ray BA to outgoing ray BC as 115°. What is the whole-circle bearing of BC?
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Explanation:
The direction BA is opposite to AB, so its bearing is 035° + 180° = 215°. The observed angle starts from the ray BA, not the forward prolongation of AB. Turning clockwise through 115° from BA gives 215° + 115° = 330°. The result is already within the whole-circle range. It represents a northwest-directed outgoing course. Subtracting the angle or adding it directly to AB would use the wrong reference ray.
వివరణ:
Question 7
ప్రశ్న 7
The measured interior angles of a four-sided closed compass traverse are 92°, 87°, 93° and 90°. If all four angle measurements are assumed equally reliable and equal angular corrections are adopted, what correction is required for EACH angle?
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Explanation:
For a four-sided simple closed polygon, the theoretical interior-angle sum is (4 − 2) × 180° = 360°. The observed angles total 92° + 87° + 93° + 90° = 362°. Therefore the angular misclosure is an excess of 2°. The required aggregate correction is −2°. With four equally reliable angles receiving equal corrections, each adjustment is −2°/4 = −0°30′. Actual adjustment should occur only after comparing the misclosure with the applicable survey tolerance.
వివరణ:
Question 8
ప్రశ్న 8
In a five-sided simple closed traverse, four measured interior angles are 109°, 122°, 96° and 118°. If the polygon's interior angles are geometrically consistent, what is its FIFTH interior angle?
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Explanation:
The interior angles of a simple five-sided polygon add to (5 − 2) × 180° = 540°. The four specified angles sum to 109° + 122° + 96° + 118° = 445°. The missing interior angle is therefore 540° − 445° = 95°. This is a geometrical angular-consistency requirement for a closed polygon. It does not by itself verify the observed course lengths. Coordinate closure must also be assessed when the traverse is surveyed.
వివరణ:
Question 9
ప్రశ్న 9
A horizontal compass-traverse course is 50 m long with reduced bearing N 36°52′ E. For this question take cos 36°52′ = 0.8 and sin 36°52′ = 0.6. What are its signed LATITUDE and DEPARTURE?
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Explanation:
Latitude is the north-south component of the horizontal traverse course. Departure is its east-west component. For an N θ E bearing, both components are positive. Latitude equals 50 × 0.8 = +40 m. Departure equals 50 × 0.6 = +30 m. The same components would be obtained from the cosine and sine of the corresponding whole-circle azimuth.
వివరణ:
Question 10
ప్రశ్న 10
A 100 m horizontal traverse leg has reduced bearing S 53°08′ E. Use cos 53°08′ = 0.6 and sin 53°08′ = 0.8. What are the signed latitude and departure?
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Explanation:
The course points south and east, placing it in the southeast quadrant. Its latitude must therefore be negative and its departure positive. The latitude magnitude is 100 × cos 53°08′ = 60 m. The departure magnitude is 100 × sin 53°08′ = 80 m. Hence latitude is −60 m and departure is +80 m. Using both components as positive would incorrectly place the line in the northeast quadrant.
వివరణ:
Question 11
ప్రశ్న 11
A compass-traverse side has a horizontal length of 75 m and reduced bearing S 36°52′ W. Use cos 36°52′ = 0.8 and sin 36°52′ = 0.6. Which signed rectangular components are correct?
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Explanation:
An S θ W direction lies in the southwest quadrant. The north-south component is southing and is written as a negative latitude. The east-west component is westing and is written as a negative departure. The latitude is −75 × 0.8 = −60 m. The departure is −75 × 0.6 = −45 m. These signs are required for correct accumulation of traverse coordinates.
వివరణ:
Question 12
ప్రశ్న 12
A 50 m horizontal traverse course bears N 53°08′ W. Use cos 53°08′ = 0.6 and sin 53°08′ = 0.8. What are its signed latitude and departure?
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Explanation:
The reduced bearing indicates a northwest-directed course. Northing is positive latitude, whereas westing is negative departure. The latitude magnitude is 50 × cos 53°08′ = 30 m. The departure magnitude is 50 × sin 53°08′ = 40 m. Therefore the latitude is +30 m and departure is −40 m. Reversing the cosine and sine factors would incorrectly interchange the component magnitudes.
వివరణ:
Question 13
ప్రశ్న 13
A compass traverse runs 80 m DUE EAST from a station. Using northing-positive and easting-positive conventions, what are the latitude and departure of this line?
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Explanation:
A due-east line has a whole-circle bearing of 090°. It has no north-south displacement. Its latitude is therefore zero. The entire eighty-metre horizontal length is an eastward displacement. Easting is taken positive, so departure equals +80 m. The corresponding due-west course would instead have a negative eighty-metre departure.
వివరణ:
Question 14
ప్రశ్న 14
A horizontal compass-traverse line has a whole-circle bearing of 126° and length 100.00 m. Use cos 126° = −0.5878 and sin 126° = +0.8090. What are its signed latitude and departure to the stated precision?
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Explanation:
A whole-circle bearing of 126° lies in the southeast quadrant. Latitude is obtained as horizontal length multiplied by the cosine of the azimuth. The supplied cosine is negative, giving 100 × (−0.5878) = −58.78 m. Departure is the length multiplied by the sine, giving 100 × 0.8090 = +80.90 m. The resulting signs correctly represent southing and easting. No additional manual quadrant sign should be applied after using signed trigonometric values.
వివరణ:
Question 15
ప్రశ్న 15
Survey station A has grid coordinates Easting = 1200 m and Northing = 850 m. Traverse leg AB is 50 m long at bearing N 36°52′ E; use cos = 0.8 and sin = 0.6 for that bearing. What are the coordinates of B?
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Explanation:
The traverse direction N 36°52′ E produces a positive latitude and a positive departure. The latitude equals 50 × 0.8 = +40 m. The departure equals 50 × 0.6 = +30 m. Easting at B is the starting easting plus departure, 1200 + 30 = 1230 m. Northing at B is the starting northing plus latitude, 850 + 40 = 890 m. Coordinates must be updated on their corresponding east-west and north-south axes rather than interchanged.
వివరణ:
Question 16
ప్రశ్న 16
For a traverse line directed from A to B, the stations have coordinates A = (E 100 m, N 200 m) and B = (E 40 m, N 120 m). What are the length and reduced bearing of AB? Use the 3:4:5 triangle relationship and give the bearing to the nearest minute.
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Explanation:
The easting change from A to B is 40 − 100 = −60 m. The northing change is 120 − 200 = −80 m. Thus B lies west and south of A. The distance is sqrt(60² + 80²) = 100 m. The reduced angle measured from south satisfies tan θ = 60/80 = 0.75, giving approximately 36°52′. Hence the bearing is S 36°52′ W.
వివరణ:
Question 17
ప్రశ్న 17
An OPEN traverse starts at A = (E 0, N 0) and follows 50 m due EAST, then 30 m due NORTH, then 20 m due WEST. What straight line, stated as length and reduced bearing from the TERMINAL station back to A, would close it geometrically? Use sqrt(2) = 1.4142.
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Explanation:
The first course adds fifty metres of easting. The second adds thirty metres of northing, and the third subtracts twenty metres of easting. The terminal station is therefore (E 30 m, N 30 m) relative to A. A direct return to A requires thirty metres west and thirty metres south. Its length is sqrt(30² + 30²) = 30sqrt(2) ≈ 42.43 m. Its reduced bearing is S 45° W, rather than the northeast bearing of the outward displacement.
వివరణ:
Question 18
ప్రశ్న 18
A proposed rectangular closed traverse travels 40 m due EAST from A to B, then 30 m due NORTH to C, then 40 m due WEST to D. What course DA is needed to return exactly to A?
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Explanation:
The first and third courses cancel their east-west components. The traverse therefore has zero remaining departure after the third course. Its second course contributes thirty metres of positive latitude. The last course must contribute thirty metres of negative latitude to return to A. A due-south direction has whole-circle bearing 180°. Consequently DA must be 30 m due south.
వివరణ:
Question 19
ప్రశ్న 19
The signed sums from a closed-loop traverse's UNADJUSTED course components are Σlatitude = +0.30 m and Σdeparture = −0.40 m. What are the magnitude of its linear misclosure and the approximate reduced bearing of the CLOSING vector from the calculated endpoint back to the start? Use the 3:4:5 ratio and 53°08′ for the associated acute angle.
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Explanation:
The algebraic component sums indicate that the calculated endpoint is 0.30 m north and 0.40 m west of the starting point. The linear misclosure magnitude is sqrt(0.30² + 0.40²) = 0.50 m. Returning to the starting point requires the opposite components, 0.30 m south and 0.40 m east. The required closing direction therefore lies in the southeast quadrant. Its acute bearing angle from south is arctan(0.40/0.30) ≈ 53°08′. The closing vector is 0.50 m at S 53°08′ E, opposite to the residual-displacement direction.
వివరణ:
Question 20
ప్రశ్న 20
A closed-loop compass traverse has a total measured horizontal perimeter of 1500 m and an unadjusted linear misclosure of 0.30 m. Under the conventional ratio of total traverse length to linear misclosure, what is its relative precision?
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Explanation:
Relative precision of a traverse is commonly expressed as one part in N. Here N is the total measured traverse perimeter divided by the linear misclosure. The provided values give N = 1500/0.30 = 5000. Therefore the relative precision is written as 1 in 5000. This ratio characterizes positional closure rather than proving that every individual course is free of error. Compliance still depends on the accuracy requirement adopted for the particular survey.
వివరణ:
Question 21
ప్రశ్న 21
For an unadjusted closed traverse, the sum of north latitudes is 75 m and the sum of south latitudes is 65 m. The sum of east departures is 120 m and the sum of west departures is 132 m. What is the LINEAR misclosure magnitude? Give the answer to two decimal places.
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Explanation:
The net latitude is +75 − 65 = +10 m, indicating a residual northing. The net departure is +120 − 132 = −12 m, indicating a residual westing. Linear misclosure is the magnitude of the residual displacement vector. It is calculated as sqrt(10² + (−12)²) = sqrt(244). The resulting magnitude is approximately 15.62 m. Adding the absolute component residuals instead would give twenty-two metres and is not the vector magnitude.
వివరణ:
Question 22
ప్రశ్న 22
The fore whole-circle bearings of consecutive compass-traverse legs AB and BC are 060° and 165°, referred to the same meridian. What is the SMALLER included angle ABC at B?
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Explanation:
The included angle at B is between the ray BA and the ray BC. The incoming fore bearing AB is 060°. Therefore the back direction BA has whole-circle bearing 240°. The outgoing direction BC has bearing 165°. Their absolute difference is 240° − 165° = 75°. This is the smaller included angle, whereas the difference of the two fore bearings is not the angle formed by BA and BC.
వివరణ:
Question 23
ప్రశ్న 23
At one compass station B, observed whole-circle bearings of BA and BC are 260° and 150°. A later check establishes a CONSTANT +4° additive magnetic bearing bias affecting BOTH observations at B. What is the corrected SMALLER included angle ABC?
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Explanation:
Both bearings were observed from the same station under an unchanged additive magnetic bias. Subtracting 4° from each yields corrected BA = 256° and corrected BC = 146°. Their difference is 110°. The uncorrected difference of 260° − 150° is also 110° because the common bias cancels. This is the smaller included angle and needs no additional 180° reversal because BA is already the backward-directed ray. The cancellation does not generally hold when comparing observed bearings taken at differently affected stations.
వివరణ:
Question 24
ప్రశ్న 24
A closed-loop traverse contains three measured courses with rectangular components (latitude, departure) in metres: AB = (+30, +40), BC = (−20, +30), and CD = (−10, −50). The final course DA must close the traverse EXACTLY. What are the required direction and length of DA?
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Explanation:
The first three latitudes sum to +30 − 20 − 10 = 0 m. Their departures sum to +40 + 30 − 50 = +20 m. To close the loop, the last course must make both total component sums zero. Hence DA needs zero latitude and a departure of −20 m. Zero latitude with negative departure describes a due-west horizontal line. Its necessary length is twenty metres.
వివరణ:
Question 25
ప్రశ్న 25
A closed compass traverse has algebraic sums of adjusted latitudes and departures equal to zero, but some independently observed check distances disagree with the resulting coordinates. Which engineering assessment is MOST defensible?
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Explanation:
The algebraic component closure conditions test consistency of the net traverse displacement. They do not individually establish the correctness of every observed direction and length. Different errors can compensate to produce small or even zero residual sums. An adjustment can also force component closure while leaving problematic observations undetected. Independent check lines and control ties supply additional evidence of accuracy. Disagreements beyond tolerance should therefore be investigated rather than concealed by the closure result.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. It is an open traverse; there is no independent end-position closure check unless further control or check observations are added. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. A connecting or link traverse, for which computed coordinate change can be compared with the known P-to-Q coordinate difference. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 105°. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 080°. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 120°. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 330°. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: C. −0°30′. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: A. 095°. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. Latitude +40 m; departure +30 m. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. Latitude −60 m; departure +80 m. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: C. Latitude −60 m; departure −45 m. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. Latitude +30 m; departure −40 m. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. Latitude 0 m; departure +80 m. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: B. Latitude −58.78 m; departure +80.90 m. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. Easting 1230 m; northing 890 m. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. 100 m, S 36°52′ W. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: D. 42.43 m, S 45° W. సమాధానం: D.
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Question 18 ప్రశ్న 18Answer: B. 30 m due SOUTH, whole-circle bearing 180°. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. 0.50 m, S 53°08′ E. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. 1 in 5000. సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: D. 15.62 m. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: B. 075°. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. 110°. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. 20 m due WEST. సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. Coordinate closure alone does not prove that individual observations are correct; investigate independent check discrepancies and possible compensating or systematic errors. సమాధానం: D.