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NV Land Surveyor practice questions

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Question 1Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A closed traverse has these departures and latitudes (in meters): AB: +100.00, +50.00; BC: +20.00, +90.00; CA: -120.00, -140.00. What is the linear misclosure of the traverse?

  1. 20.00 m
  2. 40.00 m
  3. 10.00 m
  4. 0.00 m Correct answer.

Explanation

Sum the departures: +100.00 + 20.00 - 120.00 = 0.00. Sum the latitudes: +50.00 + 90.00 - 140.00 = 0.00. Linear misclosure is the square root of the sum of the squared departure and latitude misclosures, which is zero here.

Question 2Surveying fundamentals (official NCEES FS content domains not enumerated in source)

In a total station, which measurement does the instrument make directly to determine the slope distance to a prism?

  1. The barometric pressure difference
  2. The phase shift or time of flight of a modulated signal Correct answer.
  3. The vertical angle only
  4. The horizontal angle between two prisms

Explanation

Electronic distance measurement determines slope distance by measuring either the phase difference of a modulated light or microwave signal or the travel time of a pulse. The vertical angle is measured separately by the theodolite portion and is combined with slope distance to compute horizontal and vertical components.

Question 3Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A leveling rod reading of 1.245 m is taken on a benchmark whose elevation is 120.500 m. The instrument height is therefore:

  1. 119.255 m
  2. 120.500 m
  3. 121.745 m Correct answer.
  4. 1.245 m

Explanation

Height of instrument equals benchmark elevation plus the backsight rod reading: 120.500 + 1.245 = 121.745 m. This instrument height is then used to reduce subsequent foresight readings to ground elevations.

Question 4Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A bearing of N 45° E is equivalent to which azimuth measured clockwise from north?

  1. 135°
  2. 45° Correct answer.
  3. 315°
  4. 225°

Explanation

Azimuths are measured clockwise from north, so a northeast bearing in the first quadrant equals its angle from north directly. N 45° E therefore corresponds to an azimuth of 45°.

Question 5Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A rectangular parcel measures 250.00 ft by 400.00 ft. What is its area in acres (1 acre = 43,560 ft²)?

  1. 2.30 acres Correct answer.
  2. 10.00 acres
  3. 0.43 acres
  4. 1.00 acre

Explanation

The area is 250.00 × 400.00 = 100,000 ft². Dividing by 43,560 ft² per acre gives approximately 2.296 acres, which rounds to 2.30 acres.

Question 6Surveying fundamentals (official NCEES FS content domains not enumerated in source)

Which reference surface is used as the basis for horizontal geodetic coordinates such as latitude and longitude?

  1. The mean sea level surface
  2. The reference ellipsoid Correct answer.
  3. The geoid
  4. The local tangent plane

Explanation

Latitude and longitude are defined on a reference ellipsoid, a mathematically defined surface that approximates the size and shape of the Earth. The geoid is the equipotential surface used for vertical datums.

Question 7Surveying fundamentals (official NCEES FS content domains not enumerated in source)

The primary purpose of a closed traverse that returns to its starting point is to:

  1. Provide a check on both angular and linear misclosure Correct answer.
  2. Eliminate the need for angular measurement
  3. Remove the effect of refraction from distance measurement
  4. Guarantee that all angles are exactly 90 degrees

Explanation

A closed traverse returns to a known point, so the sum of computed departures and latitudes must equal zero and the angular sum must match the theoretical value. These conditions reveal both angular and linear misclosure, allowing adjustment.

Question 8Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A total station reads a zenith angle of 90° 00' 00" to a target. The resulting vertical component of the sight line is:

  1. Zero Correct answer.
  2. Equal to the horizontal distance
  3. Half the slope distance
  4. Equal to the slope distance

Explanation

A zenith angle of 90° means the line of sight is horizontal, since zenith is measured from the vertical axis. The vertical component is slope distance times the cosine of the zenith angle, which is zero when the angle is 90°.

Question 9Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A traverse leg has a bearing of S 30° W and a length of 200.00 m. What are the departure and latitude of this leg?

  1. Departure +173.21 m, latitude -100.00 m
  2. Departure -100.00 m, latitude -173.21 m Correct answer.
  3. Departure -173.21 m, latitude -100.00 m
  4. Departure +100.00 m, latitude +173.21 m

Explanation

Departure equals length times the sine of the bearing angle: 200.00 × sin 30° = 100.00 m, negative because the direction is west. Latitude equals length times the cosine: 200.00 × cos 30° = 173.21 m, negative because the direction is south.

Question 10Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A differential leveling run of 2.4 km has a misclosure of 0.012 m. Using the common allowance of 8 mm times the square root of the distance in kilometers, is the run acceptable?

  1. No, because the allowance is about 0.0040 m
  2. Yes, because the allowance is about 0.0192 m
  3. Yes, because the allowance is about 0.0124 m Correct answer.
  4. No, because the allowance is about 0.0080 m

Explanation

The allowance is 8 mm × √2.4 = 8 × 1.549 = 12.4 mm, or about 0.0124 m. Since the observed misclosure of 0.012 m is less than the allowance, the run is acceptable.

Question 11Surveying fundamentals (official NCEES FS content domains not enumerated in source)

In a horizontal angle measurement by the repetition method, the angle is turned six times and the accumulated reading is 542° 17' 30". What is the mean angle?

  1. 45° 11' 27"
  2. 108° 27' 30"
  3. 90° 00' 00"
  4. 90° 22' 55" Correct answer.

Explanation

Divide the accumulated angle by the number of repetitions: 542° 17' 30" ÷ 6 = 90° 22' 55". Repetition averages out graduation and pointing errors, improving the precision of the result.

Question 12Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A horizontal curve has a radius of 500.00 ft and a central angle of 40°. What is the length of the curve?

  1. 349.07 ft Correct answer.
  2. 200.00 ft
  3. 436.33 ft
  4. 872.66 ft

Explanation

Arc length equals radius times the central angle in radians: 500.00 × (40 × π/180) = 500.00 × 0.6981 = 349.07 ft. The central angle must be converted from degrees to radians before multiplying.

Question 13Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A distance measured along a slope of 5° is 300.000 m. What is the horizontal distance?

  1. 26.145 m
  2. 298.858 m Correct answer.
  3. 300.000 m
  4. 299.886 m

Explanation

Horizontal distance equals slope distance times the cosine of the slope angle: 300.000 × cos 5° = 300.000 × 0.99619 = 298.858 m. The slope correction is the small difference between slope and horizontal distance.

Question 14Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A map has a scale of 1:24,000. A line measures 3.50 inches on the map. What is the corresponding ground distance in miles (1 mile = 63,360 inches)?

  1. 1.33 miles Correct answer.
  2. 2.10 miles
  3. 13.3 miles
  4. 0.53 miles

Explanation

Ground distance in inches is 3.50 × 24,000 = 84,000 inches. Dividing by 63,360 inches per mile gives 1.326 miles, which rounds to 1.33 miles.

Question 15Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A backsight reading of 1.500 m is taken on a point of elevation 100.000 m, and a foresight of 2.250 m is then read on a second point. What is the elevation of the second point?

  1. 101.750 m
  2. 99.250 m Correct answer.
  3. 100.750 m
  4. 98.750 m

Explanation

Height of instrument is 100.000 + 1.500 = 101.500 m. Subtracting the foresight gives the new elevation: 101.500 - 2.250 = 99.250 m. A larger foresight than backsight indicates a lower elevation.

Question 16Surveying fundamentals (official NCEES FS content domains not enumerated in source)

The magnetic declination at a site is 7° E. A magnetic bearing of N 20° E is observed. What is the true bearing?

  1. N 20° E
  2. N 13° E
  3. S 27° W
  4. N 27° E Correct answer.

Explanation

With east declination, magnetic north lies east of true north, so the declination is added to the magnetic bearing to obtain the true bearing. N 20° E + 7° = N 27° E.

Question 17Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A closed traverse's angular misclosure is distributed using the compass (Bowditch) rule. This method adjusts departures and latitudes proportionally to:

  1. The number of instrument setups
  2. The elevations of the stations
  3. The traverse leg lengths Correct answer.
  4. The interior angles at each station

Explanation

The compass rule distributes the linear misclosure in proportion to each leg's length relative to the total traverse length. Longer legs receive larger corrections because random errors accumulate with distance.

Question 18Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A vertical curve has a 400-ft length, an entering grade of +3.0%, and an exiting grade of -1.0%. What is the elevation of the curve at the midpoint relative to the PVI elevation?

  1. 1.00 ft below the PVI
  2. 4.00 ft below the PVI
  3. 2.00 ft above the PVI
  4. 1.00 ft above the PVI Correct answer.

Explanation

The algebraic difference in grades is -4.0%, and the external distance is L × |A| / 800 = 400 × 4 / 800 = 2.00 ft. At the midpoint the curve lies above the PVI by half the external distance, or 1.00 ft.

Question 19Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A total station is set on a point and a backsight azimuth of 120° is established. The instrument is then turned to a foresight with a horizontal angle right of 45° from the backsight. What azimuth is displayed for the foresight?

  1. 165° Correct answer.
  2. 120°
  3. 75°
  4. 45°

Explanation

Azimuths increase clockwise, so a clockwise angle right is added to the backsight azimuth. Adding 45° to 120° gives a foresight azimuth of 165°.

Question 20Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A boundary described as running "thence N 90° E" along a line is best interpreted as running:

  1. Due south
  2. Due north
  3. Due east Correct answer.
  4. Due west

Explanation

A bearing of N 90° E means the line is rotated 90° clockwise from north, which places it due east. Bearings of exactly 0° or 90° are degenerate quadrant values that correspond to cardinal directions.

Question 21Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A parcel boundary has these coordinates (N, E) in feet: A (1000.000, 1000.000); B (1000.000, 1300.000); C (700.000, 1300.000); D (700.000, 1000.000). Using the coordinate (shoelace) method, what is the area?

  1. 120,000 ft²
  2. 60,000 ft²
  3. 90,000 ft² Correct answer.
  4. 45,000 ft²

Explanation

The coordinates form a 300 ft by 300 ft square, so the area is 300 × 300 = 90,000 ft². The shoelace formula, summing cross products of successive coordinates, yields the same result.

Question 22Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A traverse has a total length of 2,000.00 m and a linear misclosure of 0.100 m. What is the relative precision?

  1. 1:200
  2. 1:200,000
  3. 1:2,000
  4. 1:20,000 Correct answer.

Explanation

Relative precision is the total traverse length divided by the linear misclosure: 2,000.00 / 0.100 = 20,000, expressed as 1:20,000. A larger second term indicates a more precise traverse.

Question 23Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A horizontal curve has a degree of curve of 4°00' by the arc definition. What is its radius, using the arc definition with a 100-ft arc?

  1. 716.20 ft
  2. 1,432.39 ft Correct answer.
  3. 2,864.79 ft
  4. 1,250.00 ft

Explanation

For the arc definition, radius equals 5729.58 divided by the degree of curve: 5729.58 / 4 = 1,432.39 ft. The constant 5729.58 derives from 180 × 100 / π.

Question 24Surveying fundamentals (official NCEES FS content domains not enumerated in source)

Two points have geodetic coordinates on a reference ellipsoid, and a surveyor computes a grid distance from the ellipsoidal distance. Which statement correctly describes the relationship between grid distance and ground distance?

  1. Grid distance is always longer than the ellipsoidal distance
  2. Grid distance differs from ground distance because of both elevation above the ellipsoid and the map projection scale factor Correct answer.
  3. Grid distance always equals ground distance measured at the surface
  4. Grid distance accounts only for curvature of the Earth, not elevation

Explanation

Converting ground distance to grid distance requires reducing to the ellipsoid using elevation (the combined factor) and then applying the projection scale factor. Both effects must be considered, and either can lengthen or shorten the result depending on location and elevation.

Question 25Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A leveling instrument has a collimation error such that the line of sight rises 0.004 m per 30 m of sight. Backsight and foresight distances are 30 m and 90 m respectively. If the true elevation difference is computed from the readings, the error in the computed difference is:

  1. 0.004 m
  2. 0.016 m
  3. 0.012 m
  4. 0.008 m Correct answer.

Explanation

The error in a rod reading grows with sight distance: the backsight is in error by 0.004 m and the foresight by 0.012 m. The elevation difference error is the difference of these, 0.012 - 0.004 = 0.008 m, which is why balancing sight distances eliminates the effect.

Question 26Surveying fundamentals (official NCEES FS content domains not enumerated in source)

In a state plane coordinate system based on the transverse Mercator projection, the central meridian is best described as the meridian along which:

  1. the scale factor equals the zone's specified central-meridian value, usually less than 1, and grid north equals true north Correct answer.
  2. the scale factor reaches its maximum value and distortion is greatest
  3. grid azimuths differ from geodetic azimuths by a nonzero convergence angle
  4. the projection cylinder is tangent to the ellipsoid at the equator

Explanation

Along the central meridian, meridian convergence is zero, so grid north coincides with true north. The scale factor equals the zone's designated central-meridian value, k₀, which state plane transverse Mercator systems generally set below unity to distribute scale distortion across the zone.

Question 27Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor measures the same baseline five times and obtains 499.992, 500.004, 499.998, 500.001, and 500.005 m. The standard deviation of this single measurement, computed with n-1 in the denominator, is closest to:

  1. 0.0050 m Correct answer.
  2. 0.0104 m
  3. 0.0047 m
  4. 0.0021 m

Explanation

The mean is 500.000 m, and the squared residuals are 64, 16, 4, 1, and 25 (in units of 10^-6 m^2), summing to 110 x 10^-6. Dividing by n-1 = 4 gives 27.5 x 10^-6, and the square root is about 0.00524 m, closest to 0.0050 m. Using n instead of n-1 would understate the scatter slightly.

Question 28Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A survey crew must establish horizontal control over a heavily wooded 40-acre tract where GPS signals are blocked by canopy and no intervisible line exceeds 150 m. Which equipment and procedure is most appropriate for the primary control?

  1. A total station running a closed-loop traverse with angular and linear misclosure adjusted by least squares Correct answer.
  2. A handheld recreational GPS receiver averaging positions for ten minutes at each corner
  3. An elevation-grade barometer read at each corner and referenced to a nearby benchmark
  4. A single differential leveling loop run around the property perimeter

Explanation

Horizontal positions require angle and distance observations, and a closed traverse with redundant measurements permits adjustment and a defensible precision statement. A total station works under canopy where satellite signals fail and over the short sight distances described. Leveling and barometry produce elevations only, not horizontal coordinates.

Question 29Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor measures a distance with an EDM and applies a correction for atmospheric conditions. If the actual temperature is lower than the instrument's reference temperature, with pressure and other atmospheric conditions unchanged, the correction applied to the measured distance will:

  1. decrease the distance because the instrument's internal reference assumed warmer air Correct answer.
  2. decrease the distance because the signal travels faster in colder air
  3. depend only on barometric pressure and not on temperature
  4. have no effect because EDM measures distance independent of the atmosphere

Explanation

At unchanged pressure, colder air has a higher refractive index, so the EDM signal travels more slowly than under the warmer reference conditions. Using the reference propagation speed therefore overstates the uncorrected distance. The atmospheric correction decreases the measured distance.

Question 30Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor sets up on a point whose grid coordinate is known and needs to stake a point 250.00 m away on a grid azimuth of 315°. What are the northing and easting changes to the stakeout point?

  1. -176.78 m northing, +176.78 m easting
  2. -125.00 m northing, +216.51 m easting
  3. +250.00 m northing, 0.00 m easting
  4. +176.78 m northing, -176.78 m easting Correct answer.

Explanation

Grid azimuth is measured clockwise from grid north, so 315° lies in the northwest quadrant. The northing change is 250.00 cos 315° = +176.78 m and the easting change is 250.00 sin 315° = -176.78 m. The equal magnitudes follow from the 45° angle with each axis.

Question 31Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A deed written in 1890 calls for a boundary to run to a white oak tree that no longer exists, but the line's bearing and distance are also given and the original stone corner is found at the computed position. Under the hierarchy of evidence commonly applied in boundary retracement, the found stone corner should be:

  1. rejected because the monument is not mentioned in the deed
  2. used only if it agrees with the record bearing
  3. held as controlling over the record bearing and distance Correct answer.
  4. ignored in favor of the deed's stated distance

Explanation

In boundary retracement, natural and artificial monuments called for in the deed generally control over courses and distances, because the monument was the object the parties actually observed. A found original corner therefore prevails even when the record bearing and distance do not match perfectly. This ordering protects the intent of the original survey.

Question 32Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor must convert a geodetic azimuth of 128° 30' to a grid azimuth in a state plane zone where the mapping angle (convergence) at the point is +1° 15'. The grid azimuth is:

  1. 129° 45'
  2. 128° 30'
  3. 127° 15' Correct answer.
  4. 126° 45'

Explanation

The mapping angle is applied as grid azimuth equals geodetic azimuth minus convergence when convergence is positive (east of the central meridian in the northern hemisphere). Subtracting 1° 15' from 128° 30' gives 127° 15'. The sign convention must be verified from the projection documentation before applying it.

Question 33Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor needs to determine the area of a parcel from coordinates using the coordinate (shoelace) method. The computed double meridian distance sum is 2,450,000 ft^2. The parcel area is:

  1. 612,500 ft^2
  2. 4,900,000 ft^2
  3. 1,225,000 ft^2 Correct answer.
  4. 2,450,000 ft^2

Explanation

In the double meridian distance method, the sum of double meridian distance times latitude equals twice the enclosed area. Dividing the accumulated total by two gives the parcel area, so 2,450,000 divided by 2 equals 1,225,000 ft^2. The sign of the total indicates the direction of traverse.

Question 34Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor measures a horizontal angle four times by repetition and obtains a total accumulated angle of 384° 00' 20". What is the most probable value of the single angle?

  1. 96° 00' 05" Correct answer.
  2. 96° 00' 20"
  3. 96° 00' 00"
  4. 128° 00' 07"

Explanation

In the repetition method the accumulated angle is divided by the number of repetitions to obtain the mean angle. Dividing 384° 00' 20" by four gives 96° 00' 05". Repetition reduces the effect of reading errors because the total is read once.

Question 35Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor computes the geoid height (geoid-ellipsoid separation) at a project site as N = -28.5 m and obtains an ellipsoidal height h = 250.0 m from GNSS. The orthometric height H is:

  1. 250.0 m
  2. -28.5 m
  3. 221.5 m
  4. 278.5 m Correct answer.

Explanation

The relationship is h = H + N, so the orthometric height equals the ellipsoidal height minus the geoid separation: H = 250.0 - (-28.5) = 278.5 m. A negative geoid height means the geoid lies below the ellipsoid at that location, which increases the orthometric value.

Question 36Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor is asked to set out a circular curve by the deflection-angle method using a total station located at the PC. The deflection angle to a point on the curve is measured from the:

  1. tangent at the PC Correct answer.
  2. long chord connecting the PC and PT
  3. radial line through the PC
  4. chord drawn from the PT to the point

Explanation

In the deflection-angle method the instrument is set at the PC with the backsight along the back tangent, and each deflection is turned from that tangent to the chord leading to the station being set. The deflection angle to any point equals half the subtended central angle. This makes the method convenient because each stake is set from a single setup.

Question 37Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor measures a slope distance of 400.000 m between two points whose elevation difference is 24.000 m. The horizontal distance between the points is closest to:

  1. 399.998 m
  2. 400.720 m
  3. 399.279 m Correct answer.
  4. 376.000 m

Explanation

The horizontal distance is the square root of the slope distance squared minus the elevation difference squared: sqrt(400.000^2 - 24.000^2) = sqrt(160000 - 576) = sqrt(159424), which is about 399.279 m. The correction is always subtracted from a slope distance.

Question 38Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor must determine the volume of earth between two parallel cross sections using the average end area method. Section 1 has an area of 120 ft^2, section 2 has an area of 180 ft^2, and the distance between them is 50 ft. The volume is:

  1. 15,000 ft^3
  2. 9,000 ft^3
  3. 6,000 ft^3
  4. 7,500 ft^3 Correct answer.

Explanation

The average end area method multiplies the mean of the two end areas by the distance between them: (120 + 180)/2 x 50 = 150 x 50 = 7,500 ft^3. This method assumes a linear variation of area between sections. For irregular terrain the prismoidal formula is preferred.

Question 39Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor observes a star at upper culmination and needs to relate the observation to azimuth. At upper culmination, the star's hour angle is:

  1. 90°
  2. 0° Correct answer.
  3. 180°
  4. equal to its declination

Explanation

Hour angle is measured westward from the observer's meridian, so a star at upper culmination lies on the meridian and its hour angle is zero. At that instant its azimuth is either 0° or 180° depending on whether it culminates north or south of the zenith. Hour angle increases at about 15° per hour thereafter.

Question 40Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor must determine the azimuth of a line with a precision of about 1:50,000 in an area with no existing control and no access to GNSS. Which field method is most suitable?

  1. Resection from two topographic map features
  2. Magnetic compass readings taken at both ends of the line
  3. Gyroscopic theodolite observation of a single sight
  4. Solar observation using the hour angle method with several timed sights Correct answer.

Explanation

The hour angle method of solar observation uses precisely timed altitude or azimuth sights to the sun and a known position to compute true azimuth to a high order of accuracy. Multiple sights reduce random timing and pointing errors, supporting the stated precision. Magnetic methods are limited by declination uncertainty and local attraction.

Question 41Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor records a backsight reading of 1.250 m on a benchmark of elevation 50.000 m, then reads an intermediate foresight of 2.400 m and a turning-point foresight of 3.100 m before moving the instrument. The elevation of the turning point is:

  1. 48.150 m Correct answer.
  2. 47.350 m
  3. 49.400 m
  4. 51.250 m

Explanation

The height of instrument is 50.000 + 1.250 = 51.250 m. The turning point elevation is 51.250 - 3.100 = 48.150 m. The intermediate foresight is used only to compute elevations of intermediate points and does not affect the turning-point elevation.

Question 42Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor prepares a plat and must show the basis of bearings used. If the bearings are referenced to grid north of a state plane zone, the plat should state that bearings are relative to:

  1. true north at the parcel's centroid
  2. the direction of the first line of the traverse
  3. the zone's central meridian as defined by the projection Correct answer.
  4. magnetic north observed at the time of the survey

Explanation

Grid bearings in a state plane system are referenced to grid north, which is parallel to the central meridian of the projection zone throughout the zone. Stating this basis allows another surveyor to reproduce the orientation unambiguously. Magnetic and assumed bases do not provide that reproducibility.

Question 43Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor observes a vertical angle of +12° 00' to a target and a slope distance of 500.000 m. The elevation difference between instrument and target is closest to:

  1. +489.07 m
  2. +106.00 m
  3. +103.95 m Correct answer.
  4. +106.43 m

Explanation

The elevation difference equals the slope distance times the sine of the vertical angle: 500.000 sin 12° = 500.000 x 0.207912 = 103.96 m, closest to 103.95 m. The horizontal distance would be 500.000 cos 12° = 489.07 m. Instrument height and target height would then be applied to obtain the target elevation.

Question 44Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor is asked to explain why a closed traverse should have its angular misclosure distributed before the latitudes and departures are adjusted. The primary reason is that:

  1. the angular adjustment eliminates the need for a linear check
  2. the angular adjustment changes the total traverse length
  3. the linear adjustment assumes the angles already satisfy geometric closure Correct answer.
  4. the compass rule requires balanced latitudes and departures

Explanation

Latitudes and departures are computed from bearings derived from the measured angles, so any angular misclosure propagates into those components. Adjusting angles first ensures the traverse geometry closes in direction before the linear misclosure is distributed. This ordering keeps the linear adjustment meaningful.

Question 45Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor computes the azimuth of a line as 305° 20'. The equivalent bearing is:

  1. N 54° 40' E
  2. S 54° 40' E
  3. N 35° 20' W
  4. N 54° 40' W Correct answer.

Explanation

An azimuth between 270° and 360° lies in the northwest quadrant, and the bearing angle equals 360° minus the azimuth: 360° 00' - 305° 20' = 54° 40'. The bearing is therefore N 54° 40' W. Azimuths are measured clockwise from north, whereas bearings are measured from the nearer north or south end of the meridian.

Question 46Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor must locate the intersection of two straight property lines whose endpoints are known by coordinates. Which computational approach yields the intersection point directly?

  1. Computing the midpoints of each line and connecting them
  2. Averaging the coordinates of the four line endpoints
  3. Solving the two line equations simultaneously for the intersection coordinates Correct answer.
  4. Rotating both lines until they become parallel

Explanation

Each line can be expressed as an equation from its endpoint coordinates, and solving the pair simultaneously gives the coordinates of their common point. This is the standard analytic solution for a line-line intersection. Averaging or midpoint constructions do not locate the crossing.

Question 47Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor is reviewing a topographic survey and must decide whether a digital terrain model is appropriate. Which statement about a TIN surface built from surveyed spot shots and breaklines is correct?

  1. A TIN interpolates elevations only at the exact surveyed points
  2. Breaklines force triangle edges to follow ridges and drainage features, improving surface fidelity Correct answer.
  3. Breaklines are used only to define the outer boundary of the model
  4. Increasing the number of triangles always increases the accuracy of the model

Explanation

Breaklines constrain triangulation so that triangle edges honor significant terrain features such as ridges, tops of banks, and flow lines. Without them the surface can interpolate across a sharp feature and misrepresent the ground. The resulting TIN better represents drainage and slope changes.

Question 48Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor must express the precision of a distance measured as 1,234.567 m with a stated uncertainty of plus or minus 0.005 m. Which statement correctly reports the result?

  1. 1,234.567 m ± 0.005 m Correct answer.
  2. 1,234.5670 m ± 0.005 m
  3. 1,234 m ± 0.005 m
  4. 1,234.57 m ± 0.005 m

Explanation

The reported value should be carried to the same decimal place as the uncertainty, so a 0.005 m uncertainty supports three decimal places. Writing 1,234.567 m with the stated tolerance is consistent. Extra or missing digits misrepresent the actual precision of the measurement.

Question 49Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor is computing the position of a point by resection from three known control points using a total station. The geometry is weak when the unknown point lies:

  1. equidistant from all three known points
  2. inside the triangle formed by the three known points
  3. at the center of the triangle formed by the known points
  4. on or near the circle passing through the three known points Correct answer.

Explanation

Resection becomes indeterminate when the instrument station lies on the circle defined by the three known points, because the observed angles no longer distinguish positions on that circle. This condition is called the dangerous circle. Good practice places the station well off that circle to keep the solution well conditioned.

Question 50Surveying fundamentals (official NCEES FS content domains not enumerated in source)

A surveyor is asked to lay out a grade of 2.0 percent on a pipeline from a point of known elevation. If the instrument height of sight is 1.500 m and the target is 60.0 m away along the line, what rod reading should be set to place the target on grade at the far point?

  1. 1.200 m
  2. 1.800 m
  3. 2.700 m
  4. 0.300 m Correct answer.

Explanation

The grade changes elevation by 0.020 x 60.0 = 1.200 m over the 60 m. Since the far point is 1.200 m higher than the instrument station, and the instrument line of sight is 1.500 m above that station, the rod reading on grade is 1.500 - 1.200 = 0.300 m. A rising grade reduces the required rod reading.