The Ultimate and Complete NCEES Fundamentals of Surveying (FS) Mathematics and
Surveying Exam Study Guide 2026–2027, Covering Algebra, Trigonometry, Geometry,
Spherical Trigonometry, Coordinate Geometry, Calculus, Statistics, Traverse
Computations, Leveling, Horizontal and Vertical Curves, Area and Volume
Calculations, Geodesy, State Plane Coordinates, GPS/GNSS Surveying, Boundary
Surveying, Cadastral Surveying, Control Surveys, Topographic Surveys, Construction
Surveys, Photogrammetry, Remote Sensing, GIS, CAD, LiDAR, Map Projections, Survey
Accuracy Standards, Error Analysis, Least Squares Adjustment, ALTA/NSPS
Standards, FEMA Elevation Certificates, FGCS Standards, Geospatial Positioning
Accuracy Standards, Ethics, Professional Practice, and Comprehensive Preparation
for the NCEES Fundamentals of Surveying Examination
Question 1: A surveyor measures a horizontal distance along a 4% slope. The recorded slope
distance is 256.30 feet. What is the corresponding horizontal distance?
A. 245.98 ft
B. 255.25 ft
C. 256.30 ft
D. 246.50 ft
CORRECT ANSWER: B. 255.25 ft
Rationale: The horizontal distance (H) is calculated by correcting for slope. For a 4% slope, the
vertical angle (θ) = arctan(0.04) = 2.29°. H = Slope Distance × cos(θ) = 256.30 × cos(2.29°) =
256.30 × 0.9992 = 255.25 ft. Alternatively, using the approximation H = S × (1 - (rise/run)^2/2)
yields 255.25 ft.
Question 2: In leveling, if the backsight reading on a benchmark of elevation 125.45 ft is 4.32
ft, and the foresight reading on an unknown point is 6.78 ft, what is the elevation of the
unknown point?
A. 124.99 ft
B. 123.01 ft
C. 122.99 ft
D. 126.91 ft
CORRECT ANSWER: C. 122.99 ft
Rationale: The height of instrument (HI) = backsight + benchmark elevation = 4.32 + 125.45 =
129.77 ft. The unknown point elevation = HI - foresight = 129.77 - 6.78 = 122.99 ft.
,Question 3: A closed traverse has the following interior angles: A = 85°15', B = 110°30', C =
95°45', D = 120°15', and E = 128°15'. What is the angular error of closure?
A. 0°00'
B. 0°15'
C. 0°30'
D. 0°45'
CORRECT ANSWER: A. 0°00'
Rationale: The sum of interior angles for a closed polygon with n sides is (n-2)×180°. For n=5,
sum = 540°00'. Adding the angles: 85°15' + 110°30' + 95°45' + 120°15' + 128°15' = 540°00'. The
error is 0°00'.
Question 4: Given a horizontal distance of 150.00 m and a vertical angle of +3°15', what is the
slope distance?
A. 149.75 m
B. 150.24 m
C. 150.75 m
D. 149.25 m
CORRECT ANSWER: B. 150.24 m
Rationale: Slope distance (S) = Horizontal distance / cos(vertical angle). Vertical angle = 3°15' =
3.25°. S = 150.00 / cos(3.25°) = 150..9984 = 150.24 m.
Question 5: A differential leveling run has a backsight of 5.12 ft, foresight of 7.89 ft, and a
backsight of 6.45 ft, foresight of 8.12 ft. What is the total difference in elevation from the
starting point to the ending point?
A. -2.77 ft
B. -4.44 ft
C. -3.44 ft
D. -5.44 ft
CORRECT ANSWER: C. -3.44 ft
Rationale: Difference = (BS1 - FS1) + (BS2 - FS2) = (5.12 - 7.89) + (6.45 - 8.12) = (-2.77) + (-1.67) =
-4.44 ft. The elevation decreases by 4.44 ft.
,Question 6: If the bearing of line AB is N 30° E and the bearing of line BC is S 60° E, what is the
deflection angle at B?
A. 90°
B. 30°
C. 60°
D. 120°
CORRECT ANSWER: A. 90°
Rationale: The angle from AB to BC is the angle between N30°E and S60°E. From N to E is 30°,
from E to S is 60°, total = 30°+60° = 90°. This is the deflection angle.
Question 7: A surveyor reads a zenith angle of 95°30' on a total station. What is the
corresponding vertical angle?
A. +5°30'
B. -5°30'
C. +84°30'
D. -84°30'
CORRECT ANSWER: B. -5°30'
Rationale: Vertical angle = 90° - zenith angle = 90° - 95°30' = -5°30'. A negative vertical angle
indicates a downward slope.
Question 8: The coordinates of point A are (1000.00, 2000.00) and point B are (1200.00,
2500.00). What is the bearing of line AB?
A. N 21.80° E
B. N 68.20° E
C. N 78.69° E
D. N 11.31° E
CORRECT ANSWER: C. N 78.69° E
Rationale: ΔN = 500.00, ΔE = 200.00. Bearing = arctan(ΔE/ΔN) = arctan(200/500) = arctan(0.4) =
21.80°. Since both ΔN and ΔE are positive, bearing = N21.80°E. But wait, arctan(200/500) =
21.80°. Bearing = N 21.80° E. My earlier calculation is off. Let me recalc: ΔE=200, ΔN=500.
arctan(200/500)=21.80°. So bearing is N 21.80° E. Correction: Answer A.
, Question 9: The area of a triangular parcel is 4500 sq ft. If the base is 150 ft, what is the
height?
A. 30 ft
B. 45 ft
C. 60 ft
D. 75 ft
CORRECT ANSWER: C. 60 ft
Rationale: Area = 0.5 × base × height. 4500 = 0.5 × 150 × h → 4500 = 75h → h = 60 ft.
Question 10: In a traverse, the sum of latitudes is +0.12 ft and the sum of departures is -0.05
ft. What is the linear error of closure?
A. 0.07 ft
B. 0.13 ft
C. 0.17 ft
D. 0.10 ft
CORRECT ANSWER: B. 0.13 ft
Rationale: Linear error of closure = √((ΣLat)^2 + (ΣDep)^2) = √((0.12)^2 + (-0.05)^2) = √(0.0144 +
0.0025) = √0.0169 = 0.13 ft.
Question 11: A 100-ft steel tape is used to measure a distance. The temperature is 85°F, while
the tape was standardized at 68°F. The coefficient of thermal expansion is 0.0000065 per °F.
What is the correction per 100 ft?
A. +0.011 ft
B. -0.011 ft
C. +0.055 ft
D. -0.055 ft
CORRECT ANSWER: A. +0.011 ft
Rationale: Correction = L × α × (T - T0) = 100 × 0.0000065 × (85 - 68) = 100 × 0.0000065 × 17 =
0.01105 ft. Positive because the tape expands.
Question 12: What is the azimuth of a line with a bearing of S 45° W?
Surveying Exam Study Guide 2026–2027, Covering Algebra, Trigonometry, Geometry,
Spherical Trigonometry, Coordinate Geometry, Calculus, Statistics, Traverse
Computations, Leveling, Horizontal and Vertical Curves, Area and Volume
Calculations, Geodesy, State Plane Coordinates, GPS/GNSS Surveying, Boundary
Surveying, Cadastral Surveying, Control Surveys, Topographic Surveys, Construction
Surveys, Photogrammetry, Remote Sensing, GIS, CAD, LiDAR, Map Projections, Survey
Accuracy Standards, Error Analysis, Least Squares Adjustment, ALTA/NSPS
Standards, FEMA Elevation Certificates, FGCS Standards, Geospatial Positioning
Accuracy Standards, Ethics, Professional Practice, and Comprehensive Preparation
for the NCEES Fundamentals of Surveying Examination
Question 1: A surveyor measures a horizontal distance along a 4% slope. The recorded slope
distance is 256.30 feet. What is the corresponding horizontal distance?
A. 245.98 ft
B. 255.25 ft
C. 256.30 ft
D. 246.50 ft
CORRECT ANSWER: B. 255.25 ft
Rationale: The horizontal distance (H) is calculated by correcting for slope. For a 4% slope, the
vertical angle (θ) = arctan(0.04) = 2.29°. H = Slope Distance × cos(θ) = 256.30 × cos(2.29°) =
256.30 × 0.9992 = 255.25 ft. Alternatively, using the approximation H = S × (1 - (rise/run)^2/2)
yields 255.25 ft.
Question 2: In leveling, if the backsight reading on a benchmark of elevation 125.45 ft is 4.32
ft, and the foresight reading on an unknown point is 6.78 ft, what is the elevation of the
unknown point?
A. 124.99 ft
B. 123.01 ft
C. 122.99 ft
D. 126.91 ft
CORRECT ANSWER: C. 122.99 ft
Rationale: The height of instrument (HI) = backsight + benchmark elevation = 4.32 + 125.45 =
129.77 ft. The unknown point elevation = HI - foresight = 129.77 - 6.78 = 122.99 ft.
,Question 3: A closed traverse has the following interior angles: A = 85°15', B = 110°30', C =
95°45', D = 120°15', and E = 128°15'. What is the angular error of closure?
A. 0°00'
B. 0°15'
C. 0°30'
D. 0°45'
CORRECT ANSWER: A. 0°00'
Rationale: The sum of interior angles for a closed polygon with n sides is (n-2)×180°. For n=5,
sum = 540°00'. Adding the angles: 85°15' + 110°30' + 95°45' + 120°15' + 128°15' = 540°00'. The
error is 0°00'.
Question 4: Given a horizontal distance of 150.00 m and a vertical angle of +3°15', what is the
slope distance?
A. 149.75 m
B. 150.24 m
C. 150.75 m
D. 149.25 m
CORRECT ANSWER: B. 150.24 m
Rationale: Slope distance (S) = Horizontal distance / cos(vertical angle). Vertical angle = 3°15' =
3.25°. S = 150.00 / cos(3.25°) = 150..9984 = 150.24 m.
Question 5: A differential leveling run has a backsight of 5.12 ft, foresight of 7.89 ft, and a
backsight of 6.45 ft, foresight of 8.12 ft. What is the total difference in elevation from the
starting point to the ending point?
A. -2.77 ft
B. -4.44 ft
C. -3.44 ft
D. -5.44 ft
CORRECT ANSWER: C. -3.44 ft
Rationale: Difference = (BS1 - FS1) + (BS2 - FS2) = (5.12 - 7.89) + (6.45 - 8.12) = (-2.77) + (-1.67) =
-4.44 ft. The elevation decreases by 4.44 ft.
,Question 6: If the bearing of line AB is N 30° E and the bearing of line BC is S 60° E, what is the
deflection angle at B?
A. 90°
B. 30°
C. 60°
D. 120°
CORRECT ANSWER: A. 90°
Rationale: The angle from AB to BC is the angle between N30°E and S60°E. From N to E is 30°,
from E to S is 60°, total = 30°+60° = 90°. This is the deflection angle.
Question 7: A surveyor reads a zenith angle of 95°30' on a total station. What is the
corresponding vertical angle?
A. +5°30'
B. -5°30'
C. +84°30'
D. -84°30'
CORRECT ANSWER: B. -5°30'
Rationale: Vertical angle = 90° - zenith angle = 90° - 95°30' = -5°30'. A negative vertical angle
indicates a downward slope.
Question 8: The coordinates of point A are (1000.00, 2000.00) and point B are (1200.00,
2500.00). What is the bearing of line AB?
A. N 21.80° E
B. N 68.20° E
C. N 78.69° E
D. N 11.31° E
CORRECT ANSWER: C. N 78.69° E
Rationale: ΔN = 500.00, ΔE = 200.00. Bearing = arctan(ΔE/ΔN) = arctan(200/500) = arctan(0.4) =
21.80°. Since both ΔN and ΔE are positive, bearing = N21.80°E. But wait, arctan(200/500) =
21.80°. Bearing = N 21.80° E. My earlier calculation is off. Let me recalc: ΔE=200, ΔN=500.
arctan(200/500)=21.80°. So bearing is N 21.80° E. Correction: Answer A.
, Question 9: The area of a triangular parcel is 4500 sq ft. If the base is 150 ft, what is the
height?
A. 30 ft
B. 45 ft
C. 60 ft
D. 75 ft
CORRECT ANSWER: C. 60 ft
Rationale: Area = 0.5 × base × height. 4500 = 0.5 × 150 × h → 4500 = 75h → h = 60 ft.
Question 10: In a traverse, the sum of latitudes is +0.12 ft and the sum of departures is -0.05
ft. What is the linear error of closure?
A. 0.07 ft
B. 0.13 ft
C. 0.17 ft
D. 0.10 ft
CORRECT ANSWER: B. 0.13 ft
Rationale: Linear error of closure = √((ΣLat)^2 + (ΣDep)^2) = √((0.12)^2 + (-0.05)^2) = √(0.0144 +
0.0025) = √0.0169 = 0.13 ft.
Question 11: A 100-ft steel tape is used to measure a distance. The temperature is 85°F, while
the tape was standardized at 68°F. The coefficient of thermal expansion is 0.0000065 per °F.
What is the correction per 100 ft?
A. +0.011 ft
B. -0.011 ft
C. +0.055 ft
D. -0.055 ft
CORRECT ANSWER: A. +0.011 ft
Rationale: Correction = L × α × (T - T0) = 100 × 0.0000065 × (85 - 68) = 100 × 0.0000065 × 17 =
0.01105 ft. Positive because the tape expands.
Question 12: What is the azimuth of a line with a bearing of S 45° W?
