LSIT Mathematics for Surveying 5 — Questions and Answers
Question 1: The tangent length (T) of a simple circular curve with radius R = 600 ft and central angle Δ = 38°20' is closest to:
- 208.8 ft (Correct answer)
- 196.4 ft
- 214.5 ft
- 202.3 ft
Correct answer: 208.8 ft
T = R × tan(Δ/2) = 600 × tan(19°10') = 600 × 0.34730 ≈ 208.8 ft.
Question 2: A surveyor uses a total station to measure an azimuth of 214°33' to a target. What is the equivalent bearing?
- S 34°33' W (Correct answer)
- S 34°33' E
- N 34°33' W
- S 55°27' W
Correct answer: S 34°33' W
Azimuths between 180° and 270° are in the SW quadrant; bearing = azimuth − 180° = 214°33' − 180° = S 34°33' W.
Question 3: In differential leveling, a backsight of 8.213 ft is taken on a benchmark (elev = 504.127 ft) and a foresight of 2.645 ft is taken on the next point. What is the elevation of the next point?
- 509.695 ft (Correct answer)
- 495.559 ft
- 506.772 ft
- 513.940 ft
Correct answer: 509.695 ft
HI = 504.127 + 8.213 = 512.340 ft; Elev = 512.340 − 2.645 = 509.695 ft.
Question 4: A rectangular parcel measures 330.0 ft × 660.0 ft. How many acres does it contain?
- 5.0 acres (Correct answer)
- 4.5 acres
- 10.0 acres
- 2.5 acres
Correct answer: 5.0 acres
Area = 330.0 × 660.0 = 217,800 sq ft; 217,800 / 43,560 = 5.0 acres.
Question 5: Using the Pythagorean theorem, if the northing difference between two points is 315.62 ft and the easting difference is 248.17 ft, the horizontal distance is approximately:
- 401.5 ft (Correct answer)
- 385.3 ft
- 412.7 ft
- 395.8 ft
Correct answer: 401.5 ft
Distance = √(315.62² + 248.17²) = √(99,616 + 61,588) = √161,204 ≈ 401.5 ft.
Question 6: A surveyor distributes a closure error of 0.96 ft over a 4-course traverse using the compass (Bowditch) rule. Course 1 has a length of 400 ft and the total traverse length is 1,600 ft. What correction is applied to Course 1's latitude?
- 0.24 ft (Correct answer)
- 0.12 ft
- 0.48 ft
- 0.06 ft
Correct answer: 0.24 ft
Correction = (course length / total length) × total error = (400/1600) × 0.96 = 0.25 × 0.96 = 0.24 ft.
Question 7: A transit-stadia measurement gives a rod intercept of 1.240 ft with a stadia constant K = 100 and C = 1.0. What is the horizontal distance (for a horizontal sight)?
- 125.0 ft (Correct answer)
- 124.0 ft
- 123.4 ft
- 126.0 ft
Correct answer: 125.0 ft
Horizontal distance = K × s + C = 100 × 1.240 + 1.0 = 124.0 + 1.0 = 125.0 ft.
LSIT Mathematics for Surveying 5
The tangent length (T) of a simple circular curve with radius R = 600 ft and central angle Δ = 38°20' is closest to: