Original question written by FE Exam AI Prep from the public FE Civil exam specification. Before release, a solver model re-derived the answer twice from the question alone, without its answer key, and a critic model checked the result, and the arithmetic was re-run as an executable calculation. No licensed engineer reviews these questions. It is not an NCEES question and does not come from any NCEES practice exam. FE Exam AI Prep is an independent study tool and is not affiliated with, endorsed by, or sponsored by NCEES.
Slope Distance to Horizontal Distance: FE Civil practice problem 1
For the Clear Lake utility relocation corridor, a design checker preparing a capacity verification worksheet reduces a slope distance of 385 ft measured at vertical angle 7 degrees. What horizontal distance is obtained?
- A423.16 ft
- B470.58 ft
- C546.29 ft
- D382.13 ft
Show the answer and worked solution
Answer: D — 382.13 ft
Worked solution
- Step 1. Use horizontal projection
H = S cos theta
H = 385 cos(7 deg) = 382.13 ft
Why the other choices are wrong
- Choice A
- This figure runs about 11% above the correct value, and swapping to sine, using the complementary angle (90 - 7 deg), or leaving 7 deg unconverted to radians all fail to reproduce it from H = 385 * cos(7 deg); the horizontal distance still equals 385 * cos(7 deg) = 382.13 ft.
- Choice B
- This total sits about 23% above the correct value, and no single sine-for-cosine swap, complementary-angle slip, or degree-radian mix-up reproduces it from H = 385 * cos(7 deg); the correct horizontal distance is simply 385 * cos(7 deg) = 382.13 ft.
- Choice C
- This result sits about 43% above the correct value, far beyond what a cosine-sine mix-up, a degree-radian slip, or a complementary-angle substitution could generate from H = 385 * cos(7 deg); the correct horizontal distance is 385 * cos(7 deg) = 382.13 ft.