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 3
For the Clear Lake utility relocation corridor, a design checker preparing a design validation memo reduces a slope distance of 385 ft measured at vertical angle 3 degrees. What horizontal distance is obtained?
- A384.47 ft
- B236.69 ft
- C286.53 ft
- D423.65 ft
Show the answer and worked solution
Answer: A — 384.47 ft
Worked solution
- Step 1. Use horizontal projection
H = S cos theta
H = 385 cos(3 deg) = 384.47 ft
Why the other choices are wrong
- Choice B
- This figure runs about 38% under S*cos(theta) = 385*cos(3 deg) = 384.47 ft, a gap that does not match using the complement of the vertical angle or a degree-radian mix-up; multiplying the slope distance by the cosine of the given vertical angle is the only operation the reduction calls for.
- Choice C
- This distance falls about 25% short of S*cos(theta) = 385*cos(3 deg) = 384.47 ft, more than substituting the tangent or the sine of the vertical angle would produce; the horizontal length equals the slope distance multiplied by the cosine of the stated vertical angle, nothing else.
- Choice D
- This horizontal distance sits about 10% above S*cos(theta) = 385*cos(3 deg) = 384.47 ft, well beyond what mixing up sine and cosine or misreading the vertical angle would cause; the slope distance times the cosine of the stated angle gives the reduced horizontal length directly.