12 multiple-choice questions, progressively harder.
Stations F, N, and tower base B lie in that order on one horizontal line. Both instruments are 1.5 m above this ground and sight the same top above them in the shared vertical plane. FN is 12.0 m, far elevation is 30.0 degrees, and near elevation is 45.0 degrees. Find NB to the nearest tenth.
Solution
Correct answer: A
Let the near run be xxx; the far run is x+12x+12x+12. Both give the same rise above the equal eye level.
Stations F, N, and the base of a vertical tower lie in that order on one horizontal line. The instruments are each 1.6 m high, separation FN is 22.0 m, and their elevations to the same top in a vertical plane are 24.0 degrees at F and 48.0 degrees at N. Find whole tower height to the nearest tenth.
Correct answer: C
Use a shared rise uuu above instrument level. Dividing it by each tangent gives the corresponding run; the far run exceeds the near run by 22 m.
Two stations A and C stand on opposite sides of vertical tower base B, with A, B, C in that order on a straight horizontal line. AC is 52.0 m. Both instruments are 1.5 m above ground, and elevations to the same top are 35.0 degrees at A and 55.0 degrees at C in one vertical plane. Find tower height to the nearest tenth.
Correct answer: B
Here the two ground runs add to 52 m. Let uuu be the top’s rise above the common instrument level.
On one straight horizontal line toward a vertical tower, an observer moves from horizontal run 60.0 m to run 30.0 m while keeping the same eye elevation. The top’s elevation was 20.0 degrees at the far point and remains above the eye. What is the near elevation, to the nearest tenth of a degree?
Correct answer: D
The same vertical rise is divided by half the run, so its tangent ratio doubles.
From one eye above a vertical tower's top, the top and base are seen at depressions 16.0∘16.0^\circ16.0∘ and 38.0∘38.0^\circ38.0∘ respectively. They share a horizontal run of 26.026.026.0 m and a vertical plane. Find the tower height HHH to the nearest tenth.
Both endpoints are below eye level. Subtract the smaller drop to the top from the larger drop to the base.
Two stations are 20 m apart and have equal eye elevations. Both sight a tower top, but nobody establishes that their station line passes through its base. Is it justified to write the far horizontal run as the near run plus 20 m?
Call the near horizontal run xxx and the proposed far run rrr. The claimed equation would be
Equal eye elevations can give a shared vertical rise, but they do not establish this horizontal station order. The equation needs collinearity through the base in the required order.
A vertical object and observer share level ground and a vertical sight plane. Horizontal run is fixed at 50.0 m and eye height at 1.6 m. The elevation angle is known only to lie from 29.5 to 30.5 degrees. What is the largest modeled whole height, to the nearest tenth?
For fixed positive run, raising the sight line within this acute range increases the vertical rise. Use the upper angle endpoint.
A straight road has a 12 percent gradient and horizontal run 80.0 m. Find the sloped road length to the nearest tenth of a meter.
The gradient gives a rise of 0.12 times horizontal run.
Two straight ramps have equal positive surface lengths. Their inclinations from horizontal are 5 degrees and 9 degrees. Which has the longer horizontal run?
Let LLL be the common surface length, with rises r5r_5r5 and r9r_9r9 and runs d5d_5d5 and d9d_9d9. The steeper ramp has larger rise. Pythagoras leaves a smaller horizontal run.
Thus the 5-degree ramp has the longer run.
An observer and vertical mast stand on the same level ground. Eye height is 1.5 m and horizontal run is 24.0 m. The angle from the upward vertical to the rising sight line is 61.0 degrees in their vertical plane. Find the whole mast height to the nearest tenth.
The vertical and horizontal references are perpendicular. The elevation is the complement of the given angle.
Two instruments at different horizontal runs have the same eye elevation relative to a target base. Their angles and runs give exactly the same rise above their eyes. The recorded common eye elevation may contain an unknown vertical offset. What has their agreement established?
Both observations test the rise uuu from their common eye level to the top. Whole height also uses the eye elevation eee.
An error in the shared eye elevation shifts both whole-height estimates by the same amount, without changing their checked rise.
Stations F, N, and vertical tower base B are in that order on one horizontal line. FN is 10 m. Instruments have equal eye heights and sight the same top above them in the common vertical plane. A record gives far elevation 40 degrees and near elevation 35 degrees. Is the record compatible with these conditions?
Let the near run be positive xxx and the common rise be positive uuu. The far run is x+10x+10x+10, so
The far acute elevation must therefore be smaller, contrary to the recorded far 40 degrees and near 35 degrees.
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