This page has some problem sets for you to practice on. It is good to use the NCEES Reference handbook when doing these.
LSIT Math 1
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Question 1 |
A traverse was run from point A to Point E and the coordinates were computed with the following results
The distance and bearing, respectively, of a straight line from Point A to Point E are most nearly:
A | 517.06 ft, S09°54’E |
B | 517.06 ft, S80° 06’ E |
C | 521.65 ft, S 12° 23’ E |
D | 521.65 ft, S 77° 37’ E |
Question 1 Explanation:
This is a standard inverse question. Remember that Y values are Northings and that you are going south from A to E
Question 2 |
What is the clockwise (angle right) angle Set on Point “A”, Backsighting Point “B” and turning to the northwest corner of the pad as shown?
A | 7° 22’ 09” |
B | 18° 26’ 06” |
C | 22° 37’ 12” |
D | 341° 33’ 54” |
E | 352° 37’ 50” |
Question 2 Explanation:
Create a Right triangle from A to NW corner (25 / 75) and use inverse tangent to determine counterclockwise angle. Subtract from 360°
Question 3 |
Two lines are defined by the following equations:
2x + y -6 = 0 4x - 3y +8 = 0The lines intersect at:
A | x = 0, y = 0 |
B | x = ½, y = 5 |
C | x = 1, y = 4 |
D | x = 2, y = 2 |
E | x = 3, y = 3 |
Question 3 Explanation:
This is linear algebra with two unknowns and two equations. You are looking to determine what x an y value will work for both equations. An easy way to do this is to rearrange the first equation to y = 6-2x and then replace y in the second equation so you have 4x-18+6x+8=0 and solve for x
Question 4 |
You are set up on a building “A” with a theodolite (HI = 5.00 feet). You measure a zenith angle to the top of a Building “B” of 62° 12’ 50” and a zenith angle to the bottom of Building “B” of 144° 32’ 25”. Building “B” is 845 feet tall. How far is building “A” from building “B”
A | 840.0 ft |
B | 746.9 ft |
C | 614.4 ft |
D | 609.4 ft |
E | 437.6 ft |
Question 4 Explanation:
Solve for the angle from top to bottom, then determine the distance from instrument to the top or bottom using law of sines. Then solve for right triangle. Don't forget about the instrument height!
Question 5 |
The value of the Products of 53.4 x 0.0023, using the correct significant figure is:
A | 0.12282 |
B | 0.1228 |
C | 0.123 |
D | 0.12 |
E | 0.1 |
Question 5 Explanation:
Significant figures when multiplying or dividing uses the least significant count. In this case the least significant is 0.0023 which has only 2 significant figures (the 0's are place holders.)
Question 6 |
One corner of a 60-ft x 120-ft lot, otherwise rectangular, is a curve with a radius of 20 ft. and a central angle of 90°. The area (sq. ft.) of the lot is most nearly:
A | 6,872 |
B | 6,886 |
C | 7,114 |
D | 7,200 |
Question 6 Explanation:
There are a number of ways to determine the area. An easy way is to determine the area of the lot without the curve (7200 sq ft) then subtract a 20 x 20 square leaving 6800 sq ft. Then determine 1/4 the area of a 20' radius curve (314.1 sq ft) and add that back in.
Question 7 |
Given:
10x + 5y = 235 21x + 9y = 405Solve for Y
A | -36 |
B | -6 |
C | 15 |
D | 45 |
E | 59 |
Question 7 Explanation:
There are a number of ways to solve using linear algebra. An easy way is to rearrange the first equation so that 10x = 235 - 5y; then x = 23.5 - 0.5y. Replace x in the second equation so that you have 21(23.5-0.5y) + 9 y = 405. THis can be reduced to 1.5y = 88.5
Question 8 |
On the lot shown, determine the buildable area which is delineated by the dashed line. (Note all dimensions are in feet).
A | 1.40 acres |
B | 1.77 acres |
C | 1.80 acres |
D | 1.84 acres |
E | 2.29 acres |
Question 8 Explanation:
You can create a rectangle and triangle to solve for the area. The only problem is determining the height and base of the triangle. It just takes some trig.
Question 9 |
Assuming the earth is a sphere and its radius is 5,631 miles, the latitude of a point that is 2,700 miles north of the equator is most nearly:
A | 21°47'26”N |
B | 27°28'22"N |
C | 21°36',46”N |
D | 21°22'03” N |
Question 9 Explanation:
Using the radius you can determine a circumference of 35,380.6 miles. Then set up a ratio of 360°/35,380.6 = X/2700. Don't forget to convert the answer to degrees minutes seconds
Question 10 |
A theodolite is used to sight on a 1 inch diameter range pole at a distance of 172 ft. If the edge of the range pole is sighted instead of the center, the angular error introduced in the direction of the line is most nearly:
A | 35” |
B | 50” |
C | 70" |
D | 100" |
Question 10 Explanation:
The angular error would approximately equal the inverse tangent of how far off you were (0.5 inches) divided by the distance from the object (172 feet)
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