# Surveying Solved Problems

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```					Surveying
Solved Problems
for the FS and PS Exams
Third Edition

Jan Van Sickle, PLS

Professional Publications, Inc. • Belmont, CA
1-4     S U R V E Y I N G       S O L V E D        P R O B L E M S   F O R   T H E    F S    A N D         P S      E X A M S

23. No point along a stationed line may be closer than                   26. Two sides of a triangle are 83.40 ft and 95.20 ft, re-
15 ft to an existing building. With the instrument oc-                   spectively. The angle between the two sides is 51◦ 15 00 .
cupying station 0+00 and backsighting station 0+40,                      Which values correctly represent the remaining three el-
an angle of 339◦ 07 44 is turned to the building corner.                 ements of the triangle?
Then, with the instrument occupying station 0+40 and
backsighting station 0+00, an angle of 104◦ 25 09 is                                                                C
turned to the same building corner. What is the mini-
mum distance between the line and the building corner?
a                       b     83.40 ft
0 40
b
10

A    4                                                                                        51 15 00
25 09
B                                                  A
building                                                                                             c       95.20 ft

(A)     77.97   ft,   56◦ 31 56         ,   72◦ 13 04
(B)     77.97   ft,   56◦ 31 56         ,   72◦ 15 07
0 00
339 0

(C)     78.23   ft,   56◦ 31 56         ,   72◦ 13 04
7
44                         (D)     85.46   ft,   49◦ 35 15         ,   79◦ 09 45

27. x is the length of one side of triangle ABC. The
(A)   16.91    ft                                                      altitude of the triangle, CD, is x − 4 ft. It bisects angle
ACB and the base AB as well. AB is equal to x. What
(B)   17.46    ft                                                      is the length of x in ft?
(C)   17.99    ft                                                                                                  C
(D)   20.00    ft

24. A three-sided gore is found between adjoining
tracts. One side of the gore is 471.67 ft, and the second                                          x
side is 450.83 ft. The angle between these two sides is
05◦ 12 40 . What is the length of the third side?                                                          x 4 ft
(A) 20.84 ft
(B) 40.95 ft
(C) 42.84 ft                                                                        A                                                   B
(D) 46.82 ft                                                                                        0.5x                   0.5x
D
(A)     12.42   ft
25. After all measurements for a survey have been com-                       (B)     18.76   ft
pleted, it is important to determine the height of the top
of a tower with respect to the instrument used to ob-                        (C)     29.85   ft
serve it. The only available information is the actual                       (D)     35.33   ft
height of the tower, 157 ft, and the two vertical angles
observed. The vertical angle to the base of the tower is                 28. The triangle shown contains 1.5 acres. It is an
01◦ 46 52 . The vertical angle to the top of the tower                   oblique triangle in which angle B is 102◦ 03 50 , angle
is 09◦ 50 33 . What is the vertical distance from the in-                A is 47◦ 56 10 , and angle C is 30◦ 00 00 . What is the
strument to the top of the tower? Ignore the eﬀects of                   length of the side opposite angle C?
the curvature and refraction in the solution.
(A) 188.08 ft                                                                                               B
(B) 191.29 ft
(C) 191.38 ft                                                                                               1.5 ac
A                            C
(D) 205.34 ft
(A)     300.00   ft
(B)     375.99   ft
(C)     460.43   ft
(D)     586.75   ft

P R O F E S S I O N A L       P U B L I C A T I O N S , I N C .
1-22      S U R V E Y I N G                    S O L V E D               P R O B L E M S   F O R         T H E       F S   A N D   P S     E X A M S

22. The triangle described by the three points is devoid                                     However, this distance is not the minimum distance be-
of angles, but all three distances are available. The                                        tween the line and the building corner. The minimum
distance and bearing from the pipe to the axle are found                                     distance must be along a line perpendicular to the sta-
by inverse, and the two other distances are taken from                                       tioned line. One side of the right triangle including that
the deed. These descriptions can be multiplied by 66 to                                      perpendicular is the distance found, 17.46 ft. One an-
convert them to feet. A form of the law of cosines can                                       gle of the same right triangle must be the supplement
be used to discover the missing angles from the available                                    of the second turned angle.
distances.
b2 + c2 − a2
cos A =
2bc
x
51
wooden stake                                                                                         75 34
t
60 f
12   44.7                                                                                                                      ft
1120.68
17.46
axle             A
0 ft

16
4
S5 1.67
1°4     8
4 5 ft
3E                       iron pipe

(1244.760)2 + (1641.678)2 − (1120.680)2                                             180◦ 00 00 − 104◦ 25 09 = 75◦ 34 51
cos A =                                                                                                                     x
(2)(1244.760)(1641.678)                                                                sin 75◦ 34 51 =
17.46 ft
= 0.73125
x = (sin 75◦ 34 51 )(17.46 ft)
A = 43◦ 00 31. 4
= 16.91 ft

The azimuth from the axle to the pipe is
◦          ◦                                ◦
180 − 51 44 53 = 128 15 07
24. This problem can be solved using the law of cosines
Therefore, the azimuth from the axle to the stake is                                         in its most familiar form.

128◦ 15 07 − 43◦ 00 31 = 85◦ 14 36                                                   a2 = b2 + c2 − 2bc cos A
= (450.83 ft)2 + (471.67 ft)2
− (2)(450.83 ft)(471.67 ft)(cos 05◦ 12 40 )
= (425,720.28 ft) − (425,285.97 ft)(cos 05◦ 12 40 )
23. In the oblique triangle shown in the problem, the
distance along the line between the two stations is 40 ft.                                       = 2192.10 ft
The two interior angles are 20◦ 52 16 and 104◦ 25 09 .                                             √
a = 2192.10 ft
First, ﬁnd the third angle, A.
= 46.82 ft
◦                         ◦                             ◦
A = 20 52 16 + 104 25 09 − 180 00 00
= 125◦ 17 25 − 180◦ 00 00                                                           The answer is (D).

= 54◦ 42 35
25. The law of sines is useful in solving this problem.
Then use the law of sines to ﬁnd b, the shortest side of                                     Two triangles are created by the problem’s situation.
the triangle.
D
a       b
=
sin A   sin B
157.00 ft

40.00 ft             b
=                                                                                                                          x
sin 54◦ 42 35    sin 20◦ 52 16
40.00 ft                                                    8°03
40
b = (sin 20◦ 52 16 )
sin 54◦ 42 35                             C
9 50 33
1 46 52                          A
14.25 ft                                                                                                 B
= 17.46 ft
sin 54◦ 42 35
(not to scale)

P R O F E S S I O N A L             P U B L I C A T I O N S , I N C .
P U B L I C   L A N D   S U R V E Y I N G        S Y S T E M   P R O B L E M S     5-15

79. A private surveyor ﬁnds that his remeasurement                       corners of section 7. What distance should the surveyor
along a section line between found original quarter cor-                 measure west along the section line from the NE corner
ner and section corner monuments diﬀers from that of                     to set the NE corner of NW 1 NE 1 ?
4    4
the oﬃcial record. The surveyor’s objective is to estab-                    (A) 1250.98 ft
lish an initial sixteenth corner monument along the line                    (B) 1285.30 ft
between the two found original corners. Which of the
following techniques is the surveyor likely to ﬁnd helpful                  (C) 1320.00 ft
in resolving the diﬀerence between the measurements?                        (D) 1349.20 ft
(A) protraction
(B) double centering                                                  80.3. Under normal circumstances and by current in-
(C) triangulation                                                     structions, would the original surveyor of section 7 have
established a monument at the SW corner of lot 1?
(D) proportionate measurement
(A) Yes, the lot corners along a township boundary
are set during the course of the original survey
80. The following three problems refer to the illustra-                           of the township boundary.
tion shown.                                                                 (B) The SW corner of lot 1 would have been monu-
Sec. 6
mented during the course of the original survey
only if the western boundary of section 7 was a
6                                                                 guide meridian.
(C) No, neither the quarter-quarter corners nor the
7                                                                 lot corner are monumented during the course of
(N89 57 W)              (83.91)                               the original survey of a township, unless man-
dated by special instructions.
N89 32 W              N88 48 W
1
2766.06 ft            2570.60 ft
(D) The SW corner of lot 1 would have been monu-
mented during the course of the original survey
only if it was a closing corner.
2
Sec. 7
81. The following two problems refer to the illustration
N89 32 W as measured                                             shown.
(83.91) record (in chains)
found original monument                                                                   Sec. 33

PLSS
(N89 57 W 79.94)

ft
12 23)
80.1. The NE, NW, and N 1 corners of section 7 are

4
N88 32 W 5137.44 ft

.5
.
(0
4                                          standard
found original corner monuments. The oﬃcial town-
parallel
.1 7)

ship plat indicates that the record bearing and distance
ft
81 1.2

(N89 58 W 78.90)
across the north line of section 7 is N89◦ 57 W 83.91
8
(

N88 32 W 5068.80 ft
chains. A private land surveyor intends to set the NE
corner of lot 1. The surveyor has measured N89◦ 32 W                                                            Sec. 4
2766.06 ft between the N 1 and the NW corners of sec-
4
tion 7. What distance should the surveyor measure west                               12.54 ft as measured
along the section line from the N 1 corner to set the NE
4
(0.23) record (in chains)
corner of lot 1?                                                                              found original monument
(A) 1259.88 ft
(B) 1320.00 ft
(C) 1383.03 ft                                                        81.1. Retracing a standard parallel, a surveyor discov-
(D) 1506.18 ft                                                        ers that the quarter corners were not established by the
original surveyor. Using the record and measured infor-
mation given, what distance should be measured along
80.2. The NE, NW, and N 1 corners of section 7 are
4                                         the standard parallel from the eastern standard corner
found original corner monuments. The oﬃcial town-                        to the northern quarter corner of section 4?
ship plat indicates that the record bearing and distance                    (A) 2521.86 ft
across the north line of section 7 is N89◦ 57 W 83.91
chains. A private land surveyor intends to set the NE                       (B) 2534.40 ft
corner of NW 1 NE 1 of section 7. The surveyor has mea-                     (C) 2603.70 ft
4    4
sured N88◦ 48 W 2570.60 ft between the NE and the N 1  4
(D) 2638.02 ft

P R O F E S S I O N A L        P U B L I C A T I O N S , I N C .
5-34      S U R V E Y I N G    S O L V E D   P R O B L E M S   F O R   T H E    F S       A N D   P S    E X A M S

measurement. The new values given to several parts of            80.3. Unless otherwise instructed, the original survey-
the line will bear the same relation to the record lengths       or does not monument the quarter-quarter corners or
as the new measurement of the whole line bears to the            lot corners of a township.
record of the whole line.
81.1. The record distance between the closing corners
80.1. The ﬁrst step in solving this problem is to ﬁnd            at the NE and NW corners of section 4 is 78.90 chains,
the record distance from the N 1 to the NE corner of
4                              or 5207.40 ft. The measured distance between the same
lot 1. Since section 7 is a closing section, the record          two corners is 5068.80 ft. The record distance between
distance 83.91 is not divided in half at the N 1 corner.
4                 the NE corner of section 4 and the N 1 corner is half
4
The record distance from the NE corner of section 7              the record distance between the NE and NW corners,
to its N 1 corner is 40.00 chains. The remainder, 43.91
4                                                       or 2603.70 ft. The quarter corner monument should be
chains, is the record distance from the N 1 corner of            placed on the standard parallel and halfway between
4
section 7 to its NW corner. This 43.91 chains is not             the NE and NW corners of section 4 by proportionate
divided in half at the NE corner of lot 1. The record            measure.
distance from the N 1 corner to the NE corner of lot 1 is
4
20.00 chains; the remainder, 23.91 chains, is the record                                  2603.70 ft       x
=
distance along the northern boundary of lot 1.                                            5207.40 ft   5068.80 ft
The proportion of the record distance between the N 1                          13,197,634.56 ft = 5207.40x
4
corner of section 7 and its NW corner, 43.91 chains, to                                                   13,197,634.56 ft
the record distance from the N 1 to the NE corner of lot
4
x=
1, 20.00 chains, must be preserved.                                                                           5207.40
= 2534.40 ft
20.00 chains = 1320.00 ft
The distance from the NE corner to the N 1 corner of
43.91 chains = 2898.06 ft                                                                      4
section 4 is 2534.40 ft. However, the eastern standard
corner is the SE corner of section 33, which lies 12.54 ft
1320.00 ft       x                               west of the NE corner of section 4.
=
2898.06 ft   2766.06 ft
2534.40 ft
3,651,199.20 ft = 2898.06x                                                             −12.54 ft
3,651,199.20 ft                                                    2521.86 ft
x=
2898.06
= 1259.88 ft
81.2. The record distance from the SE corner to the
80.2. The proportion of the record distance between              SW corner of section 33 is 79.94 chains, or 5276.04 ft.
the NE corner of section 7 and its N 1 corner, 40.00             The record distance from the SE corner to the S 1 corner
4
4                          of the section is half that, or 2638.02 ft. The measured
chains, to the record distance from the NE corner of
section 7 to the NE corner of NW 4 NE 1 , 20.00 chains,
1                              distance along the south line of section 33 is 5137.44 ft;
must be preserved.
4                          the S 1 corner must be half that distance from the SE
4
corner.
20.00 chains = 1320.00 ft                                              1
2    (5137.44 ft) = 2568.72 ft
40.00 chains = 2640.00 ft
Similarly, the measured distance from the NE corner of
1320.00 ft       x                               section 4 to its N 1 corner is half the measured distance
=                                                        4
2640.00 ft   2570.60 ft                          between the NE and NW corners.
3,393,192.00 ft = 2640.00x
1
2    (5068.80 ft) = 2534.40 ft
3,393,192.00 ft
x=
2640.00
The distance from the NE corner of section 4 to its N 1
4
= 1285.30 ft                         corner must be reduced by 12.54 ft to ﬁnd the distance
from the SE corner of section 33 to the N 1 corner of
4
section 4.

P R O F E S S I O N A L   P U B L I C A T I O N S , I N C .
9-10                S U R V E Y I N G                      S O L V E D               P R O B L E M S   F O R   T H E   F S      A N D       P S    E X A M S

60. What is the relationship between the mapping an-                                                     64. The following seven problems refer to the illustra-
gle, θ, and the distance from the central meridian for                                                   tion shown.
a station in a zone of a state plane coordinate system
based upon the Lambert projection?                                                                       Lambert conformal projection                                     grid north
(A) The absolute value of θ is always larger than the
diﬀerence in longitude from the central merid-                                                  constants for the zone                                                  N
ian.                                                                                            C 2,000,000.00 ft                                   Thornton 2
(B) θ is not always the same for a speciﬁc meridian                                                   central meridian = 105 30 00 .00                    elevation 5441 ft
Rb 26,243,052.74 ft
of longitude.
yo 461,675.83 ft
(C) The absolute value of θ is always the same as                                                           0.6306895773
1
the diﬀerence in longitude from the central                                                       2
2.359 10 10
2 o sin 1“
meridian.
(D) The absolute value of θ is always smaller than                                                                                         51
the diﬀerence in longitude from the central                                                                                             02
05
meridian.                                                                                                                                     .5

61. Which of the following expressions is equal to the
mapping angle, θ, in a state plane coordinate system

50
based on the Lambert projection?

12
(A) ∆λ                                                                                                                                                .1
26

29
45

.0
x                                                                                                                             78
(B) tan−1
Rb − y
Rb − y                                                                                    Crown
(C) sin−1                                                                                              x 2,114,700.90
R                                                                                       y 701,393.10
(D) both A and B                                                                                       elevation 5764 ft                             Capitol
latitude 39 44 21.276
longitude 104 59 03.601
elevation 5489 ft
SHORT. . GEODETIC. . .CALCULATIONS
................... ............................ ..........................................
(not to scale)

62. The geodetic azimuth (from south) from station                                                       Note that the concave lines connecting the three sta-
Telo to station Klamath is 186◦ 15 22. 21, and the map-                                                  tions indicate the curvature of long lines, such as those
ping angle at station Telo is +00◦ 40 21. 85. What is                                                    in the illustration, on the state plane. The angles shown
the grid azimuth from station Telo to station Klamath                                                    are observed values.
if the second term is not considered?
(A) 05◦ 35 00. 36
64.1. The following values are taken from the Plane
(B) 06◦ 55 44. 06
(C) 185◦ 35 00. 36                                                                                   the U.S. Government Printing Oﬃce.
(D) 186◦ 55 44. 06
tabular diﬀerence
for 1 of
63. The grid azimuth from station Exeter to station                                                               latitude            R (ft)      latitude (ft)
Abott is 212◦ 32 14. 1, and the mapping angle at station                                                            39◦ 41        25,569,283.24    101.18217
Exeter is +00◦ 27 45. 2. What is the geodetic azimuth                                                                   42        25,563,212.31    101.18283
from station Abott to station Exeter if the second term                                                                 43        25,557,141.34    101.18350
is not considered?                                                                                                      44        25,551,070.33    101.18400
(A) 32◦ 59 59. 3                                                                                                     45        25,544,999.29    101.18467
(B) 212◦ 04 28. 9
(C) 212◦ 59 59. 3                                                                                     What is the appropriate R value for the station labeled
(D) insuﬃcient information given                                                                      Capitol?
(A) 25,541,081.04 ft
(B) 25,547,152.08 ft
(C) 25,548,917.54 ft
(D) 25,554,988.58 ft

P R O F E S S I O N A L                           P U B L I C A T I O N S , I N C .
9-22                S U R V E Y I N G                      S O L V E D               P R O B L E M S   F O R   T H E     F S   A N D   P S   E X A M S

61. The following formulas are usually given in state                                                    Next, multiply this diﬀerence by the tabular diﬀerence
plane projection tables.                                                                                 for 1 of latitude to ﬁnd the diﬀerence between the
R value at latitude 39◦ 44 00. 000 and at latitude
θ                                                        39◦ 44 21. 276.
∆λ =
θ = ∆λ                                                                           (21. 276)(101.18400 ft) = 2152.79 ft
x                                                      Finally, subtract this diﬀerence from the R value given
tan θ =
Rb − y                                                   for the next smaller latitude.
x
θ = tan−1
Rb − y
39◦ 44            25,551,070.33   ft      101.18400 ft
The answer is (D).                                                                                                                   −2152.79    ft
39◦ 44 21. 276    25,548,917.54   ft
39◦ 45            25,544,999.29   ft      101.18467 ft
SHORT. . GEODETIC. . .CALCULATIONS
................... ............................ ..........................................

62. Since the geodetic azimuth is measured clockwise                                                     The answer is (C).
from south, subtract 180◦ to ﬁnd the geodetic azimuth
from north.                                                                                              64.2. There is more than one way to calculate the θ
angle from the longitude of a station. One way is to use
186◦ 15 22. 21                                                the following relationship.
−180◦ 00 00. 00
06◦ 15 22. 21
θ = ∆λ
The formula for ﬁnding the grid azimuth of short
lines is                                                                                                 The symbol ∆λ refers to the change in longitude from
the central meridian of the zone, which in this case is
grid azimuth = geodetic azimuth − θ                                                             from 105◦ 30 00 .
= 06◦ 15 22. 21 − (+00◦ 40 21. 85)
central meridian = 105◦ 30 00. 000
= 05◦ 35 00. 36
λ of Capitol = −104◦ 59 03. 601
∆λ = +00◦ 30 56. 399
The symbol indicates the sine of the apex angle for the
zone. It is shown in the illustration as 0.6306895773.
63. There is not enough information given to answer
this question. While the grid azimuth will be precisely
180◦ diﬀerent at station Exeter (212◦ 32 14. 1) from its                                                                θ = ∆λ
value at station Abott (32◦ 32 14. 1), the same cannot                                                                   = (+00◦ 30 56. 399)(0.6306895773)
be said of the geodetic azimuth. Without the value of
the mapping angle at station Abott or some information                                                                   = +00◦ 19 30. 81
from which it may be derived, the question cannot be
64.3. The value of θ for station Capitol is
+00◦ 19 30. 8115, and the value of R is 25,548,917.54 ft.
64.1. The R value for a particular station is the dis-
These two quantities are suﬃcient to ﬁnd the value of
tance from the apex of the cone to that station in feet.
the x-coordinate, using the following formula.
One way to derive its value is to use the tabular diﬀer-
ence for 1 of latitude given.
x = R sin θ + C
First, ﬁnd the diﬀerence in seconds of latitude from the
next smaller minute given.
C is the constant 2,000,000 ft between the central merid-
ian and the y-axis. The following ﬁgure illustrates these
latitude of station Capitol =       39◦ 44 21. 276                                            relationships.
−39◦ 44 00. 000
diﬀerence in seconds =         21. 276

P R O F E S S I O N A L                           P U B L I C A T I O N S , I N C .

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