This document contains several surveying problems and their solutions. Problem 18.01 asks about the length of the third side of a triangular parcel given the other two sides and the angle between them. The solution uses the Law of Cosines to calculate the length as 91.65 ft. Problem 18.02 then gives the third side and asks for the remaining angles, which the solution finds using the Law of Sines to be 49.1° and 70.9°.

1. Surveying FE Review
Spring 2018
Problem 18.01: Two sides of a triangular-shaped parcel are
80 ft. and 100 ft. with a 60° angle between them. The length
of the third side of the parcel (ft.) is most nearly:
A. 91.65 ft.
B. 86.66 ft.
C. 80.00 ft.
D. 100.00 ft.
Problem 18.01: Two sides of a triangular-shaped parcel are
80 ft. and 100 ft. with a 60° angle between them. The length
of the third side of the parcel (ft.) is most nearly:
A. 91.65 ft.
B. 86.66 ft.
C. 80.00 ft.
D. 100.00 ft.
Solution: Two triangle sides, b and c, and angle A, are
known (see figure).
Problem 18.01: Two sides of a triangular-shaped parcel are
80 ft. and 100 ft. with a 60° angle between them. The length
of the third side of the parcel (ft.) is most nearly:
A. 91.65 ft.
B. 86.66 ft.
C. 80.00 ft.
D. 100.00 ft.
Solution: Use the Law of Cosines to determine the length
of side a. The formula for the Law of Cosines is given in the
NCEES Handbook, Mathematics, page 23.
Problem 18.01: Two sides of a triangular-shaped parcel are
80 ft. and 100 ft. with a 60° angle between them. The length
of the third side of the parcel (ft.) is most nearly:
A. 91.65 ft.
B. 86.66 ft.
C. 80.00 ft.
D. 100.00 ft.
Solution: Using the Law of Cosines:
 2 2 2
a = b + c - 2bc cos A
        2 22 o
a = 80 + 100 - 2 80 100 cos 60 91.6 .5ft
Problem 18.02: Two sides of a triangular-shaped parcel are
80 ft. long and 100 ft. long, with a 60° angle between them.
The third side is 91.65 ft. long. The remaining two angles of
the parcel are most nearly:
A. 49°, 71°
B. 64°, 56°
C. 30°, 90°
D. 60°, 60°
CIVL 4197 Surveying FE REview 1/9

2. Problem 18.02: Two sides of a triangular-shaped parcel are
80 ft. long and 100 ft. long, with a 60° angle between them.
The third side is 91.65 ft. long. The remaining two angles of
the parcel are most nearly:
Solution: Three triangle sides, a, b, c, and angle A, are
known (see figure).
a = 91.65
A. 49°, 71°
B. 64°, 56°
C. 30°, 90°
D. 60°, 60°
Problem 18.02: Two sides of a triangular-shaped parcel are
80 ft. long and 100 ft. long, with a 60° angle between them.
The third side is 91.65 ft. long. The remaining two angles of
the parcel are most nearly:
Solution: Since the three sides of the triangle are given,
use the Law of Sines to determine the associated angles.
The formula for the Law of Sines is given in the NCEES
Handbook, Mathematics, page 23.
a = 91.65
A. 49°, 71°
B. 64°, 56°
C. 30°, 90°
D. 60°, 60°
Problem 18.02: Two sides of a triangular-shaped parcel are
80 ft. long and 100 ft. long, with a 60° angle between them.
The third side is 91.65 ft. long. The remaining two angles of
the parcel are most nearly:
Solution:
     
a b c
= =
sin A sin B sin C
     
91.65 80 100
= = = 105.83
sin 60 sin B sin C
a = 91.65
A. 49°, 71°
B. 64°, 56°
C. 30°, 90°
D. 60°, 60°
Problem 18.02: Two sides of a triangular-shaped parcel are
80 ft. long and 100 ft. long, with a 60° angle between them.
The third side is 91.65 ft. long. The remaining two angles of
the parcel are most nearly:
A. 49°, 71°
B. 64°, 56°
C. 30°, 90°
D. 60°, 60°
 
o100
= 105.83 C = 70.9
sin C
Solution:
a = 91.65
 
o80
= 105.83 B = 49.1
sin B

Problem 18.03: In an equilateral triangular-shaped parcel,
the height (altitude) of the triangle is 5 ft. less than its side
length. The side length (ft.) of the triangle is most nearly
equal to:
A. 10 ft.
B. 20 ft.
C. 30 ft.
D. 40 ft.
Problem 18.03: In an equilateral triangular-shaped parcel,
the height (altitude) of the triangle is 5 ft. less than its side
length. The side length (ft.) of the triangle is most nearly
equal to:
A. 10 ft.
B. 20 ft.
C. 30 ft.
D. 40 ft.
Solution: An equilateral triangle has three sides of equal
length. This implies that each internal angle is 60°. The
figure illustrates the relationship between the side and the
altitude.
CIVL 4197 Surveying FE REview 2/9

3. Problem 18.03: In an equilateral triangular-shaped parcel,
the height (altitude) of the triangle is 5 ft. less than its side
length. The side length (ft.) of the triangle is most nearly
equal to:
A. 10 ft.
B. 20 ft.
C. 30 ft.
D. 40 ft.
Solution:
     oCD a-5
sin A = sin B = = = sin 60
aAC
 a - 5 = 0.866 a
 0.134 a = 5 a = 37.31 ft.
Problem 18.04: Which of the following statements is NOT
correct?
A. A leveling staff is a crew assigned to a route
surveying task under the supervision of a licensed
professional surveyor.
B. A leveling staff is a rod used in surveying.
C. A level line is one where all points are normal to
the direction of the force of gravity.
D. A freely-suspended plumb-bob shows the
direction of the gravitational force.
Problem 18.04: Which of the following statements is NOT
correct?
A. A leveling staff is a crew assigned to a route
surveying task under the supervision of a licensed
professional surveyor.
B. A leveling staff is a rod used in surveying.
C. A level line is one where all points are normal to
the direction of the force of gravity.
D. A freely-suspended plumb-bob shows the
direction of the gravitational force.
Solution: Statements B, C and D are correct. Statement A
is incorrect.
Problem 18.05: A surveyor takes several leveling readings.
The instrument is on a known point of elevation of 123.45 ft.,
and the height of the instrument (HI) is 5.15 ft. To determine
the elevation of the underside of a beam, an inverted sight
(IS) reading of 3.13 ft. is obtained. To determine the elevation
of a point on a slope, a reading of 4.32 ft. is obtained. The
elevations of the underside of the beam and the point on the
slope (ft.) are respectively most nearly:
A. 128.60 ft., 126.58 ft.
B. 131.73 ft., 124.28 ft.
C. 125.47 ft., 132.92 ft.
D. 121.43 ft., 113.98 ft.
Problem 18.05: A surveyor takes several leveling readings.
The instrument is on a known point of elevation of 123.45 ft.,
and the height of the instrument (HI) is 5.15 ft. To determine
the elevation of the underside of a beam, an inverted sight
(IS) reading of 3.13 ft. is obtained. To determine the elevation
of a point on a slope, a reading of 4.32 ft. is obtained. The
elevations of the underside of the beam and the point on the
slope (ft.) are respectively most nearly:
Problem 18.05:
Elevation at Sta. B = Elevation at Sta. A + HI + IS
= 123.45 + 5.15 + 3.13 = 131.73 ft.
Solution:
Elevation at Sta. C = Elevation at Sta. A + HI - FS
= 123.45 + 5.15 – 4.32 = 124.28 ft.
CIVL 4197 Surveying FE REview 3/9

4. Problem 18.05:
Elevation at Sta. B = Elevation at Sta. A + HI + IS
= 123.45 + 5.15 + 3.13 = 131.73 ft.
Solution:
Elevation at Sta. C = Elevation at Sta. A + HI - FS
= 123.45 + 5.15 – 4.32 = 124.28 ft.
A. 128.60 ft., 126.58 ft.
B. 131.73 ft., 124.28 ft.
C. 125.47 ft., 132.92 ft.
D. 121.43 ft., 113.98 ft.
Problem 18.06: A closed traverse has six segments and
four interior angles measuring 90° each. The sum of the
remaining interior angles is most nearly:
A. 0o
B. 90o
C. 180o
D. 360o
Problem 18.06: A closed traverse has six segments and
four interior angles measuring 90° each. The sum of the
remaining interior angles is most nearly:
A. 0o
B. 90o
C. 180o
D. 360o
Solution: For a polygon of n sides, the internal angle is
expressed as:
  
Sum of the interior angles 180 2n
    
180 6 2 720
Sum of two unknown angles = 720o – 360o = 360o
Problem 18.07: The table below shows differential leveling
data using a transit level. The starting station is of known
elevation. Find the elevation of station D.
A. 508.40 m
B. 489.77 m
C. 510.23 m
D. 491.60 m
Station BS (m) FS (m) Elevation (m) Notes
A 3.95 500.00 Benchmark
B 2.47 6.34
C 3.81 5.51
D 6.78
Problem 18.07: The table below shows differential leveling
data using a transit level. The starting station is of known
elevation. Find the elevation of station D.
A. 508.40 m
B. 489.77 m
C. 510.23 m
D. 491.60 m
Station BS (m) FS (m) Elevation (m) Notes
A 3.95 500.00 Benchmark
B 2.47 6.34
C 3.81 5.51
D 6.78
    3.95 2.47 3.81 10.23BS
    6.34 5.51 6.78 18.63FS
  D AELE ELE BS FS
   500.00 10.23 18.63 491.60
Problem 18.07: The table below shows differential leveling
data using a transit level. The starting station is of known
elevation. Find the elevation of station D.
A. 508.40 m
B. 489.77 m
C. 510.23 m
D. 491.60 m
Station BS (m) HI(m) FS (m) Elevation (m)
A 3.95 503.95 500.00
B 2.47 500.08 6.34 497.61
C 3.81 498.38 5.51 494.57
D 6.78 491.60
CIVL 4197 Surveying FE REview 4/9

5. Problem 18.08: A pile is made of recycled aggregate
material. The pile has a diameter of 100 ft. at the base, and
25 ft. at the top. It is 25 ft. high. The volume of the pile (yd3)
is most nearly:
A. 105,000 yd3
B. 5,787 yd3
C. 3,860 yd3
D. 2,576 yd3
Problem 18.08: A pile is made of recycled aggregate
material. The pile has a diameter of 100 ft. at the base, and
25 ft. at the top. It is 25 ft. high. The volume of the pile (yd3)
is most nearly:
Solution: Generally, if the properties of two end sections are
given from which the end areas can be readily computed,
the volume equals the average cross-sectional area
multiplied by the height.
A. 105,000 yd3
B. 5,787 yd3
C. 3,860 yd3
D. 2,576 yd3
Problem 18.08: A pile is made of recycled aggregate
material. The pile has a diameter of 100 ft. at the base, and
25 ft. at the top. It is 25 ft. high. The volume of the pile (yd3)
is most nearly:
Solution: In the NCEES Handbook, Civil Engineering,
page 176, is the Average End Area method and is
expressed as:
 
 1 2
2
L A A
V
A. 105,000 yd3
B. 5,787 yd3
C. 3,860 yd3
D. 2,576 yd3
Problem 18.08: A pile is made of recycled aggregate
material. The pile has a diameter of 100 ft. at the base, and
25 ft. at the top. It is 25 ft. high. The volume of the pile (yd3)
is most nearly:
Solution:
 
2
2
21
1
100
7,854ft
4 4
d
A   

 
2
2
22
2
25
491ft
4 4
d
A   

3
3,863yd
A. 105,000 yd3
B. 5,787 yd3
C. 3,860 yd3
D. 2,576 yd3
 1 2
2
L A A
V


 2 2
25 . 7,854 491
2
ft ft ft
 3
104,312ft
Problem 18.09: What is the southern azimuth of a line with
a bearing of S12° 34' 56"E?
A. 12°34'56"
B. 77°25'04"
C. 167°25'04"
D. 347°25'04"
Problem 18.09: What is the southern azimuth of a line with
a bearing of S12° 34' 56"E?
A. 12°34'56"
B. 77°25'04"
C. 167°25'04"
D. 347°25'04"
Solution: Azimuths are generally expressed from the
north. However, there are exceptions: some navigation
systems use south as the reference plane.
The rotation of the azimuth is always clockwise. In this
case, the bearing needs to be subtracted from 360° to
determine the azimuth from the south.
CIVL 4197 Surveying FE REview 5/9

6. A. 12°34'56"
B. 77°25'04"
C. 167°25'04"
D. 347°25'04"
Problem 18.09: What is the southern azimuth of a line with
a bearing of S12° 34' 56"E?
Solution: The azimuth from the south is:
360° - 12° 34' 56" = 347° 25' 04"
Problem 18.10: Line AB bears N 12° 34' 56'' E, and line AC
bears S 12° 34' 56'' E. The deflection angle between the
lines is:
A. Straight East or 90°
B. 25° 09' 52" (Left)
C. 167° 25' 04" (Left)
D. 154° 50' 08" (Right)
Problem 18.10: Line AB bears N 12° 34' 56'' E, and line AC
bears S 12° 34' 56'' E. The deflection angle between the
lines is:
A. Straight East or 90°
B. 25° 09' 52" (Left)
C. 167° 25' 04" (Left)
D. 154° 50' 08" (Right)
Solution: A deflection angle is the
difference in angle from the prolongation
of the back line to the forward line along
a traverse.
The difference in angle between AB' and AB is the sum of
the angles 12°34'56" and 12°34'56" which is 25°09' 52".
Problem 18.10: Line AB bears N 12° 34‘ 56'‘ E, and line AC
bears S 12° 34' 56'' E. The deflection angle between the
lines is:
A. Straight East or 90°
B. 25° 09' 52" (Left)
C. 167° 25' 04" (Left)
D. 154° 50' 08" (Right)
Solution: A deflection angle is the
difference in angle from the prolongation
of the back line to the forward line along
a traverse.
Since line AC is located at the left hand side of the
prolongation line, it is deflected to the left.
Problem 18.11: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the transit
rule?
A. + 0.192 ft
B. - 0.192 ft
C. - 0.343 ft
D. + 0.343 ft
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
Problem 18.11: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the transit
rule?
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
  sinDepartures L
     
850sin 80.5 1,250sin 136.25
     
1,000sin 220.5 1,850sin 325.33
 1.004
CIVL 4197 Surveying FE REview 6/9

7. Problem 18.11: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the transit
rule?
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
 1.004CD
CD
Dep
Correction Dep ft
Departures
 

  
  
649.448
1.004
838.343 864.391 649.448 1052.282
ft
 0.192ft
Problem 18.11: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the transit
rule?
A. + 0.192 ft
B. - 0.192 ft
C. - 0.343 ft
D. + 0.343 ft
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
Problem 18.11a: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the compass
rule?
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
A. + 0.203 ft
B. - 0.203 ft
C. - 0.343 ft
D. + 0.343 ft
Problem 18.11a: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the compass
rule?
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
  sinDepartures L
     
850sin 80.5 1,250sin 136.25
     
1,000sin 220.5 1,850sin 325.33
 1.004
Problem 18.11a: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the compass
rule?
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
  1.004CD
CD
L
Correction Dep ft
perimeter
  
  
1,000
1.004
850 1,250 1,000 1,850
ft
 0.203ft
Problem 18.11a: The table below shows length and
azimuthal angles for lines in a closed traverse, ABCD. What
is the correction to the departure of CD, using the compass
rule?
A. + 0.203 ft
B. - 0.203 ft
C. - 0.343 ft
D. + 0.343 ft
Line Length(ft.) Azimuthangle
AB 850.00 80° 30'
BC 1250.00 136° 15'
CD 1000.00 220° 30'
DA 1850.00 325° 20'
CIVL 4197 Surveying FE REview 7/9

8. Problem 18.12: A 2,500-m long trapezoidal open channel is
constructed at a specified slope of 1.5%. If the base
elevation at the channel inlet is 90.357 m, the base
elevation at the channel outlet should be most nearly:
A. 127.9 m
B. 100.875 m
C. 52.8 m
D. 37.5 m
Problem 18.12: A 2,500-m long trapezoidal open channel is
constructed at a specified slope of 1.5%. If the base
elevation at the channel inlet is 90.357 m, the base
elevation at the channel outlet should be most nearly:
A. 127.9 m
B. 100.875 m
C. 52.8 m
D. 37.5 m
Solution: The problem concerns open channel flow;
therefore, the slope should be directed downwards.
2 1
1.5
2,500
100
EL EL m
 
   
 
2
1.5
90.357 2,500 52.857
100
EL m m m
 
   
 
Problem 18.13: Two cross-sections of a proposed roadway
are located at Station 1+65.00 and 3+50.00. One cross-
section needs 200 ft2 of cut and the other needs 125 ft2 of
fill. The net excavation (yd3) required between the two
sections is most nearly:
A. 30,000 yd3
B. 13,800 yd3
C. 1,100 yd3
D. 257 yd3
Problem 18.13: Two cross-sections of a proposed roadway
are located at Station 1+65.00 and 3+50.00. One cross-
section needs 200 ft2 of cut and the other needs 125 ft2 of
fill. The net excavation (yd3) required between the two
sections is most nearly:
A. 30,000 yd3
B. 13,800 yd3
C. 1,100 yd3
D. 257 yd3
Solution: The volume is calculated using the average
end area method from the Earthwork Formulas section of
the NCEES Handbook, Civil Engineering, page 176:
 
 1 2
2
L A A
V
A. 30,000 yd3
B. 13,800 yd3
C. 1,100 yd3
D. 257 yd3
Problem 18.13: Two cross-sections of a proposed roadway
are located at Station 1+65.00 and 3+50.00. One cross-
section needs 200 ft2 of cut and the other needs 125 ft2 of
fill. The net excavation (yd3) required between the two
sections is most nearly:
Solution:
   

350 165 200 125
2
V 3
6,938ft 3
257yd
cut fill
Problem 18.14: Excavated soil from a barrow pit is
stockpiled in a conical shape. The stockpile has a diameter
of 20 m at its base and a height of 15 m. Its volume in m3 is
most nearly:
A. 5,000 m3
B. 2,500 m3
C. 1,600 m3
D. 1,000 m3
CIVL 4197 Surveying FE REview 8/9

9. Problem 18.14: Excavated soil from a barrow pit is
stockpiled in a conical shape. The stockpile has a diameter
of 20 m at its base and a height of 15 m. Its volume in m3 is
most nearly:
Solution: The shape of the stockpile in this problem is
conical. As given in the NCEES Handbook, Civil Engineering,
page 176, the volume of a cone V is:
  
3
base area height
V
A. 5,000 m3
B. 2,500 m3
C. 1,600 m3
D. 1,000 m3
Problem 18.14: Excavated soil from a barrow pit is
stockpiled in a conical shape. The stockpile has a diameter
of 20 m at its base and a height of 15 m. Its volume in m3 is
most nearly:
Solution:
  
3
base area height
V
 
 
2
20
15m
4
3
m 
 
 
 

  2
314.15m 15 m
3
 3
1,571m
A. 5,000 m3
B. 2,500 m3
C. 1,600 m3
D. 1,000 m3
Problem 18.15: A 300 m long runway measures 20 mm on
an aerial photo. The scale of the photo is most nearly:
A. 1:1,500
B. 1:15,000
C. 1:15
D. 1:666
Problem 18.15: A 300 m long runway measures 20 mm on
an aerial photo. The scale of the photo is most nearly:
A. 1:1,500
B. 1:15,000
C. 1:15
D. 1:666
Solution: Photo scale:
 
Distance at ground
1:
Distance in photo
  
300 1,000
1:
20
mm
mm
mm
 1:15,000
Surveying FE Review
Spring 2018
Questions?
CIVL 4197 Surveying FE REview 9/9