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Module 4-Answers - FOS

Fundamentals of Surveying (San Jose Community College)

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SOLUTIONS TO REVIEW EXERCISES: Using Transit Rule:

𝐿𝐶 1.01
4.1. From the field notes of a closed traverse shown below, adjust the transverse 𝐾1 = ∑ 𝑁𝐿−∑ 𝑆𝐿
= +965.69−(−964.68) = 0.0005232
using.
a. Transit Rule 𝐷𝐶 1.70
𝐾2 = ∑ 𝐸𝐷−∑ 𝑊𝐷
= = 0.00073
b. Compass Rule +1,171.91−(−1,170.21)

c. Compute the linear error of closure Length Corr. Lat. 𝒄𝑳 Corr. Dep. 𝒄𝑫
d. Compute the relative error or precision Line Latitude Departure
(D) = (𝑳𝒂𝒕 × 𝑲𝟏 ) = (𝑫𝒆𝒑 × 𝑲𝟐 )
Sta. Occ. Sta. Obs. Bearings Distances AB 400.00 m +400.00 m +000.00 m 0.209 m 0.000 m
A B Due North 400.00 m BC 800.00 m +565.69 m +565.69 m 0.296 m 0.412 m
B C N 45° E 800.00 m CD 700.00 m -350.00 m +606.22 m 0.183 m 0.443 m
C D S 60° E 700.00 m DE 600.00 m -563.82 m -205.21 m 0.295 m 0.150 m
D E S 20° W 600.00 m EA 966.34 m -50.86 m -965.00 m 0.027 m 0.704 m
E A S 86°59’ W 966.34 Total 3,466.34 m 1.010 m 1.700 m
Solution: Adjusting the latitudes and departures:

Determining latitude and departures 𝐴𝑑𝑗. 𝐿𝑎𝑡. = 𝐶𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝐿𝑎𝑡. ± 𝑐𝐿

𝐿𝑎𝑡 = 𝑑 cos ∅ 𝐷𝑒𝑝 = 𝑑 sin ∅ 𝐴𝑑𝑗. 𝐷𝑒𝑝. = 𝐶𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝐷𝑒𝑝. ± 𝑐𝐷

Line Length Bearings Latitude Departure Line Lat. 𝒄𝑳 Adj. Lat. Dep. 𝒄𝑫 Adj. Dep.
AB 400.00 m Due North +400.00 m +000.00 m AB +400.00 m -0.209 +399.791 m +000.00 m -0.000 +000.00 m
BC 800.00 m N 45°00’ E +565.69 m +565.69 m BC +565.69 m -0.296 +565.394 m +565.69 m -0.412 +565.278 m
CD 700.00 m S 60°00’ E -350.00 m +606.22 m CD -350.00 m -0.183 -350.183 m +606.22 m -0.443 +605.777 m
DE 600.00 m S 20°00’ W -563.82 m -205.21 m DE -563.82 m -0.295 -564.115 m -205.21 m -0.150 -205.36 m
EA 966.34 m S 86°59’ W -50.86 m -965.00 m EA -50.86 m -0.027 -50.887 m -965.00 m -0.704 -965.704 m
Total 3,466.34 m +1.01 m +1.70 m Total -1.010 0.00 0.00

Determining total closure in latitude and departure. Determining Adjusted length and bearing of each course.
𝐷𝑒𝑝.′
∑ 𝑁𝑜𝑟𝑡ℎ 𝐿𝑎𝑡. = +400 + 565.69 = +965.69 𝐷 ′ = √(𝐿𝑎𝑡. ′)2 + (𝐷𝑒𝑝. ′)2 ∅ = tan−1 𝐿𝑎𝑡.′

∑ 𝑆𝑜𝑢𝑡ℎ 𝐿𝑎𝑡. = −350 − 563.82 − 50.86 = −964.68 Line Adjusted Length Adjusted Bearing
AB 399.791 m Due North
∑ 𝐸𝑎𝑠𝑡 𝐷𝑒𝑝. = +565.69 + 606.22 = +1,171.91
BC 799.506 m N 44°59’38.84” E
∑ 𝑊𝑒𝑠𝑡 𝐷𝑒𝑝. = −205.21 − 965 = −1,170.21 CD 699.710 m S 59°58’08.31” E
DE 600.332 m S 20°00’12.57” W
Determining correction for latitude and departure. EA 967.044 m S 86°59’01.08” W

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Using Compass Rule:

𝐿𝐸𝐶 = √𝐶𝐿 2 + 𝐶𝐷 2 = √(+1.01 𝑚)2 + (+1.70𝑚)2
𝐶𝐿 1.01
𝐾1 = 𝐷
= 3,466.34
= 0.00029
𝑳𝑬𝑪 = 𝟏. 𝟗𝟕𝟕 𝒎 (Ans.)
𝐶𝐷 1.70
𝐾2 = = = 0.00049 −𝐶 −(+1.70)
𝐷 3,466.34 ∅ = tan−1 −𝐶𝐷 = tan−1 = 59°17′ 04.94"
𝐿 −(+1.01)
Length Corr. Lat. Corr. Dep.
Line 𝑲𝟏 𝑲𝟐 Since the ∑ 𝐿𝑎𝑡 is positive the direction is North, and the ∑ 𝐷𝑒𝑝 is positive the
(D) 𝒄𝑳 = (𝑫 × 𝑲𝟏 ) 𝒄𝑫 = (𝑫 × 𝑲𝟐 )
AB 400.00 m 0.00029 0.00049 0.116 m 0.196 m direction is East
BC 800.00 m 0.00029 0.00049 0.232 m 0.392 m Bearing of the side error: N 𝟒𝟕°𝟎𝟓′ E
CD 700.00 m 0.00029 0.00049 0.203 m 0.343 m
𝐿𝐸𝐶 1.977 3 𝟏
DE 600.00 m 0.00029 0.00049 0.174 m 0.294 m 𝑅𝑃 = = 3,466.34 = 5260 say, 𝟏𝟕𝟓𝟎 (Ans.)
𝐷
EA 966.34 m 0.00029 0.00049 0.280 m 0.475 m
Total 3,466.34 m 1.01 m 1.70 m

4.2. Find the bearing of line 4-5 and the missing side 5-1 of the closed traversed
Adjusting the latitudes and departures: shown in the field notes shown.
𝐴𝑑𝑗. 𝐿𝑎𝑡. = 𝐶𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝐿𝑎𝑡. ± 𝑐𝐿 Lines Bearings Distances
1–2 S 70° 15’ E 32.20 m
𝐴𝑑𝑗. 𝐷𝑒𝑝. = 𝐶𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝐷𝑒𝑝. ± 𝑐𝐷
2–3 S 36° 30’ W 31.20 m
Line Lat. 𝒄𝑳 Adj. Lat. Dep. 𝒄𝑫 Adj. Dep. 3–4 N 66° 30’ W 17.40 m
AB +400.00 m -0.116 +399.884 m +000.00 m -0.196 -0.196 m 4–5 - 36.30 m
BC +565.69 m -0.232 +565.458 m +565.69 m -0.392 +565.298 m 5–1 N 60° 00’ E -
CD -350.00 m -0.203 -350.203 m +606.22 m -0.343 +605.877 m Solution:
DE -563.82 m -0.174 -563.994 m -205.21 m -0.294 -205.504 m
Determine the latitude and departure of the given sides.
EA -50.86 m -0.280 -51.140 m -965.00 m -0.475 -965.475 m
Total -1.01 0.00 -1.70 0.00 Line Distance Bearing Latitude Departure
Determining Adjusted length and bearing of each course. 1–2 32.20 m S 70° 15’ E -10.881 m +30.306 m
𝐷𝑒𝑝.′ 2–3 31.20 m S 36° 30’ W -25.080 m -18.558 m
𝐷 ′ = √(𝐿𝑎𝑡. ′)2 + (𝐷𝑒𝑝. ′)2 ∅ = tan−1 𝐿𝑎𝑡.′ 3–4 17.40 m N 66° 30’ W +6.938 m -15.957 m
Total -29.023 m -4.209 m
Line Adjusted Length Adjusted Bearing
AB 399.884 m N 00°01’41.10” W
BC 799.565 m N 44°59’30.81” E
CD 699.806 m S 59°58’17.96” E
DE 600.268 m S 20°01’13.29” W
EA 966.828 m S 86°58’04.61” W

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A–B S 35° 30’ W 44.37 m

B–C N 57° 15’ W 137.84 m
𝐿𝐸𝐶 = √(−29.023)2 + (−4.209)2
C–D N 1° 45’ E 12.83 m
𝐿𝐸𝐶 = 29.327 𝑚 D–E - 64.86 m
E–A - 106.72
−(−4.209)
∅𝐿𝐸𝐶 = tan−1 −(−29.023) Solution:
Determine the latitude and departure of the given sides.
∅𝐿𝐸𝐶 = 8°15′ 06.02"
Line Distance Bearing Latitude Departure
Bearing of LEC, S 8°15’06.02” W
AB 44.37 m S 35° 30’ W -36.122 m -25.766 m
BC 137.84 m N 57° 15’ W +74.568 m -115.929 m
CD 12.83 m N 1° 45’ E +12.824 m +0.392 m
∅1 = 60° − 8°15′ 06.02" Total +51.270 m -141.303 m
∅1 = 51°44′ 53.98"
𝐿𝐸𝐶 = √(+51.270)2 + (−141.303)2
Using cosine Law to determine the length at 51.
𝐿𝐸𝐶 = 150.317 𝑚
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶
−(−141.303)
(36.30)2 = ("51")2 + (29.327)2 − 2("51")(29.327) cos 51°44′ 53.98" ∅𝐿𝐸𝐶 = tan−1 −(+51.270)

5-1 = 𝟒𝟔. 𝟐𝟏𝟓 𝒎 (Ans.) ∅𝐿𝐸𝐶 = 70°03′ 26.42"

To determine the ∅4 , use cosine law, Bearing of LEC, N 70°03′ 26.42" W
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶 To determine the ∅𝐷 & ∅𝐸 , use cosine law,
(46.215)2 = (36.30)2 + (29.327)2 − 2(36.30)(29.327) cos ∅4 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶
∅4 = 88°52′ 17.03" (106.72)2 = (150.317)2 + (64.86)2 −
2(150.317)(64.86) cos ∅𝐷
Bearing of Line 4-5, 𝛽4−5 = 88°52′ 17.03" − 8°15′ 06.02" = 80°37′ 11.01"
∅𝐷 = 37°46′ 24.47"
Bearing of Line 4-5, N 𝟖𝟎°𝟑𝟕′ 𝟏𝟏. 𝟎𝟏" W (Ans.)
(150.317)2 = (106.72)2 + (64.86)2 −
2(106.72)(64.86) cos ∅𝐸
4.3. Determine the bearings of lines 4-5 and 5-1 of the closed traverse shown the ∅𝐸 = 59°37′ 46.51"
technical description of which is as follows:
Bearing of Line DE, 𝛽𝐷𝐸 = 180° − 59°37′ 46.51" − 70°03′ 26.42" = 50°18′ 47.07"
Lines Bearing Distances
Bearing of Line DE, N 𝟓𝟎°𝟏𝟖′ 𝟒𝟕. 𝟎𝟕" E (Ans.)

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Bearing of Line EA, 𝛽𝐸𝐴 = 59°37′ 46.51" − 50°18′ 47.07" = 9°18′ 59.44" Using cosine Law to determine the length at DE.

Bearing of Line EA, S 𝟗°𝟏𝟖′ 𝟓𝟗. 𝟒𝟒" E (Ans.) 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶

′
(83.60)2 = (𝐷𝐸)2 + (54.64)2 − 2(𝐷𝐸)(54.64) cos 61°56 44.02"
4.4. From the data shown below, determine the value of the unknown bearing and DE = 𝟗𝟑. 𝟗𝟗 𝒎 (Ans.)
distance.
To determine the ∅𝐸 , use cosine law,
Lines Bearings Distances
AB N 32° 27’ E 110.8 m 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶
BC - 83.6 m
(54.64)2 = (93.99)2 + (83.60)2 − 2(93.99)(83.60) cos ∅𝐸
CD S 8° 51’ W 126.9 m
DE S 73° 31’ W - ∅𝐸 = 35°13′ 30.74"
EA N 18° 44’ W 90.2 m
Solution: Bearing of Line EC, 𝛽𝐸𝐶 = 180° − 35°13′ 30.74" − 73°31′ = 71°15′ 29.26"

Determine the latitude and departure of the given sides. Bearing of Line EC, S 𝟕𝟏°𝟏𝟓′ 𝟐𝟗. 𝟐𝟔" E (Ans.)

Line Distance Bearing Latitude Departure

EA 90.20 m N 18° 44’ W +85.42 m -28.97 m
AB 110.80 m N 32° 27’ E +93.50 m +59.45 m
BC 126.90 m S 8° 51’ W -125.39 m -19.52m 4.5. In the survey of a closed lot with five sides, the following data are given where
Total +53.53 m +10.96 m in all the bearings and distances of all sides except the lengths of lines 4-5 and
5-1 were omitted. Find the lengths of these two missing lines.
𝐿𝐸𝐶 = √(+53.53)2 + (+10.96)2 Lines Bearing Distance
1–2 S 73° 21’ E 247.20 m
𝐿𝐸𝐶 = 54.64 𝑚
2–3 S 40° 10’ E 154.30 m
−(+10.96)
∅𝐿𝐸𝐶 = tan−1 −(+53.53) 3–4 S 26° 42’ W 611.90 m
4–5 N 14° 20’ W -
∅𝐿𝐸𝐶 = 11°34′ 15.98" 5–1 N 12° 20’ E -
Solution:
Bearing of LEC, N 11°34′ 15.98" E
Determine the latitude and departure of the given sides.
Line Distance Bearing Latitude Departure
∅𝐷 = 73°31′ − 11°34′ 15.98" 1–2 247.20 m S 73° 21’ E -70.83 m +236.84 m
2–3 154.30 m S 40° 10’ E -117.91 m +99.52 m
∅1 = 61°56′ 44.02" 3–4 611.90 m S 26° 42’ W -546.65 m -274.94 m
Total -735.39 m +61.42 m

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𝐿𝐸𝐶 = √(−735.39)2 + (61.42)2

𝐿𝐸𝐶 = 737.95 𝑚
−(+61.42)
∅𝐿𝐸𝐶 = tan−1 −(−735.39)

∅𝐿𝐸𝐶 = 4°46′ 27.41"

Bearing of LEC, S 4°46′ 27.41" E
To compute the distance of 4-5 and 5-1, use the
Law of Sine and Cosine.
To compute line 4-5 and 5-1, first determine ∅𝐷 ,
∅4 and ∅1 .
∅4 = 14°20′ − 4°46′ 27.41"
∅4 = 9°33′ 32.59"
∅1 = 12°20′ + 4°46′ 27.41"
∅1 = 17°06′ 27.41"
∅5 = 180° − 9°33′ 32.59" − 17°06′ 27.41"
∅5 = 153°20′
Using Sine Law:
𝐿𝐸𝐶 4−5 5−1
sin ∅5
= sin ∅ = sin ∅
1 4

737.95 4−5 737.95 5−1

sin 153°20′
= sin 17°06′ 27.41" ;sin 153°20′ = sin 9°33′ 32.59"

𝟒 − 𝟓 = 𝟒𝟖𝟑. 𝟔𝟗 𝒎 𝟓 − 𝟏 = 𝟐𝟕𝟑. 𝟎𝟔 𝒎 (Ans.)

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