Name: COMPETENTE, JOHN ANGELO C.

Course/Year: BSCE- 2A
Activity 32 - Omitted Measurement for Adjoining Courses (Case 1)

Given the following data for a closed traverse, compute the length
and bearing of unknown side BC.
Course Bearing Length

AB N23˚13'E 56.45m

BC Unknown Unknown

CD S63˚25'W 62.01m

DA N15˚54'W 46.23m
Solution:
Computation of Latitude on each course:
AB = (56.45)(cos 23˚13′) = +51.88
CD = (62.01)(cos 63˚25′) = −27.75
DA = (46.23)(cos 15˚54′) = +44.46
Computation of departure on each course"
AB = (56.45)(sin 23˚13′) = +22.25
CD = (62.01)(sin 63˚25′) = −55.45
DA = (46.23)(sin 15˚54′) = −12.67

Course Bearing Length Latitude Departure

AB N23˚13'E 56.45m +51.88 +22.25

BC Unknown Unknown - -

CD S63˚25'W 62.01m −27.75 −55.45

DA N15˚54'W 46.23m +44.46 −12.66

Sum +68.59 −45.86

Length of BC:
BC = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐵𝐶 = √(+68.59)2 + (−45.87)2 = 𝑩𝑪 = 𝟖𝟐. 𝟓𝟏 𝒎

Bearing of BC:
−𝐶 −(−45.87) +45.87
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(+68.59) = tan 𝜃 = −68.59 = tan 𝜃 = 0.6688
𝐿
𝜽 = 𝑺𝟑𝟑°𝟒𝟔′𝑬
Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

Activity 33 - Omitted Measurement for Adjoining Courses (Case 2)

Given the following data for a closed traverse, compute the length and
bearing of the unknown quantities.
Course Bearing Length

AB S72˚36'E 1,113.12m

BC S14˚12'W 1,406.23m

CD S66˚04'W 1,185.74m

DE N27˚35'W Unknown

EA Unknown 1,739.63m
Solution:
Computation of Latitude on each course:
AB = (1,113.12)(cos 72˚36′) = −332.87
BC = (1,406.23)(cos 14˚12′) = −1363.26
CD = (1,185.74)(cos 66˚04′) = −481.02
Computation of departure on each course"
AB = (1,113.12)(sin 72˚36′) = +1062.18
BC = (1,406.23)(sin 14˚12′) = −344.96
CD = (1,185.74)(sin 66˚04′) = −1083.79

Course Bearing Length Latitude Departure

AB S72˚36'E 1,113.12m −332.87 +1062.18

BC S14˚12'W 1,406.23m −1363.26 −344.96

CD S66˚04'W 1,185.74m −481.02 −1083.79

Sum −2177.15 −366.57

Length of DA:
DA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐷𝐴 = √(−2177.15)2 + (−366.57)2
𝐷𝐴 = 2207.79 𝑚

Bearing of DA:
−𝐶 −(−366.57) +366.57
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−2177.15) = tan 𝜃 = +2177.15 = tan 𝜃 = 0.1684
𝐿
𝜃 = 𝑁9°33′𝐸
𝛽 = 27˚35′ + 9°33′ = 𝛽 = 37˚8′

Used Sine Law:

𝐴𝐸 𝐴𝐷 1,739.63 2207.79
= sin 𝛼 = = = sin 𝛼 = 0.7661
sin 𝛽 sin 37˚8′ sin 𝛼
𝛼 = 50°00′

Bearing of EA = 𝑁[180° + 27˚35′ + 50°00′ ]𝑊 = 𝑵𝟐𝟓𝟕°𝟑𝟓′𝑾

𝜙 = 180° − 𝛽 − 𝛼 = 𝜙 = 180° − 37˚8′ − 50°00′ = 𝜙 = 92°52′

Determine length of DE by Sine Law:

𝐷𝐸 𝐴𝐸 DE 1,739.63
= = = = 𝑫𝑬 = 𝟐𝟐𝟔𝟖. 𝟎𝟖 𝒎
sin 𝜙 sin 𝛽 sin 92°52′ sin 50°00′
Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

Activity 34 - Omitted Measurement for Adjoining Courses (Case 3)

Given the following data for a closed traverse, compute the length
of the unknown quantities.
Course Bearing Length

AB S71˚56'E 1,113.23m

BC S15˚57'W 1,469.19m

CD S68˚54'W 1,186.47m

DE N29˚55'W Unknown

EA N61˚27'E Unknown
Solution:

Computation of Latitude on each course:

AB = (1,113.23)(cos 71˚56′) = −345.24
BC = (1,469.19)(cos 15˚57′) = −1412.63
CD = (1,186.47)(cos 68˚54′) = −427.13
Computation of departure on each course"
AB = (1,113.23)(sin 71˚56′) = +1058.34
BC = (1,469.19)(sin 15˚57′) = −403.73
CD = (1,186.47)(sin 68˚54′) = −1106.92

Course Bearing Length Latitude Departure

AB S72˚36'E 1,113.12m −345.24 +1058.34

BC S14˚12'W 1,406.23m −1412.63 −403.73

CD S66˚04'W 1,185.74m −427.13 −1106.92

Sum −2185 −452.31

Length of DA:
DA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐷𝐴 = √(−2185)2 + (−452.31)2
𝐷𝐴 = 2231.32 𝑚
Bearing of DA:
−𝐶 −(−452.31) +452.31
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−2185) = tan 𝜃 = +2185 = tan 𝜃 = 0.2070
𝐿
𝜃 = 𝑁11°41′𝐸
Computation of Angles:
Ω = 180° − 29˚55′ − 61˚27′ = Ω = 88°38′
′
ϕ = 29˚55 + 11°41′ = ϕ = 41°36′
β = 180° − 88°38′ − 41°36′ = β = 49°46′
Compute DE by Sine Law:
𝐷𝐸 𝐷𝐴 DE 2231.32
= = = = 𝑫𝑬 = 𝟏𝟕𝟎𝟑. 𝟗𝟐
sin 𝛽 sin Ω sin 49°46′ sin 88°38′

Compute EA by Sine Law:

𝐷𝐴 𝐸𝐴 2231.32 𝐸𝐴
= = = = 𝑬𝑨 = 𝟏𝟒𝟖𝟏. 𝟖𝟓
sin Ω sin ϕ sin 88°38′ sin 41°36′
Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

Activity 35 - Omitted Measurement for Adjoining Courses (Case 4)

Given the following data for a closed traverse, compute the bearing
of the unknown quantities.
Course Bearing Length

AB S74˚54'E 1,113.24m

BC S14˚51'W 1,461.37m

CD S67˚54'W 1,188.84m

DE Unknown 1,181.95m

EA Unknown 1,205.52m
Solution:

Computation of Latitude on each course:

AB = (1,113.24)(cos 74˚54′) = −290.00
BC = (1,461.37)(cos 14˚51′) = −1412.56
CD = (1,188.84)(cos 67˚54′) = −447.27
Computation of departure on each course"
AB = (1,113.24)(sin 74˚54′) = +1074.80
BC = (1,461.37)(sin 14˚51′) = −374.53
CD = (1,188.84)(sin 67˚54′) = −1101.49

Course Bearing Length Latitude Departure

AB S74˚54'E 1,113.24m −290.00 +1074.80

BC S14˚51'W 1,461.37m −1412.56 −374.53

CD S67˚54'W 1,188.84m −447.27 −1101.49

Sum −2149.83 −401.22

Length of DA:
DA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐷𝐴 = √(−2149.83)2 + (−401.22)2
𝐷𝐴 = 2186.95 𝑚
Bearing of DA:
−𝐶 −(−401.22) +401.22
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−2149.83) = tan 𝜃 = +2149.83 = tan 𝜃 = 0.1866
𝐿
𝜃 = 𝑁10°34′𝐸
By Cosine Law:
(𝐷𝐴)2 = (𝐷𝐸)2 + (𝐸𝐴)2 − 2(𝐷𝐸)(𝐸𝐴) cos 𝛺
(2186.95)2 = (1,181.95)2 + (1,205.52)2 − 2(1,181.95)(1,205.52) cos 𝛺
𝛺 = 132°41′
Compute Φ by Sine Law:
EA 𝐷𝐴 1,205.52 2186.95
= sin 𝛺 = = sin 132°41′ = 𝜙 = 23°54′
sin 𝜙 sin 𝜙

Bearing of DE = 𝑁[23°54′ − 10°34′ ]𝑊 = 𝑵𝟏𝟑°𝟐𝟎′ 𝑾

Bearing of EA = 𝑁[180° − 132°41′ − 13°20′ ]𝐸 = 𝑵𝟑𝟑°𝟓𝟗′ 𝑬

Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

Activity 36 - Omitted Measurement for Non Adjoining Courses (Case 1)

Given the following data for a closed traverse, compute the unknown quantities.
Course Bearing Length

AB S72˚54'E 1,113.52m

BC S17˚53'W 1,461.41m

CD S65˚55'W Unknown

DE N26˚27'W 1,189.12m

EA Unknown 1,210.21m
Solution:

Computation of Latitude on each course:

AB = (1,113.52)(cos 72˚54′) = −327.42
BC = (1,461.41)(cos 17˚53′) = −1390.80
CF = (1,189.12)(cos 26˚27′) = +1064.65
Computation of departure on each course"
AB = (1,113.52)(sin 72˚54′) = +1064.29
BC = (1,461.41)(sin 17˚53′) = −448.77
CF = (1,189.12)(sin 26˚27′) = −529.65

Course Bearing Length Latitude Departure

AB S72˚54'E 1,113.52m −327.42 +1064.29

BC S17˚53'W 1,461.41m −1390.80 −448.77

CF N26˚27'W 1,189.12m +1064.65 −529.65

Sum −653.57 +85.88

Length of FA:
FA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐹𝐴 = √(−653.57)2 + (+85.88)2
𝐹𝐴 = 659.19 𝑚
Bearing of FA:
−𝐶 −(+85.88) −85.88
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−653.57) = tan 𝜃 = +653.57 = tan 𝜃 = 0.1314
𝐿
𝜃 = 𝑁7°29′𝑊
Interior Angle F:
F = 180° − 7°29′ − 65°55′ = F = 106°36′
By Law of Sines Interior Angles E and A:
sin 𝐹 sin 𝐸 sin 106°36′ sin 𝐸
= = = 659.19 = 𝐸 = 31°27′
𝐸𝐴 𝐹𝐴 1,210.21

𝐴 = 180° − 31°27′ − 106°36′ = 𝐴 = 41°57′

Length of Line EF = CD
sin 𝐹 sin 𝐴 sin 106°36′ sin 41°57′
= = = = 𝑬𝑭 = 𝟖𝟒𝟒. 𝟏𝟗 𝐦 = 𝑪𝑫
𝐸𝐴 𝐸𝐹 1,210.21 𝐸𝐹

Bearing of Line EA:

𝐸𝐴 = 65°55′ − 31°27′ = 𝐄𝐀 = 𝟑𝟒°𝟐𝟖
Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

Activity 37 - Omitted Measurement for Non Adjoining Courses (Case 2)

Given the following data for a closed traverse, compute the unknown quantities.
Course Bearing Length

AB S71˚37'E 1,113.42m

BC S15˚59'W 1,465.612m

CD S65˚52'W Unknown

DE N24˚28'W 1,189.91m

EA N36˚12'E Unknown
Solution:

Computation of Latitude on each course:

AB = (1,113.42)(cos 71˚37′) = −351.14
BC = (1,465.612)(cos 15˚59′) = −1408.95
CF = (1,189.91)(cos 24˚28′) = +1083.06
Computation of departure on each course"
AB = (1,113.42)(sin 71˚37′) = +1056.60
BC = (1,465.612)(sin 15˚59′) = −403.57
CF = (1,189.91)(sin 24˚28′) = −492.82

Course Bearing Length Latitude Departure

AB S71˚37'E 1,113.42m −351.14 +1056.60

BC S15˚59'W 1,465.612m −1408.95 −403.57

CF N24˚28'W 1,189.91m +1083.06 −492.82

Sum −677.03 +160.21

Length of FA:
FA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐹𝐴 = √(−677.03)2 + (+160.21)2
𝐹𝐴 = 695.73 𝑚
Bearing of FA:
−𝐶 −(+160.21) −160.21
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−677.03) = tan 𝜃 = +677.03 = tan 𝜃 = 0.2366
𝐿
𝜃 = 𝑁13°18′𝑊
Interior Angle F, E and A:
F = 360° − ( 180° + 13°18′ + 65°52′ ) = F = 100°50′
𝐸 = 65°52′ − 36˚12′ = 𝐸 = 29˚40′
𝐴 = 180° − ( 29˚40′ + 100°50′ ) = 𝐴 = 49˚30′

By Law of Sines:
sin 𝐴 sin 𝐸 sin 49˚30′ sin 29˚40′
= = = = 𝑪𝑫 = 𝟏𝟎𝟔𝟖. 𝟖𝟔 𝒎
𝐶𝐷 𝐹𝐴 CD 695.73
sin 𝐸 sin 𝐹 sin 29˚40′ sin 100°50′
= = = = 𝑬𝑨 = 𝟏𝟑𝟖𝟎. 𝟔𝟎 𝒎
𝐹𝐴 𝐸𝐴 695.73 𝐸𝐴

Name: COMPETENTE, JOHN ANGELO C. Course/Year: BSCE- 2A

 Activity 38 - Omitted Measurement for Non Adjoining Courses (Case 3)

Given the following data for a closed traverse, compute the unknown quantities.
Course Bearing Length

AB S74˚53'E 1,114.21m

BC S15˚51'W 1,453.82m

CD Unknown 1,761.17m

DE N26˚27'W 1,158.90m

EA Unknown 1,887.74m
Solution:

Computation of Latitude on each course:

AB = (1,114.21)(cos 74˚53′) = −290.57
BC = (1,453.82)(cos 15˚51′) = −1398.55
CF = (1,158.90)(cos 26˚27′) = +1037.59
Computation of departure on each course"
AB = (1,114.21)(sin 74˚53′) = +1075.65
BC = (1,453.82)(sin 15˚51′) = −397.07
CF = (1,158.90)(sin 26˚27′) = −516.19

Course Bearing Length Latitude Departure

AB S74˚53'E 1,114.21m −290.57 +1075.65

BC S15˚51'W 1,453.82m −1398.55 −397.07

CF N26˚27'W 1,158.90m +1037.59 −516.19

Sum −651.53 +162.39

Length of FA:
FA = √(𝐶𝐿 )2 + (𝐶𝐷 )2 = 𝐹𝐴 = √(−651.53)2 + (+162.39)2
𝐹𝐴 = 671.46 𝑚
Bearing of FA:
−𝐶 −(+162.39) −162.39
tan 𝜃 = −𝐶𝐷 = tan 𝜃 = −(−651.53) = tan 𝜃 = +651.53 = tan 𝜃 = 0.2492
𝐿
𝜃 = 𝑁13°59′𝑊
By Law of Cosines for Interior Angle F:
(𝐸𝐴)2 = (𝐶𝐷)2 + (𝐹𝐴)2 − 2(𝐶𝐷)(𝐹𝐴) cos 𝐹
 (1,887.74)2 = (1,761.17)2 + (671.46)2 − 2(1,761.17)(671.46) cos 𝐹
10984.00 = −2365110.42 cos 𝐹 = cos 𝐹 = −0.0046 = 𝐹 = 90°15′
By Law of Sines for Interior Angle E:
sin 𝐹 sin 𝐸 sin 90°15′ sin E
= = = = 𝐸 = 20°50′
𝐸𝐴 𝐹𝐴 1,887.74 671.46
Bearing of Line FE = CD:
FE = 180° − ( 13°59′ + 90°15′ ) = 𝐅𝐄 = 𝐒𝟕𝟓°𝟏𝟏′ 𝑾 = 𝑪𝑫
Bearing of Line EA:
EA = 75°11′ − 20°50′ = 𝐄𝐀 = 𝐍𝟓𝟒°𝟐𝟏′ 𝑬