3. Given the following data of a closed traverse LINE DISTANCE (m) BEARING 1-2 31.20 N 15°24' E 2-3 28.50 S 82°32' E 3-4 S 33°09' E 4-5 33.00 5-1 24.23 N 65°21' W a. Determine the distance of line 3-4 b. Determine the bearing of line 4-5
3. Given the following data of a closed traverse LINE DISTANCE (m) BEARING 1-2 31.20 N 15°24' E 2-3 28.50 S 82°32' E 3-4 S 33°09' E 4-5 33.00 5-1 24.23 N 65°21' W a. Determine the distance of line 3-4 b. Determine the bearing of line 4-5
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
Note: Show how to get the values of the angles’ 3, 4, and 5 by using the sine law and by adding (refer on sample problem 2).
Transcribed Image Text:3. Given the following data of a closed traverse
LINE
DISTANCE (m)
BEARING
1-2
31.20
N 15°24' E
2-3
28.50
S 82°32' E
3-4
S 33°09' E
4-5
33.00
5-1
24.23
N 65°21' W
a. Determine the distance of line 3-4
b. Determine the bearing of line 4-5
![Sample Problem 2
A
1,738.96m
E
LATITUDES
DEPARTURES
LINE
LENGTH
BEARING
+N
-S
+E
/14°01'
1084.32 m S 75°48' E
S 15'18' W
AB
-265.9917
1051.1890
28°39
ВС
1590.51
1534.1382
419.6927
S 68*06' W
Unknown N 28 39' W
CD
1294.74
482.9222
-1201.3067
42°40'
DE
28°39'
+ 14°01'
EA
1738.96
Unknown
SUMS
2,283.0521
1,051.1890
1,620.9994
Σεο
WD
D
Determining Length of Line
DE and Bearing of Line EA:
Sum of Interior Angles for Triangle = 180°
ZA = 180° - (42°40' + 66°30')
ZA = 70°50'
ZD = 28°39' + 14°01'
ZD = 42°40'
By Sine Law:
DE
EA
At Station E:
e = 180° - (ZE + 28°39')
e = N 84°51' E
[Bearing of Side EA]
By Sine Law:
sin A
DE =
sin D
DA
E A
[1,738.96m][sin(70°50')]
sin(42°40')
DE = 2,423.6205m
sin E
sin D
sin E
[2,353.0854m][sin(42°40')]
%3D
1,738.96m
ZE = 66°30'
[Length of Side DE]
2,353.0854m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ccb8bf5-509f-4203-88ab-a015a8a5329e%2F07425a2b-6f26-468b-97d6-39112648bfe1%2Fasqgg5h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Sample Problem 2
A
1,738.96m
E
LATITUDES
DEPARTURES
LINE
LENGTH
BEARING
+N
-S
+E
/14°01'
1084.32 m S 75°48' E
S 15'18' W
AB
-265.9917
1051.1890
28°39
ВС
1590.51
1534.1382
419.6927
S 68*06' W
Unknown N 28 39' W
CD
1294.74
482.9222
-1201.3067
42°40'
DE
28°39'
+ 14°01'
EA
1738.96
Unknown
SUMS
2,283.0521
1,051.1890
1,620.9994
Σεο
WD
D
Determining Length of Line
DE and Bearing of Line EA:
Sum of Interior Angles for Triangle = 180°
ZA = 180° - (42°40' + 66°30')
ZA = 70°50'
ZD = 28°39' + 14°01'
ZD = 42°40'
By Sine Law:
DE
EA
At Station E:
e = 180° - (ZE + 28°39')
e = N 84°51' E
[Bearing of Side EA]
By Sine Law:
sin A
DE =
sin D
DA
E A
[1,738.96m][sin(70°50')]
sin(42°40')
DE = 2,423.6205m
sin E
sin D
sin E
[2,353.0854m][sin(42°40')]
%3D
1,738.96m
ZE = 66°30'
[Length of Side DE]
2,353.0854m
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