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Solve ONLY LETTER D and E.  The first three subparts are already answered in my previous question.  See attached photos. Show complete process.

Solve ONLY LETTER D and E.  The first three subparts are already answered in my previous question.  See attached photos. Show complete process.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Solve ONLY LETTER D and E. 

The first three subparts are already answered in my previous question. 

See attached photos. Show complete process. 

Given: Lines Bearing S 73°21'E S 40°10'E S 26°42'W N 14°20’W N 12°20'E Distances А-B 247.20 В-С 154.30 С-D 611.90 D-E E-A Find the following A. Illustrate the give traverse B. Length of course DE. C. Length of course E-A D. Length of the closing line E. Compute the area of the traverse using DPD method.
Transcribed Image Text:Given: Lines Bearing S 73°21'E S 40°10'E S 26°42'W N 14°20’W N 12°20'E Distances А-B 247.20 В-С 154.30 С-D 611.90 D-E E-A Find the following A. Illustrate the give traverse B. Length of course DE. C. Length of course E-A D. Length of the closing line E. Compute the area of the traverse using DPD method.
Get live help whenever you Try bartleby tutor today need from online tutors! Q SEARCH ASK 탄 CHAT Vx MAΤΗ A) A 247.20 B Grien Traverse i54.30 E 611. g0 Step 3 Lines Bearing QSB BEARING WCB DISTANCES LATITUDE DEPARTURE AB S 73 21 E 106.65 247.20 -70.83 236 84 AB S 73 21 E 106.65 247.20 -70.83 236.84 вс S40 10 E 139.83 154.30 -117.91 99.53 CD S26 42 W 206.7 611.90 -546.65 -274.94 DE N14 20 W 345.67 X 0.97 X -0.25 X EA N12 20 E 12.33 Y 0.98 Y 0.21 Y where Latitude = Lcoso Departure = L sine. = [sine. Now Know we Elatitudes = 0 %3D :. -70.83-117-91-546.65+ o-97X +0.98Y => 0.97X t0:98y = 735-39 Similarly, E Depailures 0. ;: 236.84 +99.53 -274.94-0.25X +0.2) Y = 0 -0 252 +0'21Y = - 61.43 0:25R - 0:21Y DY 61.43 Step 4 From i) and (li) we get X = 478-34s Y= 276.g3 b) Length of course DE = 478.345 C) Lengia of Course EA = 276:92 T.
Transcribed Image Text:Get live help whenever you Try bartleby tutor today need from online tutors! Q SEARCH ASK 탄 CHAT Vx MAΤΗ A) A 247.20 B Grien Traverse i54.30 E 611. g0 Step 3 Lines Bearing QSB BEARING WCB DISTANCES LATITUDE DEPARTURE AB S 73 21 E 106.65 247.20 -70.83 236 84 AB S 73 21 E 106.65 247.20 -70.83 236.84 вс S40 10 E 139.83 154.30 -117.91 99.53 CD S26 42 W 206.7 611.90 -546.65 -274.94 DE N14 20 W 345.67 X 0.97 X -0.25 X EA N12 20 E 12.33 Y 0.98 Y 0.21 Y where Latitude = Lcoso Departure = L sine. = [sine. Now Know we Elatitudes = 0 %3D :. -70.83-117-91-546.65+ o-97X +0.98Y => 0.97X t0:98y = 735-39 Similarly, E Depailures 0. ;: 236.84 +99.53 -274.94-0.25X +0.2) Y = 0 -0 252 +0'21Y = - 61.43 0:25R - 0:21Y DY 61.43 Step 4 From i) and (li) we get X = 478-34s Y= 276.g3 b) Length of course DE = 478.345 C) Lengia of Course EA = 276:92 T.
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