Question

Savannah State University Department of Engineering Technology Civil Engineering Technology CI (Dr K. Jayaraman) TEST 2 Answer ALL questions. Fall Semester, 2018 Total Marks: 50 Proceeding in a counterclockwise direction around a five-sided polygon ABCDE, the mean observed interior angles are shown in Figure below: Determine the closing error of the traverse. Adjust the interior angles of the traverse to eliminate the closing error Given that the bearing of the side AB is S50° 53' 34W, compute the bearings of all oth traverse. 11e 30 40 162 49 32 86 28 44 62 57-14 9 1315

Answer 1

we know the sum of interier angle of pentagon (5 angles) is 540⁰

sum of given traverse is = 162⁰49^'32^'' + 62⁰57^'14^'' + 109⁰13^'15^'' + 86⁰28^'44^'' + 118⁰30^'40^'' = 539⁰59^'25^''

closing error = 540⁰ - 539⁰59^'25^'' = 0⁰0^'35^''

adjustment of every angle = 0⁰0^'35^'' /5 = 0⁰0^'07^''

so we reduce 0⁰0^'07^'' on each angle

angle A = 162⁰49^'32^'' - 0⁰0^'07^'' = 162⁰49^'25^''

angle B = 62⁰57^'14^'' - 0⁰0^'07^'' = 62⁰57^'07^''

angle C = 109⁰13^'15^'' - 0⁰0^'07^'' = 109⁰13^'08^''

angle D = 86⁰28^'44^'' -0⁰0^'07^'' = 86⁰28^'37^''

angle E = 118⁰30^'40^'' - 0⁰0^'07^'' = 118⁰30^'33^''

Given

bearing of side AB is S 50⁰53^'34^'' W

azimuth = 180⁰ + 50⁰53^'34^'' = 230⁰53^'34^''

Bearing of BA is = 230⁰53^'34^'' - 180⁰ = 50⁰53^'34^''

Bearing of BC = Bearing of BA + $\angle$ B = 50⁰53^'34^'' + 62⁰57^'07^'' = 113⁰31^'41^''

Bearing of CB = Bearing of BC + 180⁰ = 113⁰31^'41^'' + 180⁰ = 293⁰31^'41^''

Bearing of CD = Bearing of CB + $\angle$ C -180 = 293⁰31^'41^'' + 109⁰13^'08^'' - 180⁰ = 222⁰44^'49^''

Bearing of DC = Bearing of CD - 180⁰ = 222⁰44^'49^'' - 180⁰ = 42⁰44^'49^''

Bearing of DE = Bearing of DC + $\angle$ D = 42⁰44^'49^'' + 86⁰28^'37^'' = 129⁰13^'26^''

Bearing of ED = Bearing of DE + 180⁰ = 129⁰13^'26^'' + 180⁰ = 309⁰13^'26^''

Bearing of EA = Bearing of ED + $\angle$ E - 180 = 309⁰13^'26^'' + 118⁰30^'33^'' - 180⁰ = 247⁰43^'59^''