Item 1Consider a five-sided closed polygon traverse. The azimuth of side AB isfixed at 218° 59'30". A = 132°47'06", B = 108°46'18",C = 107°19'37", D = 81°50′36", E = 109°16′18". (Note: line BCbears SE.) The lengths of the sides (in feet) are as follows: AB = 202.74,BC=283.87, CD = 498.37, DE = 320.33, and EA = 380.78. Thecoordinates of station A are X = 20, 000.00 ft and Y= 15, 000.00 ft.Part AUse the coordinate method to compute the area enclosed by the traverse above.Express your answer in feet squared to two significant figures. Note: Traverse does not need to be adjusted to determine the correct answer.area =SubmitProvide Feedback197| ΑΣΦRequest Answervec?ft²

Interior angleSum of the Interior Angle of the Traverse = 539°59'55"True sum of 5 sided Traverse = 540°Total correction = + 00°00'05"Correction per angle = 00°00'05" / 2= 00°00'01"Corrected Interior Angle = Interior Angle + Correction per angleFor Anticlockwise TraverseForward Bearing = corrected Interior Angle + Back Bearing of previous LineForward Bearing of line AB = 218°59'30"Back Bearing of Line AB = 218°59'30" - 180°= 38°59'30"LATITUDE = L cosθDEPARTURE = L sinθL is the length of the courseθ is the bearing of the coursearea by DMD MethodDMD of first course = DEPARTURE of the courseDMD of remaining course = DEPARTURE of the course + DEPARTURE of the previous course + Previous DMDDA = DMD * Adjusted LATITUDE of the courseArea = DA/2Area = 326250.909 / 2= 163125.45