Explanation Calculation: Calculate the sum of interior angles. The polygon is of 5 sides. Calculate the sum of interior angle. Sum of interior angle = ( 5 − 2 ) × 180 ° = 540 ° Calculate the Misclosure of the polygon. Misclosure = ( Actual reading ) − ( Calculated reading ) = ( Sum of all angles ) − ( Calculated reading ) …… (I) Calculate the sum of all angles. Sum = 57 ° 0 0 ′ 5 0 ″ + 88 ° 2 4 ′ 2 5 ″ + 126 ° 3 6 ′ 5 8 ″ + 46 ° 0 3 ′ 2 5 ″ + 221 ° 5 3 ′ 5 2 ″ = 539 ° 5 9 ′ 3 0 ″ Substitute 540 ° for calculated reading, and 539 ° 5 9 ′ 3 0 ″ for actual reading in Equation (I). Misclosure = 539 ° 5 9 ′ 3 0 ″ − 540 ° = − 3 0 ″ Calculate the correction. Correction = Misclosure Number of angles …… (II) Substitute 3 0 ″ for misclosure, and 5 for number of angles. Correction = 3 0 ″ 5 = + 6 ″ Since, misclosure is negative, therefore correction must be positive. Calculate the corrected anglesA. ∠ A corrected = ∠ A + Correction Substitute 57 ° 0 0 ′ 5 0 ″ for ∠ A , and + 6 ″ for correction. ∠ A corrected = 57 ° 0 0 ′ 5 0 ″ + 6 ″ = 57 ° 0 0 ′ 5 6 ″ Calculate the corrected anglesB. ∠ B corrected = ∠ B + Correction Substitute 88 ° 2 4 ′ 2 5 ″ for ∠ B , and + 6 ″ for correction. ∠ B corrected = 88 ° 2 4 ′ 2 5 ″ + 6 ″ = 88 ° 2 4 ′ 3 1 ″ Calculate the corrected anglesC. ∠ C corrected = ∠ C + Correction Substitute 126 ° 3 6 ′ 5 8 ″ for ∠ C , and + 6 ″ for correction. ∠ C corrected = 126 ° 3 6 ′ 5 8 ″ + 6 ″ = 126 ° 3 7 ′ 4 ″ Calculate the corrected anglesD. ∠ D corrected = ∠ D + Correction Substitute 46 ° 0 3 ′ 2 5 ″ for ∠ D , and + 6 ″ for correction. ∠ D corrected = 46 ° 0 3 ′ 2 5 ″ + 6 ″ = 46 ° 0 3 ′ 3 1 ″ Calculate the corrected anglesE. ∠ E corrected = ∠ E + Correction Substitute 221 ° 5 3 ′ 5 2 ″ for ∠ D , and + 6 ″ for correction. ∠ E corrected = 221 ° 5 3 ′ 5 2 ″ + 6 ″ = 221 ° 5 3 ′ 5 8 ″ Calculate the azimuth of lineBA. A z B A = A z A B + 180 ° …… (III) Here, azimuth atBAis, A z B A . Substitute 35 ° 0 9 ′ 3 2 ″ for A z A B in Equation (III). A z B A = 35 ° 0 9 ′ 3 2 ″ + 180 ° = 215 ° 9 ′ 3 2 ″ Calculate the azimuth of lineBC. A z B C = A z B A + ∠ B …… (IV) Here, azimuth atBCis, A z B C . Substitute 215 ° 9 ′ 3 2 ″ for A z B A , and 88 ° 2 4 ′ 3 1 ″ for ∠ B in Equation (IV). A z B C = 215 ° 9 ′ 3 2 ″ + 88 ° 2 4 ′ 3 1 ″ = 303 ° 3 4 ′ 3 ″ Calculate the azimuth of lineCB. A z C B = A z B C − 180 ° …… (V) Here, azimuth atCBis, A z C B . Substitute 303 ° 3 4 ′ 3 ″ for A z B C in Equation (V). A z B C = 303 ° 3 4 ′ 3 ″ − 180 ° = 123 ° 3 4 ′ 3 ″ Calculate the azimuth of lineCD. A z C D = A z C B + ∠ C …… (VI) Here, azimuth atCDis, A z C D . Substitute 123 ° 3 4 ′ 3 ″ for A z C B , and 126 ° 3 7 ′ 4 ″ for in Equation (VI). Calculat A z D C = A z C D − 180 ° ∠ ; C e the azimuth of lineDC. …… (VII) Here, azimuth atDCis, A z D C . Substitute 250 ° 1 1 ′ 7 ″ for A z C D in Equation (VII). A z D C = 250 ° 1 1 ′ 7 ″ − 180 ° = 70 ° 1 1 ′ 7 ″ Calculate the azimuth of lineDE. A z D E = A z D C + ∠ D …… (VIII) Here, azimuth atDEis, A z D E . Substitute 70 ° 1 1 ′ 7 ″ for A z D C , and 46 ° 0 3 ′ 3 1 ″ for ∠ D in Equation (VIII)

The departures and latitudes, linear misclosure, and relative precision for the traverse.

Elementary Surveying: An Introduction To Geomatics (15th Edition)

15th Edition

ISBN: 9780134604657

Author: Charles D. Ghilani

Publisher: PEARSON

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Chapter 10, Problem 10.10P

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The departures and latitudes, linear misclosure, and relative precision for the traverse.

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