Explanation Given Information: Coordinates of point A is ( 1000 , 1000 ) . Concept Used: X P = X A + L AP × sin ( A z i m u t h AP ) ....... (I) Here, X P and X A are the x-coordinates of point A and P respectively, L AP is the length of the line AP and A z i m u t h AP is the azimuth of line AP. Y P = Y A + L AP × cos ( A z i m u t h AP ) ....... (II) Here, Y P and Y A are the x-coordinates of point A and P respectively. Calculation: Determine the azimuth of line AB. Azimuth of line AB = 1 ° 0 ' 0 " (given in the figure). Coordinates of station B: Using equation (I) X B = X A + L AB × sin ( A z i m u t h AB ) Substitute 1000 for X A , 400.01 ft for L AB and 1 ° for A z i m u t h AB . X B = 1000 + 400.01 × sin ( 1 ° ) = 1000 + 400.01 × ( 0.017 ) = 1000 + 6.981 = 1006.981 Using equation (II) Y B = Y A + L AB × cos ( A z i m u t h AB ) Substitute 1000 for Y A , 400.01 ft for L AB and 1 ° for A z i m u t h AB . Y B = 1000 + 400.01 × cos ( 1 ° ) = 1000 + 400.01 × ( 0.999 ) = 1000 + 399.95 = 1399.95 Therefore, coordinates of point B = ( 1006.981 , 1399.95 ) . Calculate the azimuth of line BC. Azimuth of line BC = ( 180 ° − 89 ° 59 ' ) = 90 ° 1 ' Coordinates of station C, Using equation (I) X C = X B + L BC × sin ( A z i m u t h BC ) Substitute 1006.981 for X B , 430 ft for L BC and 90 ° 1 ' for A z i m u t h BC . X C = 1006.981 + 430 × sin ( 90 ° 1 ' ) = 1006.981 + 430 × ( 0.999 ) = 1006.981 + 429.999 = 1436.980 Using equation (II) Y C = Y B + L BC × cos ( A z i m u t h BC ) Substitute 1399.95 for Y A , 430 ft for L BC and 90 ° 1 ' for A z i m u t h BC . Y C = 1399.95 + 430 × cos ( 90 ° 1 ' ) = 1399.95 + 430 × ( − 0.0003 ) = 1399.95 − 0.125 = 1399.825 Therefore, coordinates of point C = ( 1436.980 , 1399.825 ) . Azimuth of line AD = ( 180 ° − 89 ° 59 ' ) = 90 ° 1 ' Coordinates of station J: Using equation (I) X J = X A + L AJ × sin ( A z i m u t h AJ ) Substitute 1000 for X A , 200 ft for L AJ and 90 ° 1 ' for A z i m u t h AJ . X J = 1000 + 200 × sin ( 90 ° 1 ' ) = 1000 + 200 × ( 0.999 ) = 1000 + 199.99 = 1199.99 Using equation (II) Y J = Y A + L AJ × cos ( A z i m u t h AJ ) Substitute 1000 for Y A , 200 ft for L AJ and 90 ° 1 ' for A z i m u t h AJ . Y J = 1000 + 200 × cos ( 90 ° 1 ' ) = 1000 + 200 × ( − 0.0003 ) = 1000 − 0.058 = 999.942 Therefore, coordinates of point J = ( 1199.99 , 999.942 ) . Coordinates of station K: Using equation (I) X K = X A + L AK × sin ( A z i m u t h AK ) Substitute 1000 for X A , 230 ft for L AK and 90 ° 1 ' for A z i m u t h AK . X K = 1000 + 230 × sin ( 90 ° 1 ' ) = 1000 + 230 × ( 0.999 ) = 1000 + 229.99 = 1229.99 Using equation (II) Y K = Y A + L AK × cos ( A z i m u t h AK ) Substitute 1000 for Y A , 230 ft for L AK and 90 ° 1 ' for A z i m u t h AK . Y K = 1000 + 230 × cos ( 90 ° 1 ' ) = 1000 + 230 × ( − 0.0003 ) = 1000 − 0.067 = 999.933 Therefore, coordinates of point K = ( 1229.99 , 999.933 ) . Coordinates of station D: Using equation (I) X D = X A + L AD × sin ( A z i m u t h AD ) Substitute 1000 for X A , 430 ft for L AD and 90 ° 1 ' for A z i m u t h AD . X D = 1000 + 430 × sin ( 90 ° 1 ' ) = 1000 + 430 × ( 0.999 ) = 1000 + 429.57 = 1429.57 Using equation (II) Y D = Y A + L AD × cos ( A z i m u t h AD ) Substitute 1000 for Y A , 430 ft for L AD and 90 ° 1 ' for A z i m u t h AD . Y D = 1000 + 430 × cos ( 90 ° 1 ' ) = 1000 + 430 × ( − 0.0003 ) = 1000 − 0.125 = 999.875 Therefore, coordinates of point D is ( 1429.57 , 999.875 ) . Calculate the azimuth of line BG. A z i m u t h B G = ( 180 ° + A z i m u t h A B ) − ∠ A B G = ( 180 ° + 1 ° ) − 43 ° 19 ' = 181 ° − 43 ° 19 ' = 137 ° 41 ' 0 " Calculate ∠ C B G . ∠ C B G = A z i m u t h B G − A z i m u t h B C = 137 ° 41 ' − 90 ° 1 ' = 47 ° 40 ' Use Angle Sum Property in Δ B C G to determine ∠ B G C . ∠ B G C + ∠ C B G + ∠ B C G = 180 ° ∠ B G C + 47 ° 40 ' + 46 ° 41 ' = 180 ° ∠ B G C = 180 ° − 47 ° 40 ' + 46 ° 41 ' = 85 ° 39 ' 0 " Use Sine Law in Δ B C G . ( B G sin ( ∠ B C G ) ) = ( B C sin ( ∠ B G C ) ) B G = ( B C sin ( ∠ B G C ) ) × sin ( ∠ B C G ) = ( 430 ft sin ( 85 ° 39 ' 0 " ) ) × sin ( 46 ° 41 ' ) = ( 430 0

The coordinates of all the labeled corners in the figure.

ELEMENTARY SURVEYING (LOOSELEAF)

15th Edition

ISBN: 9780134604701

Author: GHILANI

Publisher: PEARSON

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Area Computations

The area is the mathematical estimation of the inner space occupied by any shape or object and area computation is the evaluation of how much inner space that particular shape or object has utilized. Simple applications include calculating the area of wa…

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Chapter 11, Problem 11.18P

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The coordinates of all the labeled corners in the figure.

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Chapter 11, Problem 11.17P

Chapter 11, Problem 11.19P

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P.3.3 Oil of sp.gr. 0.9 flows through a vertical pipe (upwards). Two points A and B one above the other 40 cm apart in a pipe are connected by a U-tube carrying mercury. If the difference of pressure between A and B is 0.2 kg/cm², 1- Find the reading of the manometer. 2- If the oil flows through a horizontal pipe, find the reading in manometer for the same difference in pressure between A and B. Ans. 1- R= 0.12913 m, 2- R = 0.1575 m,

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