Solutions to Lecture Examples Directions and Traverse Relationships

Solutions to Lecture Examples Directions and Traverse

Relationships between readings FL and FR

28” 13 2 1 2 =360 0 0 +41 36 19 -249 55 02 41 20 2

Vertical angles 49 54 2 2 32 -2 34 -2

Example (1) Calculate the reduced azimuth of the lines AB and AC, then calculate the reduced azimuth (bearing) of the lines AD and AE Line AB AC AD AE Azimuth 120° 40’ 310° 30’ Reduced Azimuth (bearing) S 85 ° 10’ W N 85 ° 10’ W

Example (1)-Answer Line Azimuth AB 120° 40’ Reduced Azimuth (bearing) S 59° 20’ E AC 310° 30’ N 49° 30’ W AD 256° 10’ S 85° 10’ W AE 274° 50’ N 85° 10’ W

Example (2) Compute the azimuth of the line : - AB if Ea = 520 m, Na = 250 m, Eb = 630 m, and Nb = 420 m - AC if Ec = 720 m, Nc = 130 m - AD if Ed = 400 m, Nd = 100 m - AE if Ee = 320 m, Ne = 370 m

Example (2)-Answer Line ΔE ΔN Quad. Calculated bearing Azimuth tan-1(ΔE/ ΔN) tan-1( 1 st 32° 54’ 19” AB 110 170 AC 200 -120 2 nd -59° 02’ 11” 120° 57’ 50” AD -120 -150 3 rd 38° 39’ 35” 218° 39’ 35” AE -200 120 -59° 02’ 11” 300° 57’ 50” 4 th

Example (3) The coordinates of points A, B, and C in meters are (120. 10, 112. 32), (214. 12, 180. 45), and (144. 42, 82. 17) respectively. Calculate: a) The departure and the latitude of the lines AB and BC b) The azimuth of the lines AB and BC. c) The internal angle ABC d) The line AD is in the same direction as the line AB, but 20 m longer. Use the azimuth equations to compute the departure and latitude of the line AD.

Example (3) Answer B A C a) Dep. AB = ΔEAB = 94. 02, Lat. AB = ΔNAB = 68. 13 m Dep. BC = ΔEBC = -69. 70, Lat. BC = ΔNBC = -98. 28 m b) Az. AB = tan-1 (ΔE/ ΔN) = 54 ° 04’ 18” Az. BC = tan-1 (ΔE/ ΔN) = 215 ° 20’ 39” c) clockwise : Azimuth of BC = Azimuth of AB - The angle B +180° Angle ABC = AZAB- AZBC + 180° = = 54 ° 04’ 18” - 215 ° 20’ 39” +180 = 18° 43’ 22”

d) AZAD: The line AD will have the same direction (AZIMUTH) as AB = 54° 04’ 18” LAD = (94. 02)2 + (68. 13)2 = 116. 11 m Calculate departure = ΔE = L sin (AZ) = 94. 02 m latitude = ΔN= L cos (AZ)= 68. 13 m

Example (4) E A 115 105 110 B 120 D 30 90 C In the right polygon ABCDEA, if the azimuth of the side CD = 30° and the internal angles are as shown in the figure, compute the azimuth of all the sides and check your answer.

Example (4) - Answer A 115 E 105 CHECK : Bearing of CD = Bearing of BC + Angle C + 180 = 120 + 90 + 180 = 30 (subtracted from 360), O. K. 30 120 B 90 Bearing of DE = Bearing of CD + Angle D + 180 = 30 + 110 + 180 = 320 Bearing of EA = Bearing of DE + Angle E + 180 = 320 + 105 + 180 = 245 (subtracted from 360) Bearing of AB = Bearing of EA + Angle A + 180 = 245 + 115 + 180 = 180 (subtracted from 360) Bearing of BC = Bearing of AB + Angle B + 180 =180 + 120 + 180 = 120 (subtracted from 360) C 110 D

Coordinate Computations Point Line Azimuth DE = DN = Length (a) d sin( a) d cos( a) A N 200. 00 350. 00 AB 100. 10 0° 00'' 0. 00 100. 10 B 200. 00 BC 100. 00 90° 00' 00" 100. 00 450. 10 0. 00 C 300. 00 450. 10 CD 100. 00 180° 00'00" 0. 00 -100. 00 D 300. 00 350. 10 DA 99. 70 270° 00'00” - 99. 70 0. 00 A Sum E 200. 30 350. 10 399. 80 0. 30 0. 10

Pnt. Length Azimuth Departure = L sin (AZ) Latitude = L cos (AZ) Correction Departure (WE/ L)* L L Latitude Balanced Dep. 125. 72 255. 88 104 35’ 195 30. 1’ 590. 77 153. 74 - - 0. 13 0. 18 590. 64 358 18. 5’ - 192. 56 - 694. 27 - 0. 16 0. 21 -192. 72 54. 1 306 - 5. 99 202. 91 -0. 04 0. 06 6. 03 - 517. 4 - 388. 5 -0. 14 0. 19 -517. 54 A Sum 10102. 40 10523. 58 9408. 34 10517. 54 9611. 31 10000. 00 202. 97 388. 69 check = L 24 66. 05 10716. 29 -694. 06 E 647. 02 10255. 96 -153. 56 D 203 10125. 66 0. 08 125. 66 255. 96 =(0. 54/2466. 06)x 285. 1 =(0. 72/2466. 06)x 285. 1 C 720. 48 10000. 00 - 0. 06 B 610. 45 10000. 00 (WN/ L)* L AZ 26 10’ N Lat. A 285. 1 E W =+0. 54 E WN=0. 72 -0. 54 0. 72 0. 00