1. Solutions to Lecture Examples Directions and Traverse

2. Relationships between readings FL and FR

3. 28” 13 2 1 2 20 2 =360 0 0 +41 36 19 -249 55 02 41

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

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

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

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

8. Example (2)-AnswerLineΔEΔEΔNΔNQuad. Calculated bearing tan-1( tan-1(ΔE/ ΔN)Azimuth AB1101701st 32° 54’ 19” AC200-1202nd -59° 02’ 11” 120° 57’ 50” AD-120-1503rd 38° 39’ 35” 218° 39’ 35” AE-2001204th -59° 02’ 11” 300° 57’ 50”

9. 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 20m longer. Use the azimuth equations to compute the departure and latitude of the line AD.

10. a)Dep AB = ΔE AB = 94.02, Lat AB = ΔN AB = 68.13m Dep BC = ΔE BC = -69.70, Lat BC = ΔN BC = -98.28m 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 = AZ AB - AZ BC + 180° = = 54 ° 04’ 18” - 215 ° 20’ 39” +180 = 18° 43’ 22” Example (3) Answer A B C

11. d) AZ AD : The line AD will have the same direction (AZIMUTH) as AB = 54° 04’ 18” LAD =  (94.02)2 + (68.13)2 = 116.11m Calculate departure = ΔE = L sin (AZ) = 94.02m latitude = ΔN= L cos (AZ)= 68.13m

12. 120 E C B A 115 90 110 105 30 D Example (4) 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.

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

14. PointLineLength Azimuth  )  E = d sin(  )  N = d cos(  ) EN A 200.00350.00 AB100.100° 00' 00'' 0.00 100.10 B 200.00450.10 BC100.0090° 00' 00"100.00 0.00 C 300.00450.10 CD100.00180°00'00"0.00 -100.00 D 300.00350.10 DA99.70270°00'00”- 99.70 0.00 A 200.30350.10 Sum 399.80 0.30 0.10 Coordinate Computations

15. N E BalancedCorrectionLatitude = L cos (AZ) Departure = L sin (AZ) Azimuth Length Pnt. Lat.Dep.Latitude Departure (W E /  L)* L (W N /  L)* L AZL 10000.00 A 255.96125.660.080.06-255.88125.72 26 10 ’ 285.1 10255.9610125.66 B 153.56-590.640.180.13--153.74590.77 104 35 ’ 610.45 10102.4010716.29 C 694.06-192.72-0.210.16-- 694.27- 192.56195 30.1 ’ 720.48 9408.3410523.58 D 202.97-6.030.060.04-202.91- 5.99358 18.5 ’ 203 9611.3110517.54 E 388.69-517.540.190.14-388.5-517.4306 54.1647.02 10000.00 check A 0.00 0.72-0.54W N =0.72W =+0.54 E =  L 2466.0 5 Sum =(0.54/2466.06)x285.1=(0.72/2466.06)x285.1