Solutions to Lecture Examples Directions and Traverse

Relationships between readings FL and FR

28” =

Vertical angles

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

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

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)

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

Example (3) The coordinates of points A, B, and C in meters are (120.10, ), (214.12, ), 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.

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

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

120 E C B A 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.

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

PointLineLength Azimuth  )  E = d sin(  )  N = d cos(  ) EN A AB ° 00' 00'' B BC ° 00' 00" C CD °00'00" D DA °00'00” A Sum Coordinate Computations

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 A ’ B ’ C ’ D ’ E check A W N =0.72W =+0.54 E =  L Sum =(0.54/ )x285.1=(0.72/ )x285.1