AMPAD™ A 67° 17' B C D THE FOUR SIDED CLOSED TRAVERSE HAS THE FOLLOWING ANGLES & DISTAN CES A= 51° 23' B = 105°39' C=78° 11' D= 124°47' AB = 713.93FT, AB BC= 606.06FT CD = 391.27 FT. DA = 781.18 FT. THE AZIMUTH OF AB IS 67°17' ASSIGNING AN ORIGIN TO POINTA (N300, E100) AND USING THE LIN-ADJUSTED LATITUDES & DEPARTURES FROM TRAVERSE HOMEWORK ASSIGNMENTI. J. COMPUTE THE COORDINATES FOR POINTS B, C & D (30 POINTS).

Azimuth sides of traverse ABCDEFA are given below: AB=290°45' BC=250°48' CD=196°12' DE=175°24' EF= 112°18' FA= 30° 0O'Compute the Interior angle OA=80°45' ,OB=140°03' ,0C=125°24' , OD=159°12' ,OE=116°54' OF=97°42' OA=80°45',OB=140°00' ,eC=125°00', OD=159°10' ,OE=116°50' OF=97°40' OA=80°45' ,0B=141°03' ,0C=125°24' , OD=159°12' ,OE=116°54' OF=97°42' EA=80°45' ,OB=140°03' ,eC=125°24' , OD=159°20' ,OE=116°54' OF=97°00' EA=80°45' ,OB=140°30' ,eC=125°24', OD=159°12',OE=116°53 eF=97°42' OA=80°40',OB=140°03' ,eC=125°24', OD=159°12' ,OE=116°54' OF=97°42'

Magnetic Declination: 1. In a compass survey made twenty years ago the observed magnetic bearing of a reference line XY was N 42'21' E when the magnetic declination was 6*25' East. Lately in 2018, a new survey was made and the declination in the same survey site changed to 10'35' West. Determine the following descriptions of line XY: a.) Its true bearing twenty years ago; b.) The true azimuth from North and South; c.) Its new magnetic bearing; and d.) The magnetic azimuth from North and South in 2018.

2. From the closed compass traverse shown, if the forward bearing of line AB is N 78°9' E, compute the: LEC, the bearing of the LEC and the RP. DIST. BEARING LATTUDE DEPARTURE COURSE A-B B-C CD D-E E-A 11520 m 111.30m, 125°40 94 10 E 31 40 117.50 m. 98°20- 119.20 m.

Course Bearing  Length Interior Angle (Right) AB S 50°54'23'' E 329.722 m A = 120°07'10'' BC   210.345 m B = 59°39'10'' CD   279.330 m C = 248°00'57'' DE   283.426 m D = 86°51'04'' EF   433.007 m E = 102°09'16'' FA   307.625 m F = 103°12'41'' Compute and tabulate for the following closed-polygon traverse: (a) preliminary bearings (b) unadjusted departures and latitudes (c) linear misclosure and (d) relative precision. (Note: line BC bears NE.) If one side and/or angle is responsible for most of the error of closure, which is it likely to be?

Part A Compute the azimuth from north of line CD. Azimuth AB = 103°26'32"; angles to the right ABC = 52° 54'26", BCD= 36° 36'38". (Azimuth of AB is also from north.) O 192°25'34" O 192° 57'36" O 192°37'46" O 192° 37'46"

7. Determine whether the planes are parallel, orthogonal, or neither. 4х — 20y - 42%3D8 Зх — 15у - 3z%3D -5 parallel' orthogonal neither

Solve Problem using Cramer's Rule 2х + Зу + 2z %3D -3 y+ z= 0 %3D -х+ 2у - 3z % -1

The numerical value of departure equals the difference between the X coordinates of A and B, i.e. the increment AxAB, while the numerical value of latitude equals the difference between the Y coordinates of A and B, i.e. the increment AYAB- grid azimuth The horizontal distance DAB and the AAB can be computed. Ar - Xg - XxA Ay AB = Ys - YA D - Jar, + Ay Ay Ar AB AB = arctan AB AB

A is a point having an elevation of 125.64 m., B and C are points of unknown elevation, B is in between A andC. By means of an instrument set 1.45 m above B, vertical angles are observed, that to A being an angle of-14º45' and that to C being an angle of +8º32'. The horizontal distance AB is 547.20 and the horizontaldistance BC is 923.43 m. Making due allowance for earth’s curvature and atmospheric refraction, computethe difference in elevation between A and B.