3. DEFLECTION ANGLES TRAVERSE. Given the following traverse sketches: N68 15'E N78 55'W $56°42'w S81*47'w N76 05E 59*00' 233 00' 314 00' 315°00' 253 00' Note: In the third sketch the indicated azimuths are measured from south. Determine tabulate values accordingly. the deflection angles for each traverse and NI2 25'E S32 03'E S13*0d'w $46*17'E N23 28'E

3. DEFLECTION ANGLES TRAVERSE. Given the following traverse sketches: N68 15'E N78 55'W $56°42'w S8147'w N76 05E 5900' 233 00' 314 00' 315°00' 253 00' Note: In the third sketch the indicated azimuths are measured from south. Determine tabulate values accordingly. the deflection angles for each traverse and NI2 25'E S32 03'E S130d'w $4617'E N23 28'E

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Similar questions

A closed field traverse has the following lengthsAB = 636.45; BC = 654.49; CD = 382.85; and DA = 512.77.The interior angles are as follows (measured as angles to the right):A = 81º 23'; B = 72º 33'; C = 89º 40'; and D =116º 24'.The bearing of AB is S 33º 19' W and the side BC is in theSE quadrant.a) Which is the bearing of line CD? b) Which is the bearing of line BC?c) Which nearly gives the value of LEC?
: Determine the graph and conversion of the following angles.1. Convert the following azimuth 190º15.40'45” to heading.2. Graphic 66º7´9”3. Convert the following heading S 74º9'30” W to azimuth.4. Convert the following azimuth 200º46´7.12” to heading.5. Graph 245º52´14”
Convert the following bearings to equivalent north and south azimuths. a. AB, N 28°15' W b. BC, Due East c. EF, N 66°30' W d. CD, S 50°10' E e. DE. S 46°50' W
1Q is 18 7T is 72
determine Mn knowing that Es=0.005294
Interior Angle Traverse Given the bearing of each line of a closed traverse ABCDE, determine the interior angles A, B, C, D, and E. Line Bearing AB N 80°29’ E BC N 39°55’ W CD S 88°04’ W DE S 56°07’ W EA S 61°17’ E * Sum of Interior Angle = (s – 2)180° * Note: EXPRESS THE ANSWER UPTO 3 DECIMAL PLACES. (Please answer this problem with accurate and proven solution. Thank you so much.)
A line of level is run from B.M.I (elevation=106.40m) to B.M. 2 elevation=110.50m). The lield observations were recorded as given below:
Convert azimuth reckoned from north and south to bearings AzN = AzS = 230°30' a. N 39°30' E and S 39°30'W b. N 50°30' W and S 50°30' E c. N 50°30' E and S 50°30' W
Solve for the missing measurements (Bearings of line BC and DE) given the data for a closed traverse. Line Length Bearing AB 529.6 N 48° 20' E BC 592 ? CD 563.6 S 7° 59' E DE 753.4 ? EA 428.2 N 48° 12' W
A magnetic bearing of N5430’E was observed along a line XY in July1982. The declination for the area surveyedis found by interpolation from isogonic chartdated 1975 to be 1730’W with an annualchange of 01’ westward. 1. Compute the change of declination ofthe line.a. 007’W c. 037’Wb. 730’W d. 023’W 2. Compute the true bearing of the line.a. 7153’ c. 7207’b. 3700’ d. 3707
PROBLEM 1 Given the following interior angles (angles-to-the-right) of a five-sided closedpolygon traverse and if the azimuth of side AB is fixed at 74°31'17", determine the adjusted azimuth of line EA.A =105°13'14"; B = 92°36'06"; C = 67°15'22"; D = 217°24'30"; E = 57°30"38". (Note: line BC bears NW.) THE ANSWER HERE IS Azimuth EA=149.300278 NEEDED TO ANSWER it is connected to problem 1: Compute the relative precision for the traverse of Problem 1 if the lengths of the sides (in feet) are as follows: AB = 2157.34; BC =1722.58;CD =1318.15; DE =1536.06; and EA =1785.58. (Note: Assume units of feet for all distances.) Round off the denominator to two decimal places.
How to come up to Iz (m,n)= 0.10685