Show complete formulas and solutions. Round off to 3 decimal places for lengths.Angles must be in degrees, minutes, and seconds (° ‘ “) and seconds should not include decimals anymore (ex. 1°1’36.52” should be 1°1’37”). A surveyor set his instrument up at point A in order to observe point B. The instrument wasinclined at an angle of +30°23' from the horizontal in order to observe point B. From hisobservations, the difference between the Upper reading and the Lower reading on the rod atpoint B was 2.20m while the Middle reading on the rod was 1.95m. At point A, his instrumentwas measured to be at a height of 1.62m.The telescope used was advertised as follows:Telescope length = 20 cmFocal length = 20 cmStadia interval factor = 95,50What is the SLOPE DISTANCE between the center of the instrument to the middlereading at the rod? (answer in m)WITH FIGURE

To find :D(Inclined) = Slope distance between the center of the instrument…

A surveyor set his instrument up at point Ain order to observe point B. The instrument was inclined at an angle of +30°23' from the horizontal in order to observe point B. From his observations, the diference between the Upper reading and the Lower reading on the rod at point B was 2.20m while the Middle reading on the rod was 1.95m. At point A, his instrument was measured to be at a height of 1.62m. The telescope used was advertised as follows: Telescope length = 20 cm • Focal length = 20 cm Stadia interval factor = 95.50 What is the SLOPE DISTANCE between the center of the instrument to the middle reading at the rod? (answer in m) WITH FIGURE

A surveyor set his instrument up at point Ain order to observe point B. The instrument was inclined at an angle of +30°23' from the horizontal in order to observe point B. From his observations, the diference between the Upper reading and the Lower reading on the rod at point B was 2.20m while the Middle reading on the rod was 1.95m. At point A, his instrument was measured to be at a height of 1.62m. The telescope used was advertised as follows: Telescope length = 20 cm • Focal length = 20 cm Stadia interval factor = 95.50 What is the SLOPE DISTANCE between the center of the instrument to the middle reading at the rod? (answer in m) WITH FIGURE

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Draw the figure clearly and in detail the following data provided below.
Show complete solution and diagram
line from A to B =5 cm on a map with scale =1/2500. This line must be fixed on ground surface if the ground surface is inclined by 5° Find the measured distance using 29.95 tape (True length) to fix the line.
Show complete formulas and solutions. Round off to 3 decimal places for lengths. Angles must be in degrees, minutes, and seconds (° ‘ “) and seconds should not include decimals anymore (ex. 1°1’36.52” should be 1°1’37”).
Make a complete orthographic projection with correct measurements
Q2: A triangle has an angle of (55°50'30") and the second one is (62.4958 g), calculate the third angle with the radian system?
B: Write down all the necessary steps to draw below shape using AutoCAD (Select Any Start Point and draw all Circle with respect to Circle A)) Ø160 Ø50- Ø80 Ø 30 X6 Holes A Ø220 Ø200
W Calculate the end area of the given cross section (using sum of areas method). Use a base line of 260.000m. Keep calculations to 2 decimal points, show final answer to one decimal point. Metric: All distances are shown in meters 262.43 Distance 'A' Distance 'B' Distance 'C' Distance "D" ← 6.48 260.35 !← Distance 'C→ 2.35 Plus Calculations Subtotal → !< 264.53 Area → ← Distance 'A' 260.35 3.65 Minus Calculations drawing not to scale Subtotal →! Distance 'B'→ 262.22 Area Baseline 260.000. Total Area Total
COMPUTE THE MISSING LENGTHS OF TWO SIDES AND DRAW THE DIAGRAM
Use the method of compsite areas to calculate the centroidal coordinates of the plane regions shown Dimensions in mm -20 45 X 100 X = 48.3 mm; Y = 8.8 mm X = 53.3 mm; Y = 10.8 mm X = 58.3 mm; Y = 12.8 mm X = 63.3 mm; Y = 14.8 mm O
Show complete formulas and solutions.Round off to 3 decimal places for lengths. Angles must be in degrees, minutes, and seconds (° ‘ “) and seconds should not include decimals anymore (ex. 1°1’36.52” should be 1°1’37”) COURSE LATITUDE DEPARTURE AB 40 -80 BC 80 -40 CD unknown unknown DA -90 50 a) What is length of course CD? b) What is the bearing of course CD? c) What is the perimeter of the traverse? *PROVIDE ILLUSTRATION