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A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 What is the bearing of line 5-1?

A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 What is the bearing of line 5-1?

Structural Analysis
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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b.) RE = perimeter of all courses Perimeter = 95 + 135 + 47 + 68 Perimeter = 345 m 1.295 RE = 345 266 Technology Drven by (nn Problem 2 1 • A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 • What is the bearing of line 5-1? Technology Driven by Jnn Solution Course Bearing Distance Lat Dep 1-2 N 9.27° E 58.7 m + 57.933 + 9.456 2-3 S 88.43° E 27.3 m 0.748 + 27.290 3-4 N 86.78° E 35.2 m 1.977 + 35.144 4-5 S 5.2° E 35.0 m 34.856 3.172 + 5-1 ----- The respective sums of Departure and Latitude must be zero. Latz-1 = - (57.933 - 0.748 + 1.977 - 34.856) =- 24.307 Deps1 = - (9.456 + 27.290 + 35.144 + 3.172) = - 75.062 Distance = v24.3072 +75.0622 = 78.900 m Bearing = S arctan pe w= S arctan(25.062) w 24.307) Dep (75.062Y Lat Bearing = S 72.057° W Technolagy Driven by (nn. Compass Rule • the correction to be applied in the latitude or departure of any course is the total correction in latitude or departure multiply to the ratio length of the course is to
Transcribed Image Text:b.) RE = perimeter of all courses Perimeter = 95 + 135 + 47 + 68 Perimeter = 345 m 1.295 RE = 345 266 Technology Drven by (nn Problem 2 1 • A close traverse has the following data: Course Bearing Distance 1-2 N 9.27° E 58.7 m 2-3 S 88.43° E 27.3 m 3-4 N 86.78° E 35.2 m 4-5 S 5.2° E 35.0 m 5-1 10-11 / 27 • What is the bearing of line 5-1? Technology Driven by Jnn Solution Course Bearing Distance Lat Dep 1-2 N 9.27° E 58.7 m + 57.933 + 9.456 2-3 S 88.43° E 27.3 m 0.748 + 27.290 3-4 N 86.78° E 35.2 m 1.977 + 35.144 4-5 S 5.2° E 35.0 m 34.856 3.172 + 5-1 ----- The respective sums of Departure and Latitude must be zero. Latz-1 = - (57.933 - 0.748 + 1.977 - 34.856) =- 24.307 Deps1 = - (9.456 + 27.290 + 35.144 + 3.172) = - 75.062 Distance = v24.3072 +75.0622 = 78.900 m Bearing = S arctan pe w= S arctan(25.062) w 24.307) Dep (75.062Y Lat Bearing = S 72.057° W Technolagy Driven by (nn. Compass Rule • the correction to be applied in the latitude or departure of any course is the total correction in latitude or departure multiply to the ratio length of the course is to
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