Question
Download Solution PDFThe true reduced bearing is N 5°0’ W and the magnetic declination is 2°E. Find the true bearing in the whole circle bearing system.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
TB (True Bearing) = MB (Magnetic bearing) ± Declination
(Use + ve for eastern and – ve for western declination)
Calculation:
Given,
Declination = 2°E
NOTE: There is no use of declination in the question, as we have to just convert the TRUE QB into TRUE WCB
True bearing of Line = N 5°0’ W = 355° 00'
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.