Question

In a region with magnetic declination of $2^{\circ }$ E, the magnetic fore Bearing (FB) of a line AB was measured as $N{79}^{\circ }{50}^{\prime }E$. There was local attraction at A. To determine the correct magnetic bearing of the line, a point O was selected at which there was no local attraction. The magnetic FB of line AO and OA were observed to be $S{52}^{\circ }{40}^{\prime }EandN{50}^{\circ }{20}^{\prime }W,$ respectively. What is the true FB of line AB?

• A. $N{77}^{\circ }{50}^{\prime }E$
• B. $N{84}^{\circ }{10}^{\prime }E$
• C. $N{81}^{\circ }{50}^{\prime }E$
• D. $N{82}^{\circ }{10}^{\prime }E$

Answer 1

The correct option is B $N{84}^{\circ }{10}^{\prime }E$

Magnetic declination, $\delta =2^{\circ }E$

Magnetic F.B of AB $=N{79}^{\circ }{50}^{\prime }E\cong {79}^{\circ }{50}^{\prime }$

To find local attractions at station A

As station O is free from local attraction
Hence F.B of OA will be correct.

Correct F.B of OA = $=N{50}^{\circ }{20}^{\prime }W\simeq {309}^{\circ }{40}^{\prime }$

$\therefore \text{Correct B.B of}OA={309}^{\circ }{40}^{\prime }-{180}^{\circ }={129}^{\circ }{40}^{\prime }$

$\because \text{Obs F.B of AO = Obs B.B of OQ}$

$=S{52}^{\circ }{40}^{\prime }E={127}^{\circ }{20}^{\prime }$

Error = M.B-T.B

$={127}^{\circ }{20}^{\prime }-{129}^{\circ }{40}^{\prime }$

$=-2^{\circ }{20}^{\prime }$

$Correction=+2^{\circ }{20}^{\prime }$

$Localattraction,atStationA=+2^{\circ }{20}^{\prime }\cong 2^{\circ }{20}^{\prime }E$

$\therefore \text{Magnetic F.B of AB}=N{79}^{\circ }{50}^{\prime }E$

$\delta =2^{\circ }E,\text{L.A}=2^{\circ }{20}^{\prime }E$

$\therefore \text{True bearing of F.B of AB}=N{79}^{\circ }{50}^{\prime }E+2^{\circ }+2^{\circ }{20}^{\prime }$

$=N{84}^{\circ }{10}^{\prime }E$