Correct Answer - Option 3 : N84°10’E
Magnetic Declination, δ =2°E
Magnetic FB of AB = N 79°50’E
To find local attraction at station A
As station O is free from local attraction
Hence FB of OA will be correct
Correct FB of OA = N 50°20’W @ 309° 40’
∴ Correct BB of OA = @ 309° 40’ - 180° = 129° 40’
∵ Observed FB of AO= Observed BB of OA= S52° 40’E= 127°20’
Error = MB – TB = 127° 20’ – 129° 40’ = - 2° 20’
Correction = + 2° 20’
Local attraction at station A
= + 2°20’ @ 2°20’ E
∴ Magnetic F.B of AB = N 79° 50’ E
δ = 2° E and local attraction = 2° 20’E
∴ True Bearing of FB of AB = 79° 50’ + 2° 20’ + 2° = N84° 10’E