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Distance Formula

If the end po...

Question

If the end points of a line are free from local attraction, the difference between fore bearing and back bearing of that line should be:

A

$120^{0}$

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B

$180^{0}$

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C

90^{0}

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D

360^{0}

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Solution

The correct option is B
$180^{0}$

(a)
The difference of F.B of line is $180^{0}$ of the end point of the line are free from local attraction.

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Similar questions

Q. The fore bearing of a line AB was observed to be 245

^{o}

30', and it is known that station A is free of local attraction. The declination at the place was found to be 3

^{o}

15'W. Find the true bearing of line BA.

Q. The observed magnetic bearing of a line OE was found to be

185^{\circ}

. It was later discovered that station O had a local attraction of

+ {1.5}^{\circ}

. The true bearing of the line OE, considering a magnetic declination of

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Q. 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^{'} 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^{'} E a n d N 50^{\circ} 20^{'} W,

respectively. What is the true FB of line AB?

Q. In quadrantal bearing system, bearing of a line varies from

Q. Following bearings are observed while traversing with a compass.

Line	Fore Bearing	Back Bearing
AB	$126^{\circ} 45^{'}$	$308^{\circ} 0^{'}$
BC	$49^{\circ} 15^{'}$	$227^{\circ} 30^{'}$
CD	$340^{\circ} 30^{'}$	$161^{\circ} 45^{'}$
DE	$258^{\circ} 30^{'}$	$78^{\circ} 30^{'}$
EA	$212^{\circ} 30^{'}$	$31^{\circ} 45^{'}$

After applying the correction due to local attraction, the corrected fore bearing of BC will be:

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