Question

Following bearings are observed while traversing with a compass.

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

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

• A. ${48}^{\circ }{15}^{\prime }$
• B. ${49}^{\circ }{45}^{\prime }$
• C. ${50}^{\circ }{15}^{\prime }$
• D. ${48}^{\circ }{45}^{\prime }$

Answer 1

The correct option is D ${48}^{\circ }{45}^{\prime }$

The angle difference between DE line,

$={258.30}^{\prime }-{78}^{\circ }{30}^{\prime }$

$={180}^{\circ }$

Therefore, the station D and E free from Local Attraction,

$\therefore FBofEA={212}^{\circ }{30}^{\prime }\left(correct\right)$

$\therefore BBofEA,$

$=({212}^{\circ }{30}^{\prime }-180)$

$={32}^{\circ }{30}^{\prime }$

But the observed BB of EA,

$=({32}^{\circ }{30}^{\prime }-{31}^{\circ }{45}^{\circ })$

$=+{45}^{\prime }\left(correctionistobeappliedatA\right)$

$\therefore \text{Correct FB of AB,}$

$=({126}^{\circ }{45}^{\circ }+{45}^{\prime })$

$={127}^{\circ }{30}^{\prime }$

$\therefore \text{Correct BB of AB,}$

$={307}^{\circ }{30}^{\prime }$

So, correction applied,

$={307}^{\circ }{30}^{\prime }-{308}^{\circ }{0}^{\circ }$

$=-{30}^{\prime }\text{which is to be applied at B}$

$\therefore \text{Correct FB of BC}$

$=({49}^{\circ }{15}^{\prime }-{30}^{\prime })={48}^{\circ }{45}^{\prime }$