Question

Following are the bearing of a closed triangular traverse

Line	FB	BB
AB	$S{46}^{\circ }{30}^{\prime }E$	$N{46}^{\circ }{30}^{\prime }W$
BC	$S{50}^{\circ }{10}^{\prime }W$	$N{50}^{\circ }{45}^{\prime }E$
CA	$N{10}^{\circ }{30}^{\prime }E$	$S{11}^{\circ }{05}^{\prime }W$

The correct value of included angle B is

• A. ${298}^{\circ }{15}^{\prime }{00}^{\prime \prime }$
• B. ${83}^{\circ }{20}^{\prime }{00}^{\prime \prime }$
• C. ${277}^{\circ }{03}^{\prime }{20}^{\prime \prime }$
• D. ${276}^{\circ }{40}^{\prime }{00}^{\prime \prime }$

Answer 1

The correct option is C ${277}^{\circ }{03}^{\prime }{20}^{\prime \prime }$

Line	FB	BB
AB	${133}^{\circ }{30}^{\prime }$	${313}^{\circ }{30}^{\prime }$
BC	${230}^{\circ }{10}^{\prime }$	${50}^{\circ }{45}^{\prime }$
CA	${10}^{\circ }{30}^{\prime }$	${191}^{\circ }{05}^{\prime }$

$Includedangle,\angle A=F.BofAB-B.BofCA$

$={133}^{\circ }{30}^{\prime }-{191}^{\circ }{05}^{\prime }+{360}^{\circ }={302}^{\circ }{25}^{\prime }$

$\angle B=F.BofBC-B.BofAB$

${230}^{\circ }{10}^{\prime }-{313}^{\circ }{30}^{\prime }+{360}^{\circ }={276}^{\circ }{40}^{\prime }$

$\angle C=F.BofCA-B.BofBC$

${10}^{\circ }{30}^{\prime }-{50}^{\circ }{45}^{\prime }+{360}^{\circ }={319}^{\circ }{45}^{\prime }$

$Sumofincludedangle={898}^{\circ }{50}^{\prime }$

$Theoreticalsum=(2\times 3+4)\times {90}^{\circ }={900}^{\circ }$

$Error=MV-TV$

$Error={898}^{\circ }{50}^{\prime }-{900}^{\circ }=-1^{\circ }{10}^{\prime }$

$Correctionperangle=+\frac{1^{\circ }{10}^{\prime }}{3}={23}^{\circ }{20}^{\prime \prime }$

$\therefore \angle B={276}^{\circ }{40}^{\prime }+{23}^{\prime }{20}^{\prime \prime }={277}^{\circ }{03}^{\prime }{20}^{\prime \prime }$