A Close traverse conducted with five station A,B,C,D and E taken in anticlockwise order in the form of regular pentagon if the F.B of AB is 30o find the F.B of other side.

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written 6.3 years ago by

saxena_archit • 760

modified 6.1 years ago by

sanketshingote • 100

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Interior angle of pentagon$=\frac{(2N-4)\times90^\circ}{5}=\frac{540}{5}=$108$^\circ$

$\text{F.B of AB}=30^\circ$

$\text{F.B of BC}=\text{BB of AB} + \angle B\\ \phantom{\text{F.B of BC}}=(30^\circ0'+180^\circ0')+108^\circ0'\\ \phantom{\text{F.B of BC}}=210^\circ0'+108^\circ0\\ \phantom{\text{F.B of BC}}=318^\circ0'$

$\text{F.B of CD}=\text{BB of BC} + \angle C\\ \phantom{\text{F.B of BC}}=(318^\circ0'-180^\circ0')+108^\circ0'\\ \phantom{\text{F.B of BC}}=138^\circ0'+108^\circ0\\ \phantom{\text{F.B of BC}}=246^\circ0'$

$\text{F.B of DE}=\text{BB of CD} + \angle D\\ \phantom{\text{F.B of BC}}=(246^\circ0'+180^\circ0')+108^\circ0'\\ \phantom{\text{F.B of BC}}=66^\circ0'+108^\circ0\\ \phantom{\text{F.B of BC}}=174^\circ0'$

$\text{F.B of EA}=\text{BB of DE} +\text{Exterior} \angle E\\ \phantom{\text{F.B of BC}}=(174^\circ0'+180^\circ0')-(360^\circ0'-108^\circ0')\\ \phantom{\text{F.B of BC}}=354^\circ0'+252^\circ0\\ \phantom{\text{F.B of BC}}=102^\circ0'$

$\text{F.B of AB}=\text{BB of EA} +\text{Exterior} \angle A\\ \phantom{\text{F.B of BC}}=(102^\circ0'+180^\circ0')-(360^\circ0'-108^\circ0')\\ \phantom{\text{F.B of BC}}=282^\circ0'+252^\circ0\\ \phantom{\text{F.B of BC}}=30^\circ0'$

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