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the following bearing were observed while traversing with compass and tape check the bearing for local attraction and correct the bearing by the method of including angle.
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Line F.B B.B
AB 188$^\circ$45' 7$^\circ$45'
BC 118$^\circ$15' 298$^\circ$15'
CD 346$^\circ$35' 166$^\circ$30'
DE 337$^\circ$05' 158$^\circ$10'
EA 293$^\circ$30' 113$^\circ$00'

To find the correct including angle

$\angle$A = F.B of AB – B.B of EA = 188$^\circ$45’ - 113$^\circ$00’= 75$^\circ$45’

$\angle$B = F.B of BC – B.B of AB = 118$^\circ$15’- 7$^\circ$45’= 110$^\circ$30’

$\angle$C = F.B of CD – B.B of BC = 346$^\circ$35’- 298$^\circ$15’= 48$^\circ$20’

$\angle$D = F.B of DE – B.B of CD = 337$^\circ$05’-166$^\circ$30’= 170$^\circ$35’

$\angle$E = F.B of EA – B.B of DE = 293$^\circ$30’-158$^\circ$10’= 135$^\circ$20’

Sum of all included angle = $\angle$A + $\angle$B + $\angle$C + $\angle$D + $\angle$E = 75$^\circ$45’ + 110$^\circ$30’ + 48$^\circ$20’ + 170$^\circ$35’ + 135$^\circ$20’ = 540$^\circ$30’

Theoretical sum= (2n – 4) × 90$^\circ$ = (2 × 5 – 4) × 90$^\circ$ = 540$^\circ$

Error = 540$^\circ$ - 540$^\circ$30’ = - 30’ (-ve error)

Error divide into five station = (- 30’ )/5 = - 6’ at each station

Find correct angle

$\angle$A = 75$^\circ$45’ - 6’ = 75$^\circ$39’

$\angle$B = 110$^\circ$30’ - 6’ = 110$^\circ$24’

$\angle$C =48$^\circ$20’ - 6’ = 48$^\circ$14’

$\angle$D = 170$^\circ$35’ - 6’ = 170$^\circ$29’

$\angle$E = 135$^\circ$20’ - 6’ = 135$^\circ$14’

To find correct bearing

1. difference between F.B and B.B of BC is 180$^\circ$

BB Of AB = F.B of BC – $\angle$B (corrected angle) = 118$^\circ$15’ – 110$^\circ$24’ = 7$^\circ$51’ F.B Of AB = 7$^\circ$51’ + 180$^\circ$ = 187$^\circ$51’

BB Of EA = F.B of AB – $\angle$A (corrected angle ) = 187$^\circ$51’– 75$^\circ$39’= 112$^\circ$12’ F.B Of EA = 112$^\circ$12’+ 180$^\circ$ = 292$^\circ$12’

BB Of DE = F.B of EA – $\angle$E (corrected angle ) = 292$^\circ$12’– 135$^\circ$14’= 156$^\circ$58’

F.B Of DE = 156$^\circ$58’+ 180$^\circ$= 336$^\circ$58’

BB Of CD = F.B of DE – $\angle$D (corrected angle ) = 336$^\circ$58’– 170$^\circ$29’= 166$^\circ$29’

F.B Of CD = 166$^\circ$29’+ 180$^\circ$ = 346$^\circ$29’

BB Of BC = F.B of CD– $\angle$C (corrected angle ) = 346$^\circ$29’– 48$^\circ$14’= 298$^\circ$15’

Line Incorrect Incorrect included angle -ve error correct included error Corrected
F.B B.B F.B B.B
AB 188$^\circ$45' 7$^\circ$45’ 75$^\circ$45’ 6' $\angle$A = 75$^\circ$39’ 187$^\circ$51’ 7$^\circ$51’
BC 118$^\circ$15’ 298$^\circ$15’ 110$^\circ$30’ 6' $\angle$B= 110$^\circ$24’ 110$^\circ$24’ 298$^\circ$15’
CD 346$^\circ$35’ 166$^\circ$30’ 48$^\circ$20’ 6' $\angle$C=48$^\circ$14’ 48$^\circ$14’ 166$^\circ$29’
DE 337$^\circ$05’ 158$^\circ$10’ 170$^\circ$35’ 6' $\angle$D= 170$^\circ$29’ 336$^\circ$58’ 156$^\circ$58’
EA 293$^\circ$30’ 113$^\circ$00’ 135$^\circ$20’ 6' $\angle$E= 135$^\circ$14’ 292$^\circ$12’ 112$^\circ$12’
$\sum$540$^\circ$30' $\sum$30' $\sum$540$^\circ$