Topographic survey is carried out to delineate natural features.

Length of curve, L = (π/180)× ∆×R

= (π/180)×40×600 = 418.88m

Tangent distance, T = R tan(∆/2)

= 600tan(40/2) = 218.38m

Since the latitude of line is positive and departure is negative, the line lies in
Fourth quadrant.

l cos θ = 78; l sin θ = -45.1

tanθ = -0.578

θ = -30o

WCB of AB = 360o-30o = 330o

As per clause 3.7.2 of IS 800:2007

Applying correction due to local attraction, the correct bearing of line BC will be

The difference between back bearing and fore bearing of line DE

= 258^{o} 30' -78^{o} 30' = 180^{o}

∴ Station D and E are free from local attraction

Fore bearing of line o EA = 212^{o} 30'

∴Correct back bearing of line EA = 216^{o} 30' - 180^{o} = 36^{o} 30'

Error at A = 31^{o} 45' - 36^{o} 30' = -4^{o} 45'; Hence correction at A = +4^{o} 45'

Correct fore bearing of line AB = 126^{o} 45' + 4^{o} 45' = 131^{o} 30'

Correct back bearing of AB = 131^{o} 30 + 180^{o} = 311^{o} 30'

Error at B = -3^{o} 30'; Correction at B = +3^{o} 30'

Hence correct bearing of BC = 45^{o} 15' + 3^{o} 30' = 48^{o} 45'

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