**1 Answer**

written 5.7 years ago by |

Total error in latitude $= -173.20 -314.50+86.60+250+154.90 =+3.8$

Total error in departure $= 100+128.40+50+0.00-280.00 = -1.6$

**Transit rule =**

Correction of latitude at any side $= \text{total error in latitude} \times \frac{\text{latitude of that side}}{∑ \text{all latitude}}$

Correction of departure at any side $= \text{total error in latitude} \times \frac{\text{departure of that side}}{∑ \text{all latitude}}$

**Correction in latitude**

1.Correction in line $AB = 3.8 \times \frac{173.2}{979.2} = 0.6721$

Corrected latitude of line AB = -173.20 – 0.6721 = -173.87

2.Correction in line $BC = 3.8 \times \frac{314.5/}{979.2} = 1.2204$

Corrected latitude of line BC = -314.50 – 1.2204= -315.7204

3.Correction in line $CD = 3.8 \times \frac{86.60}{979.2} = 0.3360$

Corrected latitude of line CD = 86.60 – 0.3360= 86.264

4.Correction in line $DE = 3.8 \times \frac{250}{979.2} = 0.9701$

Corrected latitude of line DE = 250 – 0.9701=249.0299

5.Correction in line $EA = 3.8 \times \frac{154.90}{979.2} = 0.6011$

Corrected latitude of line EA = 154.90 – 0.6011=154.2989

**Correction in departure**

1.Correction in line $AB = 1.6 \times \frac{100}{558.4} = 0.2865$

Corrected departure of line AB =100 – 0.2865=100.2865

2.Correction in line $BC = 1.6 \times \frac{128.4}{558.4} = 0.3679$

Corrected departure of line BC = 128.4 – 0.3679= 128.7679

3.Correction in line $CD = 1.6 \times \frac{50}{558.4} = 0.1432$

Corrected departure of line CD = 50 – 0.1432= 50.1432

4.Correction in line $DE = 1.6 \times \frac{0.00}{558.4} = 00$

Corrected departure of line DE = 00

5.Correction in line $EA = 1.6 \times \frac{280}{558.4} = 0.8022$

Corrected departure of line EA = -280 – 0.8022=-279.1978