1. **Problem Statement:** Calculate the coordinates of new points K', K'', and L' using the traverse method given initial coordinates and bearings/distances. 2. **Traverse Method Formula:** To find new coordinates, use: $$\Delta N = D \times \cos(\theta)$$ $$\Delta E = D \times \sin(\theta)$$ where $D$ is distance and $\theta$ is bearing converted to decimal degrees. 3. **Convert Bearings to Decimal Degrees:** - $300^\circ 20' 45'' = 300 + \frac{20}{60} + \frac{45}{3600} = 300.3458^\circ$ - $82^\circ 14' 42'' = 82 + \frac{14}{60} + \frac{42}{3600} = 82.245^\circ$ - $185^\circ 16' 58'' = 185 + \frac{16}{60} + \frac{58}{3600} = 185.2828^\circ$ - $50^\circ 17' 30'' = 50 + \frac{17}{60} + \frac{30}{3600} = 50.2917^\circ$ 4. **Calculate Coordinate Changes for Each Line:** - For $K - t_1$: $$\Delta N_1 = 159.13 \times \cos(300.3458^\circ) = 159.13 \times 0.5001 = 79.57$$ $$\Delta E_1 = 159.13 \times \sin(300.3458^\circ) = 159.13 \times (-0.8659) = -137.77$$ - For $t_1 - t_2$: $$\Delta N_2 = 121.10 \times \cos(82.245^\circ) = 121.10 \times 0.1391 = 16.85$$ $$\Delta E_2 = 121.10 \times \sin(82.245^\circ) = 121.10 \times 0.9903 = 119.87$$ - For $t_2 - t_3$: $$\Delta N_3 = 208.36 \times \cos(185.2828^\circ) = 208.36 \times (-0.9967) = -207.68$$ $$\Delta E_3 = 208.36 \times \sin(185.2828^\circ) = 208.36 \times (-0.0813) = -16.94$$ - For $t_3 - L$: $$\Delta N_4 = 500.50 \times \cos(50.2917^\circ) = 500.50 \times 0.6391 = 319.77$$ $$\Delta E_4 = 500.50 \times \sin(50.2917^\circ) = 500.50 \times 0.7691 = 384.88$$ 5. **Calculate Coordinates of Traverse Points:** Given initial coordinates for K: Northing = 1538.10, Easting = 125° 30' (convert to decimal degrees for Easting if needed, but likely Easting is a distance, so treat as 125.5 for calculation) - $t_1$: $$N_{t1} = 1538.10 + 79.57 = 1617.67$$ $$E_{t1} = 125.5 - 137.77 = -12.27$$ - $t_2$: $$N_{t2} = 1617.67 + 16.85 = 1634.52$$ $$E_{t2} = -12.27 + 119.87 = 107.60$$ - $t_3$: $$N_{t3} = 1634.52 - 207.68 = 1426.84$$ $$E_{t3} = 107.60 - 16.94 = 90.66$$ - $L$: $$N_L = 1426.84 + 319.77 = 1746.61$$ $$E_L = 90.66 + 384.88 = 475.54$$ 6. **Bearing and Distance of Line $t_1 - t_2$ After Adjustment:** - Distance is given as 121.10 m. - Bearing recalculated from coordinate differences: $$\Delta N = 1634.52 - 1617.67 = 16.85$$ $$\Delta E = 107.60 - (-12.27) = 119.87$$ - Bearing $\theta = \arctan\left(\frac{\Delta E}{\Delta N}\right) = \arctan\left(\frac{119.87}{16.85}\right) = 82.25^\circ$ 7. **Definitions in Compass Traversing:** (i) True bearing: The angle measured clockwise from the north direction to the line, relative to true north. (ii) Magnetic bearing: The angle measured clockwise from magnetic north to the line. (iii) Forward bearing: The bearing of a line when moving from the starting point to the endpoint. (iv) Backward bearing: The bearing of the same line when moving from the endpoint back to the starting point, differing by 180° from the forward bearing. Final answers: - Coordinates: $t_1(1617.67, -12.27)$, $t_2(1634.52, 107.60)$, $t_3(1426.84, 90.66)$, $L(1746.61, 475.54)$ - Bearing $t_1 - t_2 = 82.25^\circ$, Distance = 121.10 m - Definitions as above.