To find the latitude (L) and departure (D) of location P, we need to use the whole circle bearings and the path length (5 km each) to calculate the components of each path in terms of latitude and departure. The formulas for calculating latitude and departure changes are:

$$ \Delta L = D \cdot \cos(\theta) \\ \Delta D = D \cdot \sin(\theta) $$

where $ \theta $ is the bearing in degrees and $ D $ is the distance (5 km in this case). Note that angles should be converted from degrees to radians for trigonometric calculations.

Let's calculate the latitude and departure for each path:

Path RC:

$$ \theta = 120^\circ \\ \Delta L_{RC} = 5 \cdot \cos(120^\circ) = 5 \cdot (-0.5) = -2.5 \text{ km} \\ \Delta D_{RC} = 5 \cdot \sin(120^\circ) = 5 \cdot (\frac{\sqrt{3}}{2}) = 5 \cdot 0.866 \approx 4.33 \text{ km} $$

Path CA:

$$ \theta = 0^\circ \\ \Delta L_{CA} = 5 \cdot \cos(0^\circ) = 5 \\ \Delta D_{CA} = 5 \cdot \sin(0^\circ) = 0 $$

Path AB:

$$ \theta = 90^\circ \\ \Delta L_{AB} = 5 \cdot \cos(90^\circ) = 0 \\ \Delta D_{AB} = 5 \cdot \sin(90^\circ) = 5 $$

Path BP:

$$ \theta = 240^\circ \\ \Delta L_{BP} = 5 \cdot \cos(240^\circ) = 5 \cdot (-0.5) = -2.5 \text{ km} \\ \Delta D_{BP} = 5 \cdot \sin(240^\circ) = 5 \cdot (-\frac{\sqrt{3}}{2}) = 5 \cdot (-0.866) \approx -4.33 \text{ km} $$

Now let's sum up the latitude and departure changes:

Total Latitude Change (L):

$$ L = \Delta L_{RC} + \Delta L_{CA} + \Delta L_{AB} + \Delta L_{BP} \\ L = -2.5 + 5 + 0 - 2.5 \\ L = 0 \text{ km} $$

Total Departure Change (D):

$$ D = \Delta D_{RC} + \Delta D_{CA} + \Delta D_{AB} + \Delta D_{BP} \\ D = 4.33 + 0 + 5 - 4.33 \\ D = 5 \text{ km} $$

Therefore, the latitude and departure of P are:

$$ L = 0 \text{ km} \\ D = 5 \text{ km} $$

Hence, the correct option is:

Option B: L = 0.0; D = 5.0

Traversing · Geomatics Engineering Or Surveying · GATE CE

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