Question:

The calculated bearings for all traverse lines of the polygon are summarized in the table below. The calculations proceed sequentially, starting with the given bearing of line AB and using the measured interior angle at each station. Summary of Traverse Bearings | Traverse Line | Bearing (Forward Bearing) | |---|---| | AB | \mathbf{N \ 41^\circ 35' \ E} (Given) | | BC | \mathbf{N \ 9^\circ 14' \ W} | | CD | \mathbf{S \ 79^\circ 21' \ W} | | DE | \mathbf{S \ 31^\circ 51' \ W} | | EF | \mathbf{S \ 12^\circ 27' \ E} | | FA | \mathbf{S \ 73^\circ 35' \ E} | Detailed Calculation Steps The bearing of the next line is found by: \text{Bearing}_{\text{Next Line}} = \text{Back Bearing}_{\text{Previous Line}} \pm \text{Interior Angle} We use Whole Circle Bearings (WCB) for easy calculation, where \text{WCB} = \text{Quadrant Bearing (QB)} for N/E lines, \text{WCB} = 180^\circ - \text{QB} for S/E lines, etc. The Back Bearing (BB) is always \text{FB} \pm 180^\circ. Since the angles are interior and measured to the right (clockwise) from the back line, we use: \text{WCB}_{\text{Next}} = \text{WCB}_{\text{Back Line}} + \text{Interior Angle} 1. Line AB \to BC * Given \text{FB}_{AB}: \mathbf{N \ 41^\circ 35' \ E} (\text{WCB}_{AB} = 41^\circ 35') * Back Bearing \text{BB}_{AB} at B (\text{WCB}_{BA}): 41^\circ 35' + 180^\circ 00' = 221^\circ 35' * Angle \angle B: 129^\circ 11' * \text{WCB}_{BC}: 221^\circ 35' + 129^\circ 11' = 350^\circ 46' * \text{FB}_{BC} (QB): 360^\circ 00' - 350^\circ 46' = 9^\circ 14' \implies \mathbf{N \ 9^\circ 14' \ W} 2. Line BC \to CD * \text{FB}_{BC}: \mathbf{N \ 9^\circ 14' \ W} * Back Bearing \text{BB}_{BC} at C (\text{WCB}_{CB}): 360^\circ 00' - 9^\circ 14' = 350^\circ 46', then 350^\circ 46' - 180^\circ 00' = 170^\circ 46' * Angle \angle C: 88^\circ 35' * \text{WCB}_{CD}: 170^\circ 46' + 88^\circ 35' = 259^\circ 21' * \text{FB}_{CD} (QB): 259^\circ 21' - 180^\circ 00' = 79^\circ 21' \implies \mathbf{S \ 79^\circ 21' \ W} 3. Line CD \to DE * \text{FB}_{CD}: \mathbf{S \ 79^\circ 21' \ W} * Back Bearing \text{BB}_{CD} at D (\text{WCB}_{DC}): 259^\circ 21' - 180^\circ 00' = 79^\circ 21' * Angle \angle D: 132^\circ 30' * \text{WCB}_{DE}: 79^\circ 21' + 132^\circ 30' = 211^\circ 51' * \text{FB}_{DE} (QB): 211^\circ 51' - 180^\circ 00' = 31^\circ 51' \implies \mathbf{S \ 31^\circ 51' \ W} 4. Line DE \to EF * \text{FB}_{DE}: \mathbf{S \ 31^\circ 51' \ W} * Back Bearing \text{BB}_{DE} at E (\text{WCB}_{ED}): 211^\circ 51' - 180^\circ 00' = 31^\circ 51' * Angle \angle E: 135^\circ 42' * \text{WCB}_{EF}: 31^\circ 51' + 135^\circ 42' = 167^\circ 33' * \text{FB}_{EF} (QB): 180^\circ 00' - 167^\circ 33' = 12^\circ 27' \implies \mathbf{S \ 12^\circ 27' \ E} 5. Line EF \to FA * \text{FB}_{EF}: \mathbf{S \ 12^\circ 27' \ E} * Back Bearing \text{BB}_{EF} at F (\text{WCB}_{FE}): 167^\circ 33' + 180^\circ 00' = 347^\circ 33' * Angle \angle F: 118^\circ 52' * \text{WCB}_{FA}: 347^\circ 33' + 118^\circ 52' = 466^\circ 25' * 466^\circ 25' - 360^\circ 00' = 106^\circ 25' * \text{FB}_{FA} (QB): 180^\circ 00' - 106^\circ 25' = 73^\circ 35' \implies \mathbf{S \ 73^\circ 35' \ E} 6. Line FA \to AB (Closure Check) * \text{FB}_{FA}: \mathbf{S \ 73^\circ 35' \ E} * Back Bearing \text{BB}_{FA} at A (\text{WCB}_{AF}): 106^\circ 25' + 180^\circ 00' = 286^\circ 25' * Angle \angle A: 115^\circ 10' * Calculated \text{WCB}_{AB}: 286^\circ 25' + 115^\circ 10' = 401^\circ 35' * 401^\circ 35' - 360^\circ 00' = 41^\circ 35' * Calculated \text{FB}_{AB} (QB): 41^\circ 35' \implies \mathbf{N \ 41^\circ 35' \ E} * The calculated bearing matches the given bearing, confirming the accuracy of the calculations.