The angular misclosure of the closed-loop traverse shown in the figure is ______ ° (Answer in decimal degrees and rounded off to three decimal places).
Show Hint
Sum the six interior angles of the closed traverse and compare with the theoretical sum (n-2) x 180 degrees for an n-sided polygon.
Step 1: Set up the closure check using the general polygon-angle rule. For a closed-loop traverse with $n$ interior angles, the geometrically required sum is $(n-2)\times 180^\circ$, because a closed figure can always be split into $(n-2)$ triangles, each contributing $180^\circ$. Here the traverse has $n=6$ marked angles (one of them, $227^\circ26'15''$, is a reflex angle bigger than $180^\circ$ at the concave corner, and it must still be included as measured). So the required sum is \[ (6-2)\times180^\circ = 720^\circ \] Step 2: Sum the field angles in seconds, then convert back. List the six angles in seconds only and add: $30''+0''+15''+45''+0''+30''=120''=2'$. Now add minutes: $45'+0'+26'+35'+35'+40'=181'$, plus the carried $2'$ gives $183'=3^\circ03'$. Now add whole degrees: $132+64+227+97+131+66=717^\circ$, plus the carried $3^\circ$ gives $720^\circ$. So the field sum is $\Sigma = 720^\circ03'00''$. Step 3: Find the misclosure and convert it to decimal form. Misclosure $= \Sigma_{field} - \Sigma_{required} = 720^\circ03'00'' - 720^\circ00'00'' = 0^\circ03'00''$. Since $1^\circ = 60'$, $3' = 3/60^\circ = 0.05^\circ$, exactly at three-decimal precision $0.050^\circ$. Step 4: Report the final closure error. \[ \boxed{0.050^\circ} \]
