The forebearing of four sides of a closed quadrilateral ABCDA are AB = 60°, BC = 149.87°, CD = 269.5°, DA = 20.17°. The calculated value of interior angle D is _____° (Answer in decimal degrees and rounded off to two decimal values).
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Use back bearing of the incoming line minus fore bearing of the outgoing line at D, then check that all four interior angles of the quadrilateral sum to 360 degrees.
Step 1: Draw the situation at station D mentally, using included-angle reasoning.
At D, two survey lines meet, the line coming in from C along CD, and the line going out towards A along DA. The interior angle $\angle CDA$ measured inside the quadrilateral is simply the angle that would be physically read on a theodolite set up at D, backsighted to C and foresighted to A, measured clockwise from the backsight to the foresight (for a traverse run in the ABCDA sense).
Step 2: Express both sight directions as azimuths measured from D.
The foresight direction, D to A, has azimuth equal to the given fore bearing of line DA, that is $20.17^\circ$. The backsight direction, D to C, is the reverse of the given fore bearing of CD (which was measured from C towards D), so it equals $FB(CD) - 180^\circ = 269.5^\circ - 180^\circ = 89.5^\circ$.
Step 3: Take the clockwise angle from the backsight (DC) to the foresight (DA).
$\angle D = \text{azimuth}(D \to C) - \text{azimuth}(D \to A) = 89.5^\circ - 20.17^\circ = 69.33^\circ$. Since this comes out positive and less than $180^\circ$, no further $360^\circ$ adjustment is needed.
Step 4: Cross check with the other three angles and the geometric closing condition.
Using the same "azimuth of backsight minus azimuth of foresight" rule at every station: at A, backsight is towards D ($20.17^\circ+180^\circ=200.17^\circ$) and foresight is towards B ($60^\circ$), giving $\angle A = 200.17^\circ - 60^\circ = 140.17^\circ$; at B, backsight towards A ($60^\circ+180^\circ=240^\circ$) and foresight towards C ($149.87^\circ$), giving $\angle B = 240^\circ-149.87^\circ = 90.13^\circ$; at C, backsight towards B ($149.87^\circ+180^\circ=329.87^\circ$) and foresight towards D ($269.5^\circ$), giving $\angle C = 329.87^\circ - 269.5^\circ = 60.37^\circ$. Adding all four, $140.17^\circ+90.13^\circ+60.37^\circ+69.33^\circ = 360.00^\circ$, exactly matching the required interior angle sum of a quadrilateral, $(4-2)\times180^\circ=360^\circ$, confirming $\angle D = 69.33^\circ$ is correct.
\[ \boxed{\angle D = 69.33^\circ} \]
