1. The problem involves finding missing azimuths and distances for a polygon with 4 points and known azimuths and distances for three sides: line 2-3 with azimuth $326^\circ00'$ and distance 116.40 meters, line 3-4 with azimuth $88^\circ00'$ and distance 174.40 meters, and line 3-1 with azimuth $112^\circ00'$ and distance 216.60 meters. Lines 1-2 and 4-1 lack given azimuths and distances. 2. Since the polygon has 4 corners (1, 2, 3, 4), the sum of the exterior angles must be $360^\circ$ and the polygon is closed, so the traverse must return to the starting point. 3. Using the azimuths and distances, calculate the coordinates of points 2, 3, and 4 relative to point 1 set at the origin: - Convert azimuths to radians for calculation: $\theta = \text{azimuth} \times \frac{\pi}{180}$. - Calculate $\Delta x = d \sin \theta$, $\Delta y = d \cos \theta$ for each known side. 4. From point 1 at $(0,0)$: - Point 2 unknown. - Point 3 relative to point 2 using line 2-3: $$\theta_{2-3} = 326^\circ = 326 \times \frac{\pi}{180} = 5.692 \text{ radians}$$ $$\Delta x_{2-3} = 116.40 \times \sin(5.692) = 116.40 \times (-0.5446) = -63.38$$ $$\Delta y_{2-3} = 116.40 \times \cos(5.692) = 116.40 \times 0.8387 = 97.62$$ So from point 2 to 3: $(x_3, y_3) = (x_2 - 63.38, y_2 + 97.62)$. - Point 4 relative to point 3 using line 3-4: $$\theta_{3-4} = 88^\circ = 1.535 \text{ radians}$$ $$\Delta x_{3-4} = 174.40 \times \sin(1.535) = 174.40 \times 0.9994 = 174.30$$ $$\Delta y_{3-4} = 174.40 \times \cos(1.535) = 174.40 \times 0.0349 = 6.09$$ So from point 3 to 4: $(x_4, y_4) = (x_3 + 174.30, y_3 + 6.09)$. - Point 1 relative to point 3 using line 3-1: $$\theta_{3-1} = 112^\circ = 1.956 \text{ radians}$$ $$\Delta x_{3-1} = 216.60 \times \sin(1.956) = 216.60 \times 0.9272 = 200.82$$ $$\Delta y_{3-1} = 216.60 \times \cos(1.956) = 216.60 \times (-0.3746) = -81.15$$ So from point 3 to 1: $(x_1, y_1) = (x_3 + 200.82, y_3 - 81.15)$ but since point 1 is at origin $ (0,0)$, we get: $$x_3 = -200.82, \quad y_3 = 81.15$$ 5. Substitute $x_3$ and $y_3$ into previous equations: From point 2 to 3: $$x_3 = x_2 - 63.38 = -200.82 \Rightarrow x_2 = -200.82 + 63.38 = -137.44$$ $$y_3 = y_2 + 97.62 = 81.15 \Rightarrow y_2 = 81.15 - 97.62 = -16.47$$ From point 4: $$x_4 = x_3 + 174.30 = -200.82 + 174.30 = -26.52$$ $$y_4 = y_3 + 6.09 = 81.15 + 6.09 = 87.24$$ 6. Now find missing line 4-1: $$\Delta x_{4-1} = x_1 - x_4 = 0 - (-26.52) = 26.52$$ $$\Delta y_{4-1} = y_1 - y_4 = 0 - 87.24 = -87.24$$ Distance: $$d_{4-1} = \sqrt{26.52^2 + (-87.24)^2} = \sqrt{703.21 + 7611.82} = \sqrt{8315.03} = 91.21 \text{ meters}$$ Azimuth: $$\theta_{4-1} = \arctan\left( \frac{26.52}{-87.24} \right) = \arctan(-0.304) = -16.9^\circ$$ Since $\Delta y$ is negative and $\Delta x$ is positive, azimuth is in quadrant IV: $$\text{Azimuth} = 360^\circ - 16.9^\circ = 343.1^\circ$$ 7. Find missing line 1-2: $$\Delta x_{1-2} = x_2 - x_1 = -137.44 - 0 = -137.44$$ $$\Delta y_{1-2} = y_2 - y_1 = -16.47 - 0 = -16.47$$ Distance: $$d_{1-2} = \sqrt{(-137.44)^2 + (-16.47)^2} = \sqrt{18889.4 + 271.3} = \sqrt{19160.7} = 138.44 \text{ meters}$$ Azimuth: $$\theta_{1-2} = \arctan\left( \frac{-137.44}{-16.47} \right) = \arctan(8.34) = 83.2^\circ$$ Both $\Delta x$ and $\Delta y$ are negative, so azimuth in quadrant III: $$\text{Azimuth} = 180^\circ + 83.2^\circ = 263.2^\circ$$ **Final answers:** - Line 1-2: Azimuth $263^\circ12'$, Distance 138.44 meters - Line 4-1: Azimuth $343^\circ06'$, Distance 91.21 meters All points and sides satisfy the closure and given data.