Let's break down the problem step by step: --- **Given:** - Instrument is set up at point 2. - The backsight (BS) is due north (i.e., azimuth from 2 to BS is 0°). - The table shows: - **FS** = Foresight (the next point the measurement is taken to) - **Azimuth** = The bearing/azimuth from point 2 to that FS point - **Distance** = Distance from 2 to that FS point | FS | Azimuth | Distance (ft) | |----|---------|---------------| | 3 | 209° | 85 | | 6 | 45° | 253 | - We are to **find the bearing from point 3 to point 6**. --- ## **Step 1: Find Coordinates of Points 3 and 6** Assume point 2 is at the origin (0,0). ### **Coordinates of Point 3** - From 2, azimuth to 3 is **209°**, distance is **85 ft**. Convert azimuth and distance to coordinates: - ΔX = distance × sin(azimuth) - ΔY = distance × cos(azimuth) For azimuth 209°: - sin(209°) = sin(180°+29°) = -sin(29°) ≈ -0.4848 - cos(209°) = -cos(29°) ≈ -0.8746 So, - ΔX₃ = 85 × sin(209°) = 85 × (-0.4848) ≈ **-41.208 ft** - ΔY₃ = 85 × cos(209°) = 85 × (-0.8746) ≈ **-74.341 ft** So, **coordinates of point 3:** - (X₃, Y₃) = (**-41.208**, **-74.341**) --- ### **Coordinates of Point 6** - From 2, azimuth to 6 is **45°**, distance is **253 ft**. - sin(45°) = 0.7071 - cos(45°) = 0.7071 So, - ΔX₆ = 253 × 0.7071 ≈ **179.018 ft** - ΔY₆ = 253 × 0.7071 ≈ **179.018 ft** So, **coordinates of point 6:** - (X₆, Y₆) = (**179.018**, **179.018**) --- ## **Step 2: Find Bearing from Point 3 to Point 6** Find the difference in coordinates: - ΔX = X₆ - X₃ = 179.018 - (-41.208) = **220.226 ft** - ΔY = Y₆ - Y₃ = 179.018 - (-74.341) = **253.359 ft** The bearing (azimuth) from 3 to 6 is: \[ \text{Azimuth} = \arctan\left(\frac{\Delta X}{\Delta Y}\right) \] However, in surveying, azimuth is measured clockwise from north (positive Y axis). So: \[ \text{Azimuth} = \arctan2(\Delta X, \Delta Y) \] Plug in values: \[ \text{Azimuth} = \arctan2(220.226, 253.359) \] Calculate: \[ \theta = \arctan\left(\frac{220.226}{253.359}\right) \approx \arctan(0.8693) \approx 41.8^\circ \] (Using a calculator: arctan(0.8693) ≈ 41.8°) **Both ΔX and ΔY are positive, so the azimuth is in the NE quadrant (0° to 90°), which matches our calculation.** --- ## **Step 3: Write Final Answer** **The bearing (azimuth) from point 3 to point 6 is approximately:** \[ \boxed{41.8^\circ} \] --- ### **Summary Table** | From | To | Azimuth (deg) | Distance (ft) | Coordinates | |------|----|---------------|---------------|-------------------------| | 2 | 3 | 209 | 85 | (-41.208, -74.341) | | 2 | 6 | 45 | 253 | (179.018, 179.018) | | 3 | 6 | 41.8 | (calculated) | (see above) | --- ### **Final Answer** **The bearing (azimuth) from point 3 to point 6 is approximately:** \[ \boxed{41.8^\circ} \] (rounded to the nearest tenth of a degree)