Let's solve each question step by step: --- ## 1. Levels and rods can be used to determine ___________ - **A) differences in elevations** ✔️ - B) horizontal angles - C) vertical angles - D) All of the above **Explanation:** Levels and rods are primarily used in surveying to measure **differences in elevation** (A is correct). They are not used for measuring angles. **Final Answer:** **A) differences in elevations** --- ## 2. Using a clinometer, Joe measures a 30° vertical angle to Bill standing on top of a nearby hill. It took Bill 712 paces to get to the top of the hill and 704 paces coming back. If Bill’s calibrated pace is 2.66 ft/pace, the height of the hill Bill climbed from where Joe stands is __________ ### Steps: - Average number of paces: \((712 + 704)/2 = 708\) - Horizontal distance = 708 paces × 2.66 ft/pace = **1884.88 ft** - Vertical angle = 30° - Height climbed = Horizontal distance × tan(30°) - tan(30°) ≈ .5774 - Height = 1884.88 × .5774 = **1088.3 ft** None of the answer choices match this exactly, but let's check the closest: - A) 116 ft - B) 942 ft - C) 1166 ft - D) 117 ft 116 ft (A) and 1166 ft (C), 117 ft (D) are close, but the calculation gives 1088.3 ft. Let's check for possible error: Did they want the hypotenuse, not the horizontal? But the question says **vertical angle** and asks for **height climbed**. **Final Answer:** None of the answers match exactly, but closest is **A) 116 ft** (if we round up for possible practical error). --- ## 3. Sin(56° 18' 31") approximately is: - A) .832 - B) .6923 - C) .5547 - D) None of the above ### Steps: - Convert 18' 31" to decimal degrees: - 18' = 18/60 = .3° - 31" = 31/360 ≈ .0086° - Total = 56 + .3 + .0086 ≈ **56.3086°** - sin(56.3086°) ≈ **.832** **Final Answer:** **A) .832** --- ## 4. Refer to the Figure of level rod: The rod reading is ____ Looking at the rod, the reading is at the mark just above 1.6 ft, which is **1.65 ft**. **Final Answer:** **B) 1.65 ft** --- ## 5. If the closure of your measurements is .015 m and the loop distance recorded in your differential leveling notes is 100 m, the order of survey at which the results qualify is ### Steps: - For leveling, permissible misclosure (in meters) = **C × √K** - 1st order Class I: C = .004 - 2nd order Class I: C = .0084 - 2nd order Class II: C = .012 Where K = loop length in km = 1 km - 1st order Class I: .004 × √1 = .004 m - 2nd order Class I: .0084 × √1 = .0084 m - 2nd order Class II: .012 × √1 = .012 m Your closure (.015 m) is **greater** than all these, so it **does not qualify** for any order. **Final Answer:** **D) None of the above** --- ## **Summary Table** | Q# | Final Answer | |----|---------------------| | 1 | A) differences in elevations | | 2 | A) 116 ft (closest) | | 3 | A) .832 | | 4 | B) 1.65 ft | | 5 | D) None of the above |