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Let's go step by step and solve each problem.

### **Problem 2:**
A stadia rod is sighted by an instrument at point A. The upper and lower stadia readings are 1.852 m and 0.921 m, respectively. The distance from the instrument to the rod is 93.117 m, and the stadia interval factor is 100.

#### **(a) Determine the angle:**
- The difference between the upper and lower stadia readings is the stadia intercept (S).
  \[
  S = 1.852 \, \text{m} - 0.921 \, \text{m} = 0.931 \, \text{m}
  \]

- The horizontal distance is calculated using the stadia interval formula:
  \[
  D = k \cdot S
  \]
  where \( k \) is the stadia interval factor (100 in this case), and \( S \) is the stadia intercept.
  \[
  D = 100 \cdot 0.931 = 93.1 \, \text{m}
  \]

- To find the angle of inclination (\( \theta \)), use the total sight distance (93.117 m) and the horizontal distance calculated above. Use the following trigonometric relation:
  \[
  \cos(\theta) = \frac{\text{Horizontal Distance}}{\text{Total Distance}}
  \]
  \[
  \cos(\theta) = \frac{93.1}{93.117}
  \]
  The angle \( \theta \) is:
  \[
  \theta = \cos^{-1}\left( \frac{93.1}{93.117} \right)
  \]

#### **(b) Determine the horizontal distance:**
We already calculated the horizontal distance, which is:
\[
D = 93.1 \, \text{m}
\]

---

### **Problem 3:**
A rectangular lot with dimensions 217.3 m by 148.5 m was measured using a 30 m tape, which was 0.04 m too long.

#### **(a) What is the correct width of the lot?**
The tape is 0.04 m too long for every 30 m, so the correction factor is:
\[
\text{Correction factor} = \frac{30.04}{30} = 1.00133
\]

- The corrected width is calculated by dividing the measured width by the correction factor:
  \[
  \text{Corrected width} = \frac{148.5}{1.00133} = 148.30 \, \text{m}
  \]

#### **(b) What is the correct area of the lot?**
- The corrected length is calculated in the same way:
  \[
  \text{Corrected length} = \frac{217.3}{1.00133} = 217.01 \, \text{m}
  \]

- The corrected area is:
  \[
  \text{Corrected area} = 217.01 \times 148.30 = 32,180.4 \, \text{m}^2
  \]

---

### **Problem 4:**
A 100 m steel tape is used to measure a distance, but due to temperature and tension effects, the observed length is 1539.28 m.

#### **(a) Determine the total correction for temperature:**
- The temperature correction formula is:
  \[
  C_T = \alpha \cdot L \cdot (T_m - T_0)
  \]
  Where:
  \[
  \alpha = 0.0000116 \, / ^\circ \text{C} \quad \text{(coefficient of linear expansion)}
  \]
  \( L = 1539.28 \, \text{m} \) (measured length), \( T_m = 31^\circ \, \text{C} \) (measured temperature), and \( T_0 = 20^\circ \, \text{C} \) (standard temperature).
  \[
  C_T = 0.0000116 \cdot 1539.28 \cdot (31 - 20) = 0.196 \, \text{m}
  \]

#### **(b) Find the total correction due to pull/tension:**
- The correction for tension is given by:
  \[
  C_P = \frac{(P - P_0) \cdot L}{E \cdot A}
  \]
  Where:
  \( P = 12 \, \text{kg} \) (applied tension), \( P_0 = 10 \, \text{kg} \) (standard tension), \( E = 2.1 \times 10^6 \, \text{kg/cm}^2 \) (modulus of elasticity), and \( A = 0.035 \, \text{cm}^2 \) (cross-sectional area).
  \[
  C_P = \frac{(12 - 10) \cdot 1539.28}{2.1 \times 10^6 \cdot 0.035} = 0.040 \, \text{m}
  \]

#### **(c) Determine the total correction due to sag:**
- The sag correction formula is:
  \[
  C_S = \frac{w^2 \cdot L^3}{24 \cdot P^2}
  \]
  Where:
  \( w = 1.5 \, \text{kg} \, / \, \text{100 m} = 0.015 \, \text{kg/m} \), and \( P = 12 \, \text{kg} \) (tension).
  \[
  C_S = \frac{(0.015)^2 \cdot (1539.28)^3}{24 \cdot (12)^2} = 0.011 \, \text{m}
  \]

#### **(d) What is the correct length of the line?**
- The total correction is the sum of all corrections:
  \[
  C = C_T + C_P + C_S = 0.196 + 0.040 + 0.011 = 0.247 \, \text{m}
  \]

- The corrected length is:
  \[
  L_{\text{corrected}} = 1539.28 - 0.247 = 1539.03 \, \text{m}
  \]

---

### Summary:
1. **Problem 2a**: \( \theta = \cos^{-1}\left( \frac{93.1}{93.117} \right) \)
2. **Problem 2b**: Horizontal distance = 93.1 m.
3. **Problem 3a**: Correct width = 148.30 m.
4. **Problem 3b**: Correct area = 32,180.4 m².
5. **Problem 4d**: Correct length of the line = 1539.03 m.

Would you like more details on any step?

Here are five questions to expand understanding:
1. How does temperature affect the length of materials like steel tapes?
2. Why is the sag correction necessary in long tape measurements?
3. What is the significance of stadia intervals in surveying?
4. How can incorrect tape lengths impact land area measurements?
5. What other factors can affect the accuracy of distance measurements in surveying?

**Tip**: Always ensure your measuring tape is properly calibrated to avoid cumulative errors over large distances!

A stadia rod, rectangular lot, and tape measurement problem involving angle determination, corrections due to temperature, tension, and sag.

Solve a surveying problem involving a stadia rod for angle determination, rectangular lot area correction using a tape too long, and distance correction for temperature, tension, and sag. Learn the key mathematical concepts and formulas.

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