Let's solve Q7 step by step: --- ## **Given Data** - Stadia constant, \( K = 100 \) - Upper stadia hair reading = **1.72 m** - Lower stadia hair reading = **1.15 m** - **Stadia interval (S)** = Upper - Lower = **1.72 m - 1.15 m = .57 m** --- ## **a. Horizontal Sight** The horizontal distance (\( D \)) is calculated as: \[ D = K \times S \] \[ D = 100 \times .57 = 57 \text{ m} \] **Final Answer (a):** \[ \boxed{57 \text{ m}} \] --- ## **b. Sight Inclined at \( 20^\circ \) from Horizontal** When the line of sight is inclined, the horizontal distance (\( D_H \)) is given by: \[ D_H = K \times S \times \cos^2 \theta \] Where \( \theta = 20^\circ \). \[ D_H = 100 \times .57 \times (\cos 20^\circ)^2 \] First, calculate \( \cos 20^\circ \): \[ \cos 20^\circ \approx .9397 \] \[ (\cos 20^\circ)^2 \approx (.9397)^2 \approx .883 \] Now compute: \[ D_H = 100 \times .57 \times .883 \] \[ D_H \approx 50.13 \text{ m} \] **Final Answer (b):** \[ \boxed{50.13 \text{ m}} \] --- ## **Summary Table** | Case | Formula | Distance (m) | |-----------------------------|-----------------------------------|--------------| | a. Sight is horizontal | \( D = K \times S \) | 57 | | b. Sight inclined \(20^\circ\) | \( D_H = K \times S \times \cos^2 \theta \) | 50.13 | --- **If you need further breakdown or explanation, let me know!**