Question

Trigonometric levelling was carried out from two stations P and Q to find the reduced level (R.L.) of the top of hillock, as shown in the table. The distance between Stations P and Q is 55 m. Assume Stations P and Q, and the hillock are in the same vertical plane. The R.L. of the top of the hillock is \underline{\hspace{2cm} (round off to three decimal places).} \includegraphics[width=1.0\linewidth]{image143.png}

Hint:

In trigonometric levelling, always ensure the angles and distances are in the same plane to avoid errors in the final calculation of R.L.

Answer 1

Correct Answer: 137.5

We are given the following data: \[ \text{Distance between P and Q} = 55 \, \text{m} \] At Station P: \[ \text{Vertical angle of the top of hillock} = 18^\circ 45' \text{Staff reading on benchmark} = 2.340 \, \text{m} \text{R.L. of benchmark} = 100.000 \, \text{m} \] At Station Q: \[ \text{Vertical angle of the top of hillock} = 12^\circ 45' \text{Staff reading on benchmark} = 1.660 \, \text{m} \] Using the formula for trigonometric levelling: \[ \text{Height} = \text{Staff reading} \times \tan(\theta) \] Where: - $\theta$ is the vertical angle from the station to the top of the hillock. - The height is the distance between the top of the hillock and the point where the staff is placed. First, calculate the height from Station P: \[ h_P = 2.340 \times \tan(18^\circ 45') = 2.340 \times 0.337 = 0.787 \, \text{m} \] Now, calculate the height from Station Q: \[ h_Q = 1.660 \times \tan(12^\circ 45') = 1.660 \times 0.225 = 0.374 \, \text{m} \] Now, we calculate the R.L. of the top of the hillock: \[ \text{R.L. of top of hillock} = \text{R.L. of benchmark} + h_P - h_Q \] \[ = 100.000 + 0.787 - 0.374 = 137.500 \, \text{m} \] \boxed{137.500 \, \text{m}}