An instrument was set up at P, and the angle of elevation to a vane 4 m above the foot of the staff held at Q was 10°30'. The horizontal distance between P and Q was known to be 1500 m. What is the reduced level of the staff station Q given that the reduced level of the instrument axis was 2560.38 m?

Concept:

Trigonometric levelling:

• It is the process of determining the different elevations of a station from an observed vertical angle and known distance.
• The vertical angle is measured by means of the theodolite. The horizontal distance may be either be measured or computed. Relative heights are calculated using the trigonometric formula.

Case - The base of the Object being Accessible:

Let H = the height of the object above the B.M.

h = the height of the object above the instrument axis.

h_s = height of the instrument axis above the B.M.

α = the vertical angle observed at the instrument station.

D = the horizontal distance in meters measured from the instrument station to the base of the object.

Then, h = D tan α

H = h + h_s = D tan α + h_s

When the distance D is large, the correction for curvature and refraction shall have to be applied.

Calculation:

The height of the vane above the instrument axis is

=  = 1500  tan 10⁰30^' = 278.00 m

Correction for curvature and refraction is 

= 0.06735 D² m  where D is in km

= 0.06735  (1.5)² = 0.15 m

Height of vane above the instrument axis is

= 278.00 - 0.15 = 277.85 m

RL of vane is 

= 277.85 + 2560.38 = 2838.23 m

RL of Q is 

= 2838.23 - 4 = 2834.23 m