Question

An angle of $90^\circ$ is to be laid out with a theodolite having a least count of $30''$. The angle was measured by repetition method and was found to be $90^\circ\,00'\,25''$. The offset value at a distance of $300$ m from the theodolite to set out the correct angle is ................. m. (Rounded off to 3 decimal places).

Hint:

Small-angle field corrections: \(\delta \approx L\,\Delta\theta_{\text{rad}}\). Convert arc-seconds to radians via \(1''=\pi/648000\) rad.

Answer 1

Measured angle has an excess of $25''$ over $90^\circ$. Required lateral correction at distance \(L=300\) m: \[ \delta = L \tan(\Delta\theta) \approx L\,\Delta\theta_{\text{rad}}\!, \] with \(\Delta\theta=25''=\dfrac{25}{3600}^\circ=\dfrac{25\pi}{648000}\ \text{rad}\approx 1.212\times10^{-4}\). \[ ⇒\ \delta \approx 300 \times 1.212\times10^{-4} = 0.03636\ \text{m}\ \approx \boxed{0.036\ \text{m}}. \] (The least count is immaterial to the offset computation.)