Question

A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m. and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO but he finds the line of sight of C shifted from the original line of sight by an angle $\theta ={40}^{\circ }$ ($\theta$ is known as parallax) estimate the distance of the lower C from his original position A.

Answer 1

According to figure-
$\tan 40=\frac{AC}{AB}$
i.e.,
$AC=AB\times \tan 40=100\times 0.8390=83.90m$